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MATHEMATICS IN THE MODERN WORLD
Chapter 1: THE NATURE OF MATHEMATICS
 Learning Objectives
At the end of this chapter, the student is expected to:
 identify patterns in nature and regularities in the world;
 explain the importance of mathematics in one’s life; and
 express appreciation for mathematics as a human endeavor.
Introduction
Mathematics relies on both logic and creativity, and it is pursued both for a variety of
practical purposes and for its intrinsic interest. For some people, and not only professional
mathematicians, the essence of mathematics lies in its beauty and its intellectual challenge. For
others, including many scientists and engineers, the chief value of mathematics is how it applies
to their own work. Because mathematics plays such a central role in modern culture, some basic
understanding of the nature of mathematics is requisite for scientific literacy. To achieve this,
students need to perceive mathematics as part of the scientific endeavor, comprehend the nature
of mathematical thinking, and become familiar with key mathematical ideas and skills.
1.1 Mathematics in Our World
Have you ever wondered how well jeepney drivers give you your change when you hand
your fare? How about when you buy street food? Most food vendors do not make a mistake in
giving you your change after buying a grilled hotdog on a stick for example, without even using
calculators. Routine transactions like these, knowingly or unknowingly, are mathematics at work
because they involve computing numbers most of the time. How much time do you allot travelling
to avoid getting late for class? Before that, do you track every second you spend taking the shower,
eating breakfast, changing into school clothes, or preparing your things for school? Most
importantly, do you check if you still have enough money for fare, food, and other expenses for
school? Just like budgeting allowance, time is also mathematics at work. Are you watching your
weight and your food caloric intake? Do you read the nutrition information from the packages of
chocolates, cookies, candies, and drinks you buy? Consciously or unconsciously, all of these
activities engage some form of mathematics.
The heart of mathematics is more than just numbers, numbers which many supposed to be
meaningless and uninteresting. Have you ever gone for beach trips or did mountain climbing
perhaps and noticed in awe the beautiful world around you? The different shapes you see around,
the changing hues of the sky from sunrise to sunset, the clouds transforming from stratus to
cumulus, the contour of the rainbow in the horizon are all beautiful because of harmony. The
degree of changing hues of color has to be of exact measurement to appear pleasing and
harmonious to the human eye. “And it is mathematics that reveals the simplicities of nature, and
permits us to generalize from simple examples to the complexities of the real world. It took many
people from many different areas of human activity to turn a mathematical insight into a useful
product” (Stewart, 1995, pp. 71-72).
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If you count the number of petals of most flowers, notice that they are either of one petal,
two petals, three petals, five petals, eight petals, or thirteen. This sequence of numbers form the
set {1, 1, 2, 3, 5, 8, 13, …} whose pattern was discovered by Fibonacci, a great European
mathematician of the Middle Ages. His full name in Italian is Leonardo Pisano, which means
Leonardo of Pisa, because he was born in Pisa, Italy around 1175. Fibonacci is the shortened word
for the Latin term “filius Bonacci,” which stands for “son of Bonaccio.” His father’s name was
Guglielmo Bonaccio.
The German mathematician and astronomer Johannes Kepler (known for his laws of
planetary motion) observed that dividing a Fibonacci number by the number immediately before
it in the ordered sequence yields a quotient approximately equal to 1.618. This amazing ratio is
denoted by 𝜑 called the Golden Ratio. Kepler once claimed that “[g]eometry has two great
treasures; one is the Theorem of Pythagoras; the other, the division of a line into extreme and mean
ratio. The first we may compare to a measure of gold, the second we may name a precious jewel”
(Stakhov and Olsen, 2009).
The Golden Ratio is so fascinating that proportions of the human body such as the face
follows the so called Divine Proportion. The closer the proportion of the body parts to the Golden
Ratio, the more aesthetically pleasing and beautiful the body is. Many painter, including the
famous Leonardo da Vinci were so fascinated with the Golden Ratio that they used it in their works
art.
The world and the whole universe is imbued with mathematics. “The Pythagoreans
believed that the nature of the universe was directly related to mathematics and that the whole
numbers and the ratios formed by the whole numbers could be used to describe and represent all
natural events’ (Aufmann, 2014). Can the course of natural events such as winning in a contest or
in a game of chance be actually explained? What is your chance of winning the lottery? Have you
ever heard of probabilities? Johann Carl Friedrich Gauss (1777-1855) was a remarkable
mathematician who made many contributions to the mathematics of probabilities. An important
aspect of studying probabilities is the so called combinatorics, a mathematical fields pioneered by
Blaise Pascal, the mathematician whose famous Pascal’s Triangle finds useful application in
algebra and statistics.
Nature has its laws. These laws, such as the law of freely falling bodies, were laid down
by Isaac Newton. Newton and Gottfried Leibniz developed modern calculus in the 17 th century.
This development would not have been possible without the Cartesian coordinate system-- the
fusion of geometry and algebra by Rene Descartes (1596-1650). Albert Einstein (1879-1955), who
made a name for his mass and energy equation, E = mc2, would not have gone farther in his theory
of relativity without mathematics. Marie Sklodowska Curie (1867-1934) a Polish chemist and
mathematician received the 1911 Nobel Prize in chemistry for developing techniques of isolating
radioactive elements. Biological scientists have also recently used mathematics extensively to
theoretically investigate treatment procedures by modeling and simulating biological processes.
Without mathematics, all these inventions and discoveries are not possible.
The 20th century saw many breakthroughs in the fields of sciences and engineering which
creatively and critically employed mathematics. From the first personal computer named
Programma 101 that was released in 1965, to the first landing of man on the moon on July 20,
1969, and to the first Global Positioning System (GPS) satellite launched in 1989 for military use,
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all of these show that Mathematics plays a vital role in the affairs of humanity. Today, there are
24 GPS satellite in orbit helping people locate their travel destinations such as Google Maps or
Waze on personal computers, tablets, or smartphones.
Despite all these scientific achievements, many “millennials” are hesitant in taking science
courses partly because they feel anxious of anything intimately connected with mathematics. The
interests of millennials in gadgets, games, and technologies that appeal to their senses have also
interfered with the study of mathematics and the sciences. Unknowingly however, these
technologies employ gadgets, to the instructions one places on the newly bought device which
operate based on mathematical logic.
Finally, Ian Stewart (1995) explains in his book Nature’s Numbers that mathematics is a
systematic way of digging out the rules and structures that lie behind some observed pattern or
regularity, and using these rules and structures to explain what is going on. Now, think of one of
the most loved animation characters Dora the Explorer. When Dora gets lost in the jungle, what
does she needs? A map. Thanks to Rene Descartes, who made the Cartesian map for without it,
Dora will may never find her way.
Mathematics is everywhere because it finds many practical applications in daily life. God,
the Mathematician Architect, designs everything in this universe to follow rules or formulas.
Whether following regular or irregular patterns, His creation benefits humankind, His greatest
masterpiece. As Johannes Kepler wrote, “Those laws [of nature] are within the grasp of the human
mind; God wanted us to recognize them by creating us after his own image so that we could share
in His own thoughts” (Stewart, 2020).
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MATHEMATICS IN THE MODERN WORLD
EXERCISES 1.1
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. Write an essay about how you use Mathematics in our world using the following guided
questions: (at least 150 words)
o What is mathematics for you?
o Where do you apply the principles of mathematics?
o Do you need mathematics every day? Why?
o What have you learned from school on mathematics so far?
o Do you appreciate mathematics? Why or why not?
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1.2 Fibonacci Numbers
Fibonacci observed numbers in nature. His most popular contribution perhaps is the
number that is seen in the petals of flowers. A calla lily flower has only 1 petal, euphorbia has 2,
trillium has 3, hibiscus has 5, cosmos flower has 8, corn marigold has 13, some asters have 21, and
a sunflower can have 34, 55, or 89 petals. Surprisingly, these petal counts represent the first ten
numbers of the Fibonacci sequence.
Calla Lily
Euphorbia
Marigold
Trillium
Hibiscus
Cosmos
Aster
Not all petal numbers of flowers, however follow this pattern discovered by Fibonacci.
Some examples include the Brassicaceae family having four petals. Remarkably, many of the
flowers abide by the pattern observed by Fibonacci.
The principle behind the Fibonacci sequence is as follows:




Let xn be the nth integer in the Fibonacci sequence, the next (n + 1)th term xn +1 is
determined by adding nth and the (n – 1)th integers.
Consider the first few terms below: Let x1 = 1 be the first term, and x2 = 1 be the second
term, the third term x3 is found by x3 = x1 + x2 = 1 + 1 = 2.
The fourth term x4 is 2 + 1 = 3, the sum of the third and the second term.
To find the new nth Fibonacci number, simply add the two numbers immediately preceding
this nth number.
n = 3: x3 = 1 + 1 = 2
n = 6: x6 = 3 + 5 = 8
n = 9: x9 = 13 + 21 = 34
n = 4: x4 = 1 + 2 = 3
n = 7: x7 = 5 + 8 = 13
n = 5: x5 = 2 + 3 = 5
n = 8: x8 = 8 + 13 = 21
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These numbers arranged in increasing order can be written as the sequence {1, 1, 2, 3, 5,
8, 13, 21, 24, 55, 89, …}.
Fibonacci Spirals in Sunflowers
Pineapples Grow in a Numerical Sequence
Similarly, when we count the clockwise and counterclockwise spirals in the sunflower
seed, it is interesting to note that the numbers 34 and 5 occur—which are consecutive Fibonacci
numbers. Pineapples also have spirals formed by their hexagonal nubs. The nubs on many
pineapples form eight spirals that diagonally upward to the left and 13 that rotate diagonally
upward to the right, again these are consecutive Fibonacci numbers (Aufmann, 2015). The same
is also observed in the clockwise and counterclockwise spirals of a pine cone.
Pine Cone
Honeycomb
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Another interesting pattern in nature is the honeycomb. According to Merriam-Webster
dictionary, “a honeycomb is a mass of hexagonal wax cells built by honeybees in their nest to
contain their brood and stores of honey.” But why build hexagonal cells? Why not squares? Jin
Akiyama, a Japanese mathematician, explains it well in an experiment made on his regular TV
show Jinjin Math. In the experiment, a student is asked to step on one mass made up of square
cells and the result is unbelievable! The mass with hexagonal cells resisted the weight of the
student while the mass with square cells was completely destroyed. It is amazing to know that the
mass made up of hexagonal cells is stronger than the one made up of square cells. Moreover, these
patterns exist naturally in the world.
Another interesting observation is
the rabbit population beginning from a
baby pair of the first generation. Since it
takes the first generation to mature before
giving birth to an offspring, there is an
adult pair for the second generation,
which is ready for reproduction. So, there
are two rabbit pairs, the parents and baby
pairs, of the third generation. Next, the
adult pair begets a baby pair but the
previous baby pair simply matures, so a
family of three rabbit pairs for the fourth
generation exists, and so on. The number
of total rabbit pairs at each generation
constitutes a Fibonacci sequence.
Denoting by Fn the Fibonacci sequence of n generations is the set of Fibonacci numbers {Fn}, that
is:
{Fn} = {1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …}.
In particular, denote F1 = 1 for the 1st generation, F2 = 1 for the 2nd generation, F3 = 2 for the 3rd
generation, F4 = 3 for the 4th generation, and so on. It is interesting to point out that the Fibonacci
numbers Fn obey the following relationship:
𝐹1 = 𝐹2 = 1
{
𝐹𝑛 = 𝐹𝑛−1 + 𝐹𝑛−2 , 𝑛 ≥ 3
That is, Fn is given by the sum of the two previous Fibonacci numbers, 𝐹𝑛−1 and 𝐹𝑛−2 , 𝑛 ≥ 3.
For example: F3 = F2 + F1
F3 = 1 + 1
F3 = 2
It is also seen that F4 = F3 + F2 = 2 + 1 = 3 and F10 = F9 + F8 = 34 + 21 = 55.
Let us investigate the ratio of two adjacent Fibonacci numbers as n becomes large.
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𝐹
The following tables gives values of the ratio 𝐹 𝑛 as n approaches ∞.
n
3
4
5
6
7
8
9
𝐹𝑛
𝐹𝑛−1
2/1 = 2
3/2 = 1.5
5/3 = 1.6666667
8/5 = 1.6
13/8 = 1.625
21/13 = 1.615384615
34/21 = 1.619047619
𝑛−1
n
10
11
12
13
14
15
16
𝐹𝑛
𝐹𝑛−1
55/34 = 1.617647059
89/55 = 1.618181818
144/89 = 1.617977528
233/144 = 1.61805556
377/233 = 1.618025751
610/377 = 1.618037135
987/610 = 1.618032787
It is interesting to note that the ratio of two adjacent Fibonacci numbers approaches the
𝐹
golden ratio; that is, 𝑛 = 1.6180339887 … as n becomes large. This is indeed a mystery. What
𝐹𝑛−1
does the golden ratio have to do with a rabbit population method?
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EXERCISES 1.2
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
Identify at least 10 patterns and regularities in your surroundings by taking photos, describe each
by applying principles of mathematics.
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1.3 The Golden Ratio
The ratio of two consecutive Fibonacci numbers as n
𝐹
becomes large, approaches the golden ratio; that is, lim 𝑛 =
𝑛→∞ 𝐹𝑛−1
1.6180339887 … This can be verified by measuring some parts
of the human body: the length of the arm, height, the distance of
the fingertips to the elbow.
According to Markowsky (1992), “the ratio of a person’s
height to the height of his or her navel is roughly the golden ratio.
We are not told why this is significant; the navel is a scar of no
great importance in an adult human being.” You may verify this
for yourself. Did you get a value close to 1.6180339887 … ?
The ratio between the forearm and the hand also yields a value close to the golden ratio!
Another name for golden ratio is divine proportion. This must be so because human beauty is
based on the divine proportion. The photo on the next page illustrates the following golden ratio
proportions in the human face:





center of pupil: bottom of teeth:
bottom of chin
outer and inner edge of eye: center of
nose
outer edges of lips: upper ridges of
lips
width of center tooth: width of
second tooth
width of eye: width of iris
The golden ratio denoted by 𝜑 is sometimes called the golden mean or golden section:
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𝜑=
1 + √5
= 1.6180339887 …
2
The golden ratio can be expressed as the ratio between two numbers, if the latter is also the
ratio between the sum and the larger of the two numbers. Geometrically, it can also be visualized
as a rectangle perfectly formed by a square and another by a square and another rectangle, which
can be repeated infinitely inside each section.
Golden Rectangle with the Golden Spiral
a
b
a+b
Suppose that a line segment is cut into two pieces of length: a and b. Below it is shown
that a is longer than b. Clearly, the length of the original segment is a + b.
𝑎
Now, two ratios are formed: 𝑏 and
𝑎+𝑏
𝑎
. The first is the ratio of the longer piece a to the
shorter piece b, and the second ratio is the whole length to the longer piece a. It is now ideal to
ask, when are the two ratios equal? This is an algebraic question that can be solved by equating
the two ratios:
𝑎
𝑏
𝑎+𝑏
=
(1)
𝑎
𝑎
𝑏
Simplifying the right side of equation (1), we get 𝑏 = 1 + 𝑎.
𝑎
Denoting the ratio 𝑏 by 𝜑, we end up with
1
𝜑 =1+𝜑
(2)
On the other hand, dividing both the numerator and denominator of the right side if
equation (1) by b, we get
𝑎
𝑎
𝑎
+1
= 𝑏𝑎
𝑏
which by writing 𝑏 = 𝜑 becomes 𝜑 =
𝑏
𝜑+1
𝜑
(3)
1
or 𝜑 = 1 + 𝜑 as in (2). Now, by multiplying both
sides of equation (2) by 𝜑, we get a quadratic equation
𝜑2 − 𝜑 − 1 = 0
(4)
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Using quadratic root formula, we find two roots of equation (4); the first root is 𝜑 =
1.6180339887 …, and the second root is 𝜑 = −1.6180339887 ….The first root is the ratio of
𝜑 = 1.618 that we are looking for, and we ignore the second root because it is negative number.
Hence, the value we are looking for is
𝜑=
1 + √5
= 1.6180339887 …
2
The golden ratio 𝜑 = 1.6180339887 … is a strange number. It is the only number that if
1
you subtract one from it, 𝜑 − 1= 0.6180339887 …, you end up with its own reciprocal 𝜑 =
0.6180339887 …
Shapes and figures that bear in the golden rectangle are generally considered to be
aesthetically pleasing. As such, the ratio is visible in many works of art and architecture such as in
the Mona Lisa, the Notre Dame Cathedral, and the Parthenon. In fact, the human DNA molecule
also contains Fibonacci umbers, being 34 ångstroms long by 21 ångstroms wide for each full cycle
of the double helix spiral. It is also visible the patterns of golden spiral in our nature.
Mona Lisa
The Parthenon
DNA molecule of a human
Notre Dame Cathedral
Golden Spirals in Nature
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EXERCISE 1.3
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. With your partner, measure each of the following:
o
o
o
o
o
Height and height of navel
Foot and hand length
Length of forearm and length of hand
Width of center tooth and width of second tooth
Shoulder length and waistline
Are the results roughly the golden ratio? If not, what must be the ratios to get the golden ratio?
2. A wood that is 12 feet in length is needed to be cut into two parts such that the ratio of the
parts constitutes the golden ratio. What must be the lengths of the wood?
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Chapter 2: MATHEMATICAL LANGUAGE AND SYMBOLS
 Learning Objectives
At the end of this chapter, the student is expected to:
 discuss the language, symbols, and conventions used in mathematics;
 explain the nature of mathematics as a language;
 evaluate mathematical expressions correctly; and
 recognize that mathematics is a useful language.
Introduction
Mathematics has its own language, much of which we are already familiar with. For
example, the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 are part of our everyday lives. Whether we refer to 0
as ‘zero’, ‘nothing’, or ‘O’ as in a telephone number, we understand its meaning. There are many
symbols in mathematics and most are used as a precise form of shorthand. We need to be confident
when using these symbols, and to gain that confidence we need to understand their meaning. To
understand their meaning there are two things to help us—context - this is the context in which we
are working, or the particular topics being studied, and convention - where mathematicians and
scientists have decided that particular symbols will have particular meaning.
2.1 The Language of Mathematics
Can you imagine how would you be able to communicate with a seatmate in the bus who
speaks an entirely different language from yours? You may be able to tell him or her to watch over
your bag as you get off the bus for a while to buy something through certain nonverbal gestures.
That can be done with sign language. Language facilitates communication and meaning-making.
It allows people to express themselves and maintain their identity. Likewise, language bridges the
gap among people from various cultural origins without prejudice to their background and
upbringing. If you plan to marry someone with different language and culture, you need to know
his or her language and culture to be able to live with him or her as a spouse.
Have you seen the characters of Mandarin language? The Mandarin language has different
characters for sun, moon, stars, things like house, chair, table, furniture, trees, plants, flowers, and
relationships like grandfather, grandmother, father, mother, sister, brother etc. These unfamiliar
characters in the written Mandarin language may make learning Mandarin more difficult than the
Greek language even if Greek letters are different from the English alphabet. Mathematics is also
a language. It has its own symbol system; the same way the English or Greek languages have their
own alphabet.
Characteristics of Mathematical Language
Mathematical language is precise which means it is able to make very fine distinctions or
definitions among a set of mathematical symbols. It is concise because a mathematician can
express otherwise long expositions or sentences briefly using the language of mathematics. The
mathematical language is powerful, that is, one can express complex thoughts with relative case.
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For example, consider the sentence “The sum of any two real numbers is also a real number.” In
mathematical notation, this declarative sentence can be written as:
∀𝑎, 𝑏 ∈ ℝ, 𝑎 + 𝑏 ∈ ℝ
Mathematics is a symbolic language. Some of the symbols you may encounter as you read this
book are the following:
Σ
∃
∀
∈
∞
⊆
The sum of
There exists
For all/for any
Element of/member of
infinity
Subset of
⇒
⟺
ℝ
ℕ
ℤ
ℚ
If …, then
If and only if
Set of real numbers
Set of natural numbers
Set of integers
Set of rational numbers
Mathematical language can describe a subset of the real world using only the symbols
above. Problems in physics like freely falling bodies, speed, and acceleration; quantities like the
chemical content of vegetables; the use of mathematical modeling in biological disease modelling;
and the formulas employed in the social sciences can all be expressing using mathematical
sentences or formulas. Mathematics describes abstract structures as well. There are areas of Pure
Mathematics which deal with abstract algebra, linear algebra, topology, real analysis, and complex
analysis.
Mathematics, therefore, is a language of Sciences, business, economics, music,
architecture, arts, and even politics. There is an intimate connection between the language of
Mathematics and the English language. The left brain hemisphere which is responsible for
controlling language is also the same part of the brain in charge of tasks involving Mathematics.
It is the left brain hemisphere that coordinates logical and analytical thinking while the right brain
hemisphere is responsible for creative thinking.
Chinese, Greek, and English languages are the same because they communicate ideas
through symbols that feed the mind with information. More often however, a Chinese word or
symbol may mean differently in the Greek or English language resulting in confusion.
Mathematics tries to avoid this difficulty by adopting a universally understood symbolic system
for its language. Thus, the language of Mathematics can be considered a common language of the
world. Any student learning Mathematics in all parts of the globe should be able to understand
Mathematics even if he or she does not understand English or Filipino.
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EXERCISES 2.1
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
Write an essay about the Language of Mathematics using the following guided questions: (at least
150 words)
o Is language of Mathematics important to you? Why or why not?
o When do you use the language of Mathematics?
o Can you live without it? Why or why not?
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2.2 Expressions vs. Sentences
ENGLISH
NOUN
(name of given to object of interest)
SENTENCE
(must state a complete thought)
PERSON
PLACE
THING
ANIMAL
TRUE (T)
FALSE (F)
SOMETIMES TRUE (ST)/
SOMETIMES FALSE (SF)
Cardo
Manila
gun
dog
The capital of
the Philippines
is Manila.
Philippines
has four major
islands.
The dog is
black.
MATHEMATICS
EXPRESSION
(name of given to mathematical
object of interest)
NUMBER
SET
FUNCTION
2
{1, 2}
f(x)
MATRIX
[
1 4
]
−2 3
SENTENCE
(must state a complete thought)
ORDERED
PAIR
TRUE (T)
FALSE (F)
SOMETIMES TRUE (ST)/
SOMETIMES FALSE (SF)
(x,y)
2+5=7
1 + 1 = 11
3x + 5 = -2
A sentence must contain a complete thought. In the English language, an ordinary sentence
must contain a subject and a predicate. The subject contains a noun or a whole clause. “Manila”
for example is a proper noun but is not in itself a sentence because it does not state a complete
thought. Similarly, a mathematical sentence must state a complete thought. An expression is a
name given to a mathematical expression but not a mathematical sentence.
Types of Mathematical Sentences
A mathematical sentence is one in which a fact or complete idea expressed. Because a
mathematical sentence states a fact, many of them can be judged to be “true” or “false”. Questions
and phrases are not mathematical sentences since they cannot be judged to be true or false.
Examples:
a. “An isosceles triangle has two congruent sides.” is a true mathematical sentence.
b. “10 + 4 = 15” is a false mathematical sentence.
c. “Did you get that one right?” is NOT a mathematical sentence – it is a question.
d. “All triangles” is NOT a mathematical sentence – it is a phrase.
There are two types of mathematical sentences:

Open Sentence
An open sentence is a sentence which contains a variable. It can be either true or false
depending on what values are used.
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Examples:
1. A triangle has n sides.
2. z is a positive number.
3. 3y = 4x + 2
4. a + b = c + d

Closed Sentence
A closed sentence is a sentence which can be judged to be always true or always false and
has no variables.
Examples:
1. A square has four corners.
2. 6 is less than 5.
3. −3 is a negative number.
4. 3 + 5 = 8
5. 9 is an even number.
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EXERCISE 2.2
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
Tell whether if each of the following sentences is an open sentence or a closed sentence an. Write
OS if a sentence is open and CS if it is closed. If CS, determine if it is true or false. If OS, identify
the expression that will make the sentence always true.
_________________________ 1. Nine is an even number.
_________________________ 2. 4x – 2 = 5
_________________________ 3. Zero is an even number.
_________________________ 4. 2 + 5 = 2x
_________________________ 5. 1⁄2 > 2⁄3
_________________________ 6. n is a composite number.
_________________________ 7. 2n < 5
_________________________ 8. – 0.5 is an integer.
_________________________ 9. 𝜋 is a variable.
_________________________ 10. 0 is not an integer.
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2.3 Translating Mathematical Sentences to English Sentences (or vice versa)
There is no single strategy for translating Mathematical sentences into English sentences
(or vice versa). As long as you can remember the basics, you should be able to tackle the more
challenging ones. Just make sure that you can justify how you come up with your own translation,
and more importantly that it makes sense to you.
To build your skills in writing Mathematical sentences to English sentences (or vice versa),
we will go over different ways of how each operation may show up as a word or phrase in the
problem. The four arithmetic operations involved are addition, subtraction, multiplication, and
division. You can use also the different mathematical symbols as stated in a sentence.
Example 1: Write the Mathematical sentences into English sentences
a. ∀𝑥 ∈ ℝ, 𝑥 2 ≥ 0
b. ∀𝑥, 𝑦 ∈ ℝ, (𝑥 + 𝑦)2 = 𝑥 2 + 2𝑥𝑦 + 𝑦 2
c. ∃𝑚, 𝑛 ∈ ℤ, 𝑚 − 𝑛 ≤ 𝑚 + 𝑛
d. ∀𝑎, 𝑏 ∈ ℚ, 𝑎𝑏 = 0 ⇒ 𝑎 = 0 ∨ 𝑏 = 0
Solution:
a. For any real number, its square is greater than or equal to zero.
b. For any real numbers x and y, the square of their sum is equal to the sum of their squares
plus twice their product.
c. There exist integers m and n such that m minus n is less than or equal to m plus n.
d. For any rational numbers a and b, if their product is zero, then either a or b equals zero.
Example 2: Write the English sentences into Mathematical sentences.
a. Ten is the square root of one hundred.
b. Ten is greater than nine.
c. Ten is an even number.
d. Ten is a multiple of 5.
Solution:
a. √100 = 10
b. 10 > 9
c. 10 ∈ {2𝑛, 𝑛 ∈ ℕ}
d. 10 ∈ {5𝑛, 𝑛 ∈ ℕ}
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EXERCISE 2.2
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
A. Translate each of the following English sentences into Mathematical sentences.
1. The square of the difference of x and y is not more than 10.
______________________________________
2. The square of a number is positive.
______________________________________
3. Four is an even number.
______________________________________
4. One-fourth is a rational number.
______________________________________
5. Six is the principal square root of 36.
______________________________________
B. Translate each of the following Mathematical sentences into English sentences.
1. ∀𝑥 ∈ ℝ, ∃𝑦 ∈ ℝ, 𝑥 + 𝑦 = 10
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
2. ∀𝑥 ∈ ℤ+ , ∃𝑦 ∈ ℝ, 𝑦 2 = 𝑥
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
3. 𝑥 + 12 = 8
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
4. 2(𝑥 − 3) = 12
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
5. 2𝑥 − 6 = 45
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
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Chapter 3: THE FUNDAMENTALS OF LOGIC
 Learning Outcomes
At the end of this chapter, the students is expected to:
1.
2.
3.
4.
Define a propositional logic and its categories;
Explain logical connectives and exemplify truth values status of a proposition;
Transform logical statements into symbolic form (or vice versa); and
Construct combined truth table of propositions
Introduction
Why do most people argue over some issue and never get to the bottom of it? Sometimes
people in dispute say that “they do not see eye to eye.” This expression means that the people
invloved in an argument never get to agree on the issues at hand. In many cases, the disagreement
lies on not being able to present sound arguments based on facts, or the failure to convince the
conteding party using logical arguments. To avoid such a scenario in Mathematics and to uphold
certainly in the validity of mathemtical statements, mathematics employs the powerful language
of logic in asserting truths of statements. The use of logic illustrates the importance of precision
and conciseness in communicating Mathematics.
4.1Propositional Logic
A propositional logic, also known as statement logic, is the branch of mathematical logic
that studies the truth and falsity of propositions. In propositional logic, the simplest statements are
considered as indivisible units, and hence, propositional logic does not study those logical
properties and relations that depend upon parts of statements that are not themselves statements on
their own, such as the subject and predicate of a statement.
A proposition is a declarative sentence subject for affirmation or denial. It is a statement
with truth value; either true (T) or false (F), but not both.
Examples:
Determine if each sentence is a proposition or not.
a. All parallelograms are quadrilaterals.
b. Rhombuses are squares.
c. Is an equilateral triangle an isosceles triangle?
d. Triangle ABC is a right triangle.
e. Draw two parallel lines that are cut by a transversal.
f. 3 + 4 = 7
g. The sum of two prime numbers is even.
h. x > 10
i. n is a prime number.
j. 2 + 5 = 5
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Answers:
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
Proposition
Proposition
Not a Proposition
Proposition
Not a Proposition
Proposition
Proposition
Proposition
Proposition
Proposition
Categories of Propositions
 Qualitative Categories of Propositions
Propositions are categorized as affirmative or negative. The following sentences are
examples of affirmative propositions:
1. A quadrilateral has four sides.
2. The Philippines is a member of the ASEAN.
3. Whales are mammals.
The following are examples of negative propositions:
1. A right triangle has no obtuse angle.
2. Tomato is not a fruit.
3. Parallel lines never intersect.
 Quantitative Categories of Propositions
Propositions are further classified according to quantity or the different possible
extensions of their subject-terms.
Type of Categorical
Proposition
Universal proposition
Description
The subject term is taken in
full extension.
Examples
All quadrilaterals are
polygons.
No parallel lines meet at a
point.
Particular Proposition
Singular Proposition
The subject term is taken
only in particular extension.
The subject term denotes a
single person or thing.
Every integer is a real
number.
Some algebraic expressions
are polynomials.
A prime number has only two
factors.
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When quality and quantity are combined, propositions may be classified based on its mood
as follows:
Affirmative
Negative
Universal
A
Particular
I
All x is y.
Some x is y.
E
O
All x is not y. Some x is not y.
The letters A, E, I and O can be used to refer to propositions universal affirmative,
universal negative, particular affirmative and particular negative, respectively.
Examples:
Determine whether each statement is A, E, I, or O proposition.
a. There are snakes in every forest.
b. Some crocodiles are found in the city.
c. All lambs are not tame.
d. Some men are never as free as a bird.
Answers:
a.
b.
c.
d.
A
I
E
O
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EXERCISES 3.1
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
A. Determine if each statement is a proposition or not.
____________ 1. Every triangle is a polygon.
____________ 2. All right angles are congruent.
____________ 3. x is greater than or equal to -2.
____________ 4. If x + 2 = 4, is x = 2?
____________ 5. The sum of the interior angles in a triangle.
____________ 6. Some rectangles are not parallelograms.
____________ 7. Each equilateral triangle is an isosceles triangle.
____________ 8. For all values of a and b, (a + b)(a – b) = a2 – b2
____________ 9. If a is a real number, a2> 2.
____________ 10. Bisect an angle.
B. Determine whether each statement is A, E, I, or O proposition.
_______ 1. Some variables are fractions.
_______ 2. Each scalene triangle has no equal sides.
_______ 3. Some rectangles are parallelograms.
_______ 4. All right angles are congruent.
_______ 5. Every triangle is not a polygon.
_______ 6. Few rational numbers are integers.
_______ 7. Every odd number is prime.
_______ 8. Some irrational numbers are not terminating decimals.
_______ 9. All mathematicians are males.
_______ 10. Some polynomials are not congruent sides.
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3.2 Logical Connectives
Logical connectives are words or symbols used to connect two or more sentences in a
logical and grammatically valid way to produce one compound statement that takes its meaning
from the original sentences as well as the logical connective used. They can be used as a means to
connect two or more ideas, to compare and contrast different ideas, or to state certain conditions.
Examples:
Statement
Connective Symbol Type of Logical Statement
Not p
not
Negation
¬p
p and q
And
p˄q
Conjunction
p or q
Or
p˅q
Disjunction
If p, then q
If –then
Conditional
p→𝑞
p if and only if q If and only if p ↔ 𝑞
Biconditional
To understand the use of symbols in logic, consider the following simple statements.
Let p: The Earth is round.
q: The Sun is cold.
r: It rains in Spain.
The following compound statements can be written in symbolic form.
1. The Earth is round and the sun is cold.
Symbolic Form: p ˄ q
2. Either the Earth is round or the Sun is not cold.
Symbolic Form: p ˅ ( ¬ q )
3. The Earth is round and either the Sun is not cold or it rains in the Spain.
Symbolic Form: p ˄ (q ˅ r)
4. If the Earth is round, then it rains in Spain.
Symbolic Form: p → r
5. The Earth is round if and only if it rains in Spain.
Symbolic Form: p ↔ r
The following symbolic forms can be written in compound statement.
1. ¬r ↔ (¬p ˄ q)
Compound Statement: It does not rain in Spain if and only if the earth is not round and
the sun is cold.
2. (q ˄ p) → r
Compound Statement: If the sun is cold and the earth is round, then it rains in Spain.
3. (¬q ˅ r) → ¬p
Compound Statement: If either the Sun is not cold or it rains in Spain, then the Earth is
not round.
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4. r ˅ (¬ q ˄¬p)
Compound Statement: Either it rains in Spain or the Sun is not cold and the Earth is not
round.
5. (¬p ˅ r) ↔ ¬q
Compound Statement: Either the Earth is not round or it rains in Spain if and only if the
Sun is not cold.
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EXERCISES 3.2
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
I. Write each statement in symbolic form using connectives ¬, ˅ , ˄, →, or ↔.
_______________________ 1. If today is Friday (p), then tomorrow is Saturday (q).
_______________________ 2. I went to the registrar’s office (p) and I ate lunch at the
canteen (q).
_______________________ 3. A triangle is an equilateral triangle (p) if and only if it is an
equiangular triangle (q).
_______________________ 4. If it is bird (p), then it has feathers (q).
_______________________ 5. If either x is a fraction or y is a decimal (p), then it is not a
rational number (¬q).
_______________________ 6. The moon is flat (p) if and only the sun rises at the south
(q) and the dog is flying (r).
_______________________ 7. If either a frog is an amphibian (p) or a jelly fish is not a
fish (¬q), then 1 + 2 = 3 (r).
_______________________ 8. Online games are not bad to students (¬p) if and only if it
will not destroy their studies (¬q) and they sleep at least 8 hours a day (r).
_______________________ 9. Cigarette smoking is dangerous to your health (p) and it
gives bad breath (q), or it can kill you (r).
_______________________ 10. If x + 5 = 7 (p) and 2x – 7 = 6 (q), then x + 8 = 2 (r) or 6x
– 2 = 4 (s).
II. Write each symbolic statement as an English sentence. Use p, q, r, s, and t as defined below.
p: Sarah Geronimo is a singer.
q: Sarah Geronimo is not a songwriter.
r: Sarah Geronimo is an actress.
s: Sarah Geronimo plays guitar.
t: Sarah Geronimo is a dancer.
1. (p ˅ r) ˄ q
_____________________________________________________________________
2. p → (q ˄¬r)
_____________________________________________________________________
3. (r ˄ p) ↔ q
_____________________________________________________________________
4. ¬s → (p ˄¬q )
_____________________________________________________________________
5. 𝑡 ↔ (¬r ˄¬p)
_____________________________________________________________________
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3.3 Truth Table
The truth table is a table that shows the truth values of a compound statement for all
possible truth values of its simple statements.
Negation Statement
A statement is a negation of another if the word is not introduced in the negative
statement. Let p be a proposition. The negation of p is “not p” or ¬p.
The following is its truth table:
p
T
F
¬p
F
T
Examples:
What is the negative of the following statements?
a. p: √2 is a rational number.
b. q: 6 is an odd number.
Solution:
a. √2 is not a rational number. or √2 is an irrational number.
In symbols, ¬p.
b. 6 is not an odd number. or 6 is an even number.
In symbols, ¬q.
Conjunction Statement
The conjunction of p and q, denoted p ∧ q, is a statement that is true if both p and q are
true, and is false otherwise. We read p ∧ q as “p and q”
This definition can be represented by the “truth table”:
p
q
p∧q
T
T
T
T
F
F
F
T
F
F
F
F
Note: This truth table shows whether the new statement is true or false for each possible
combination of the truth or falsity of each p and q.
Example 1
Let
p: It is raining.
q: The streets are wet.
Then, the statement p ∧ q is “It is raining and the streets are wet.”
Example 2
Let
r: 9 is an even number.
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s: Ten is greater than 9.
Then, the statement r ∧ s is “9 is an even number and ten is greater than 9.”
Disjunction Statement
The disjunction of p and q, denoted p ∨ q, is a statement that is true if either p is true or q
is true or both are true, and is false otherwise. We read p ∨ q as “p or q”.
The following truth table presents this definition:
p
q
p∨q
T
T
T
T
F
T
F
T
T
F
F
F
Note: The truth of the statement p ∨ q means that at least one of p or q is true.
Example 1
Let
p: It is raining.
q: The streets are wet.
Then, the statement p ∨ q is “It is raining or the streets are wet.”
Example 2
Let
r: 9 is an even number.
s: Ten is greater than 9.
Then, the statement r ∨ s is “9 is an even number or ten is greater than 9.”
Conditional Statement
The conditional from p to q, denoted p → q, is a statement that is true if it is never the
case that p is true and q is false. We read p → q as “if p then q”. p is called the “antecedent or
hypothesis” and q is called the “consequent or conclusion”.
The following truth table presents this definition:
p
q
T
T
T
F
F
T
F
F
p →q
T
F
T
T
Notice that the conditional is a new example of a binary logical operator – it assigns to
each pair of statements p and q the new statement p→q.
Consider the following statement: "If you earn an A in logic, then I'll buy you a
Yellow Mustang." It seems to be made up out of two simpler statements:
p: "You earn an A in logic," and q: "I will buy you a Yellow Mustang."
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What the original statement is then saying is this: if p is true, then q is true, or,
more simply, if p, then q. We can also phrase this as p implies q, and we write p→q.
Now let us suppose for the sake of argument that the original statement: "If you
earn an A in logic, then I'll buy you a Yellow Mustang," is true. This does not mean that
you will earn an A in logic; all it says is that if you do so, then I will buy you that Yellow
Mustang. If we think of this as a promise, the only way that it can be broken is if you do
earn an A and I do not buy you a Yellow Mustang. In general, we use this idea to define
the statement p→q.
Notes:
1. The only way that p→q can be false is if p is true and q is false—this is the case of the
"broken promise."
2. If you look at the truth table again, you see that we say that "p→q" is true when p is
false, no matter what the truth value of q. This again makes sense in the context of the
promise — if you don't get that A, then whether or not I buy you a Mustang, I have not
broken my promise. However, it goes against the grain if you think of "if p then q" as
saying that p causes q. The problem is that there are really many ways in which the English
phrase "if ... then ..." is used. Logicians have simply agreed that the meaning given by the
truth table above is the most useful for mathematics, and so that is the meaning we shall
always use. Shortly we'll talk about other English phrases that we interpret as meaning the
same thing.
Here are some examples that will help to explain each line in the truth table.
Example 1 (True Implies True) is True
If p and q are both true, then p→q is true. For instance:
If 1+1 = 2 then the sun rises in the east.
Here p: "1+1 = 2" and q: "The sun rises in the east."
Notice that the statements p and q need not have anything to do with one another. We are
not saying that the sun rises in the east because 1+1 = 2, simply that the whole statement is
logically true.
Example 2 True Can't Imply False
If p is true and q is false, then p→q is false. For instance:
When it rains, I carry an umbrella.
Here p: "It is raining," and q: "I am carrying an umbrella." In other words, we can rephrase
the sentence as: "If it is raining then I am carrying an umbrella." Now there are lots of days
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when it rains (p is true) and I forget to bring my umbrella (q is false). On any of those days
the statement p→q is clearly false.
Notice that we interpreted "When p, q" as "If p, then q."
Example 3 False Implies Anything
If p is false, then p→q is true, no matter whether q is true or not. For instance:
If the moon is made of green cheese, then I am the King of England.
Here p: "The moon is made of green cheese," which is false, and q: "I am the King of
England." The statement p→q is true, whether or not the speaker happens to be the King
of England (or whether, for that matter, there even is a King of England).
Biconditional Statement
The biconditional from p to q, denoted p ↔ q, is a statement that is true if p and q are both
true or both false, and is false otherwise. We read p ↔ q as “p if and only if q” or “p iff q”.
The following truth table presents this definition:
p
T
T
F
F
q
T
F
T
F
p↔q
T
F
F
T
Note that, from the truth table, we see that, for p↔q to be true, both p and q must have the
same truth values; otherwise it is false.
Each of the following is equivalent to the biconditional p↔q.
p if and only if q.
p is necessary and sufficient for q.
p is equivalent to q.
Examples:
a. True or false? "1+1 = 3 if and only if Mars is a black hole."
b. Rephrase the statement: "I teach math if and only if I am paid a large sum of money."
Solution:
a. True. The given statement has the form p↔q, where p: "1+1=3" and q: "Mars is a
black hole." Since both statements are false, the biconditional p↔q is true.
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b. Here are some equivalent ways of phrasing this sentence:
"My teaching math is necessary and sufficient for me to be paid a large sum of
money."
"For me to teach math it is necessary and sufficient that I be paid a large sum of
money."
Sadly, for our finances, none of these statements are true.
More Examples:
I. Write the following in symbolic form using p, q, and r for statements and the symbols
¬, ⋀, ⋁, →, ↔ where
p: Pres. Duterte is a good president.
q: Government officials are corrupt.
r: People are happy.
a. If Pres. Duterte is a good president, then government officials are corrupt.
b. If government officials are not corrupt, then the people are happy.
c. If Pres. Duterte is a good president and people are happy, then the government officials are
not corrupt.
d. Pres. Duterte is not a good president if and only if government officials are corrupt and the
people are not happy.
Answers:
a. p→ (¬𝑞)
b. ¬𝑞 → 𝑟
c. (𝑝 ∧ 𝑟) → (¬𝑞)
d. ¬𝑝 ↔ (𝑞 ∧ (¬𝑟))
In constructing truth tables for a statement that involves a combination of conjunctions,
disjunctions, and/or negations, the illustrative examples will be shown below.
Illustrative Examples:
a. Construct a truth table for ¬(¬ p˅q)˅q.
b. Use the truth table in the previous discussion to determine the truth value of ¬(¬ p˅q)˅q,
given that p is true and q is false.
Solution:
a. Start with the standard truth table form and then include a ¬ p column.
p
T
T
F
F
q
T
F
T
F
¬p
F
F
T
T
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Now use the truth values from the ¬ p and q columns to produce the truth values for ¬p˅q, as
shown in the rightmost column of the following table.
p
q
¬ p ¬p˅q
T
T
F
T
T
F
F
F
F
T
T
T
F
F
T
T
Negate the truth values in the ¬p˅q column to produce the following.
p
q
¬ p ¬p˅q ¬(¬p˅q)
T
T
F
T
F
T
F
F
F
T
F
T
T
T
F
F
F
T
T
F
As the last step, form the disjunction of ¬(¬ p˅q)˅q with q and place the results in the rightmost
column of the table. See the following table. The shaded column is the truth table for ¬(¬ p˅q)˅q.
p
T
T
F
F
q
T
F
T
F
¬p
F
F
T
T
¬p˅q
T
F
T
T
¬(¬p˅q)
F
T
F
F
¬(¬ p˅q)˅q
T
T
T
F
b. In row 2 of the above truth table, we see that when p is true, and q is false, the statement
¬(¬ p˅q)˅q in the rightmost column is true.
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EXERCISES 3.3
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
I.
Write each sentence in symbolic form. Use p, q, r, and s as defined below. Tell whether
if each statement is true or false by applying the truth table.
p: Stephen Curry is a football player. (False)
q: Stephen Curry is a basketball player. (True)
r: Stephen Curry is a rock star. (False)
s: Stephen Curry plays for the Warriors. (True)
__________________________ 1. Stephen Curry is a football player or a basketball
player, and he is not a rock star.
__________________________ 2. Stephen Curry is a rock star, and he is not a
basketball player or a football player.
__________________________ 3. If Stephen Curry is a basketball player and a rock
star, then he is not a football player.
__________________________ 4. Stephen Curry is a basketball player, if and only if
he is not a football player and he is not a rock star.
__________________________ 5. If Stephen Curry plays for the Warriors, then he is a
basketball player and he is not a football player.
II.
Construct a truth table for [(¬𝑝 ∨ 𝑞 ∨ 𝑟)] ∧ [𝑠 ∧ (¬𝑞 ∨ ¬𝑟)]
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3.4 The Inverse, the Converse, the Contrapositive
Every conditional statement has three related statements. They are called the inverse, the
converse, and the contrapositive.
Suppose 𝑝 and 𝑞 are propositions. Given the implication 𝑝 → 𝑞. Its inverse is ¬p→ ¬q,
its converse is q→p, and its contrapositive is ¬q→ ¬p.
That is,
Given:
If p, then q.
Inverse:
If not p, then not q.
Converse:
If q, then p.
Contrapositive:
If not q, then not p.
To determine whether the conditional statement is true or false, we come up with the
following truth table. Referring to the truth table of the implication statement p→q, we then create
the truth table for the inverse, converse, and contrapositive statements.
p
T
T
F
F
q p →q ¬p→ ¬q q→p ¬q→ ¬p
T
T
T
T
T
F
F
T
T
F
T
T
F
F
T
F
T
T
T
T
Examples:
Give the converse, inverse, and contrapositive of the following implications:
a. If this movie is interesting, then I am watching it.
b. If p is prime number, then it is odd.
Answers:
a. Inverse: If this movie is not interesting, then I am not watching it.
Converse: If I am watching this movie, then it is interesting.
Contrapositive: If I am not watching this movie, then it is not interesting.
b. Inverse: If p is not a prime number, then it is not odd.
Converse: If p is an odd number, then it is prime.
Contrapositive: If p is not an odd number, then it is not a prime.
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EXERCISES 3.4
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
Give the inverse, converse, and contrapositive of the following implications and tell whether if the
statement is true or false.
1. If 𝜋 is an irrational number, then it is a number that goes on forever.
Inverse:
________________________________________________________________________
________________________________________________________________________
Converse:
________________________________________________________________________
________________________________________________________________________
Contrapositive:
________________________________________________________________________
________________________________________________________________________
2. If x is an even number, then x + 1 is even.
Inverse:
________________________________________________________________________
________________________________________________________________________
Converse:
________________________________________________________________________
________________________________________________________________________
Contrapositive:
________________________________________________________________________
________________________________________________________________________
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Chapter 4: PROBLEM SOLVING AND REASONING
 Learning Objectives
At the end of this chapter, the student is expected to:
 apply inductive and deductive reasoning to solve problems;
 solve problems involving patterns and recreational problems following Polya’s
Problem Solving Strategy; and
 organize one’s methods and approaches for proving and solving problems.
Introduction
Most occupations require good problem-solving skills. For instance, architects and
engineers must solve may complicated problems as they design and construct modern buildings
that are aesthetically pleasing, functional, and that meet stringent safety requirements. Two goals
of his chapter are to help you become a better solver and to demonstrate that problem solving can
be an enjoyable experience.
One example of this is the movie Die Hard: with a Vengeance (1995) starring Bruce Willis
and Samuel Jackson. In one of the action scenes, McClane and Carver (portrayed by Willis and
Jackson, respectively) were caught in a breathtaking scenario where they needed to keep a bomb
from exploding, and the only way to prevent the explosion is to put exactly four gallons of water
on a scale. How would they do it if they only have a five-gallon and a three-gallon jug?
5 gal
3 gal
4 gal
In this movie, the bomb did not explode, thanks to McClane’s quick reasoning ability and
mathematical strategy.
A good problem solver is the one who can find a resolution of which the path to the answer
is not immediately known. McClane epitomizes a good problem solver by using a strategy which
cannot be learned through school drills.
In the real world, decision-making and problem-solving are two key areas that one should
be good at in order to survive. In this chapter, you will learn to organize your own methods and
approaches to solve mathematical problems.
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4.1 Inductive and Deductive Reasoning
Inductive Reasoning
The type of reasoning that forms a conclusion based on the examination of specific
examples to reach a general conclusion of something is called inductive reasoning. The conclusion
formed by using inductive reasoning is called a conjecture. A conjecture is an idea that may or
may not be correct.
INDUCTIVE REASONING
Inductive Reasoning is the process of reaching a general conclusion by examining specific
examples.
When you examine a list of numbers and predict the next number in the list according some pattern
you have observed, you are using inductive reasoning.
Example 1: Use inductive reasoning to predict the next number in each of the following lists.
a. 5, 10, 15, 20, 25, ?
b. 1, 4, 9, 16, 25, ?
Solution:
a. Each successive number is 5 units larger than the preceding number. Thus, it can be
predicted that the next number in the list is 5 units larger than 25, which is 30.
b. Observe that all numbers are perfect squares. 1 = 12, 4 = 22, 9 = 32, 16 = 42, 25 = 52. Thus,
it can be predicted that the next number is 36, since 36 = 6 2.
Inductive reasoning is not just used only to predict number in a list. In Example 2, we use
inductive reasoning to make a conjecture about an arithmetic procedure.
Example 2: Use Inductive Reasoning to make a conjecture.
Consider the following procedure:
1. Pick a number.
2. Multiply the number by 10.
3. Add 8 to the product.
4. Divide the sum by 2.
5. And subtract 4.
Repeat the procedure for several different numbers. Make a conjecture between the
relationship of the size of the resulting number and the size of the original number using inductive
reasoning.
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Solution:
Suppose we pick 3 as our original number. Then the procedure would produce the
following results:
Original number:
3
Multiply 3 by 10:
3 x 10 = 30
Add 8 to the product:
8 + 30 = 38
Divide the sum by 2:
38 ÷ 2 = 19
Subtract the quotient by 4: 19 – 4 = 15
We started with 3 and the procedure produces 15. Starting with 2 as our original number
and the procedure produces 10. Starting with 5 as our original number and the procedure produces
25. Starting with 10 as our original number and the procedure produces 50. In each of these cases
the procedure produces a number that is five times larger than the original number. Thus, it is
conjectured that the given procedure produces a number that is five times larger than the original
number.
Example 3: Use the data in the table and by inductive reasoning, answer the following questions
below.
Earthquake
Max. Tsunami
(in Magnitude)
Height (in meters)
7.5
5
7.6
9
7.7
13
7.8
17
7.9
21
8.0
25
8.1
29
8.2
33
8.3
37
a. If the earthquake magnitude is 8.5, how high (in meters) can the tsunami be?
b. Can a tsunami occur when the earthquake magnitude is less than 7? Explain your answer.
Solution:
a. In the table, for every 0.1 increase in earthquake magnitude, the maximum tsunami height
increases by 4 meters. Thus, it is conjectured that the maximum tsunami height for the
earthquake magnitude of 8.5 is 45 meters.
b.
c. No, because when the earthquake magnitude is 7.4, the maximum tsunami height is only 1
meter. Hence, a tsunami does not occur when the earthquake magnitude is less than 7.
Conclusions based on inductive reasoning may not always be true. In other words, a
conjecture formed by using inductive reasoning may be incorrect. To illustrate this, consider the
circles on the next page. For each circle, all possible line segments have been drawn to connect
each dot on the circle with all the other dots on the circle.
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2
1
1
15
9
7
4
5
8
6
3
1 3
3
1
1
10 11
6
4
5
2
2
12
7
4
16
8
13
14
2
The maximum number of regions formed by connecting dots on a circle
Take Note: To produce the maximum number of regions, the dots on a circle must be placed so
that no three line segments that connect the dots intersect at a single point.
For each circle, count the number of regions formed by the line segments that connect the
dots on the circle. Your results should agree with the results on the table below.
Number of Dots
1 2 3 4 5 6
Maximum Number of Regions 1 2 4 8 16 ?
There appears to be a pattern. Each additional dot seems to double the number of regions.
Guess the maximum number of regions you expect for a circle with six dots. Check your guess
by counting the maximum number of regions formed by the line segments that connect six dots
on a large circle.
26
27
19
28
14
7
15
9
21
11
3
31
30
22
23
15
10
1
29
20
8
17
12
4
13
5
18
24
25
6
2
The line segments connecting six dots on a circle yield a maximum of 31 regions
Your drawing will show that for six dots, the maximum number of regions is 31 (see the
figure above), not 32 as you may have guessed. With seven dots the maximum number of regions
is 57. This is good example to keep in mind. Just because a pattern holds true for a few cases, it
does not mean the pattern will continue. When you use inductive reasoning, you have no guarantee
that your conclusion is correct.
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Counterexamples
A statement is a true statement provided that it is true in all cases. If you can find one case
for which a statement is not true, called counterexamples, then the statement is a false statement.
In Example 4, we verify that each statement is a false statement by finding a counterexample for
each.
Example 4: Find a Counterexample
Verify that each of the following statements is a false statement by finding a
counterexample.
For all numbers x:
a. |𝑥 | > 0
b. 𝑥 2 > 𝑥
c. √𝑥 2 = 𝑥
Solution:
a. Let 𝑥 = 0. Then |0| = 0. Because 0 is not greater than 0, we have found a
counterexample. Thus, “For all numbers 𝑥, |𝑥 | > 0” is a false statement.
b. For 𝑥 = 1, we have 12 = 1. Since 1 is not greater than 1, we have found a
counterexample. Thus, “For all numbers 𝑥, 𝑥 2 > 𝑥” is a false statement.
c. Consider 𝑥 = −3. Then √(−3)2 = √9 = 3. Since 3 is not equal to −3, we have found
a counterexample. Thus, “For all numbers 𝑥, √𝑥 2 = 𝑥” is a false statement.
Take Note: A statement may have many counterexamples, but we need only one counterexample
to verify that the statement is false.
Deductive Reasoning
Another type of reasoning is called inductive reasoning. Deductive reasoning is
distinguished from the inductive reasoning that uses general procedures and principles to reach a
conclusion.
DEDUCTIVE REASONING
Deductive Reasoning is the process of reaching a conclusion by applying general assumptions,
procedures, or principles.
Example 5: Use Deductive Reasoning to Establish a Conjecture
Consider the following procedure: Pick a number. Multiply the number by 10, add 8 to the
product, divide the sum by 2, and subtract 4.
Solution:
Let n represent the original number.
Multiply n by 10:
10n
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Add 8 to the product:
8 + 10n
Divide the sum by two:
(8 + 10n) ÷ 2 = 4 + 5n
Subtract the quotient by 4: 4 + 5n – 4 = 5n
We started with n and ended with 5n after the following given procedure. This means that
the given procedure produces a number that is five times larger than the original number.
Example 6: Solve a Logic Puzzle
Each of the four friends Donna, Sarah, Nikkie, and Xhanelle, has a different pet (fish, cat,
dog, and snake). From the following clues, determine the pet of each individual.
1. Sarah is older than her friend who owns the cat and younger than her friend who owns the
dog.
2. Nikkie and her friend who owns the snake are both of the same age and are the youngest
members of their group.
3. Donna is older than her friend who owns the fish.
Solution:
From Clue 1, Sarah does not own a cat nor a dog. In the following chart, write X1 (which
stands for “ruled out by clue 1”) in the cat and dog column for Sarah.
Fish
Donna
Sarah
Nikkie
Xhanelle
Cat
X1
Dog
Snake
X1
From Clue 2, Nikkie does not own a snake and a dog and being the youngest. And since
Sarah is not the youngest from Clue 1, then Sarah does not own a snake as well. Write X2 (ruled
out by clue 2) in snake column for Nikkie and X1 in snake column for Sarah. There are now Xs in
t he 3 pets in Sarah’s row, therefore Sarah owns the fish. Put a check ( ) which means Sarah’s
pet is a fish. So, Donna, Nikkie, and Xhanelle do ot own the fish.
Fish
X2
Cat
Fish
X2
Cat
X3
Dog
Snake
Donna
Sarah
X1
X1
X1
Nikkie
X2
X2
X2
Xhanelle
X2
From the Clue 3, Donna is older than Sarah, hence, Donna owns the dog. Write X3 (ruled
out by clue 3) in cat and snake column for Donna. There are now Xs in snake column for Donna,
Sarah, and Nikkie; therefore, Xhanelle owns the snake. Put a check in the box. Write X3 in the cat
column for Xhanelle; hence, Nikkie owns the cat. Put a check in the box.
Donna
Dog
Snake
X3
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Sarah
Nikkie
Xhanelle
X1
X2
X2
X3
X1
X2
X3
X1
X2
Thus, Sarah owns the fish, Donna owns the dog, Xhanelle owns the snake, and Nikkie
owns the cat.
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EXERCISES 4.1
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
A. Use inductive reasoning to predict the next number in each of the following lists.
1. 3, 6, 9, 12, 15, ?
2. 1, 3, 6, 10, 15, ?
3. 2, 4, 8, 16, 32, 64, ?
4. 1, 8, 27, 64, 125, ?
5. 2, 5, 10, 17, 26, ?
B. Use Inductive Reasoning to make a conjecture. Complete the procedure for several
different numbers.
Consider the following procedure:
1. Pick a number.
2. Multiply the number by 9.
3. Add 15 to the product.
4. Divide the sum by 3.
5. And subtract 5.
C. Verify that each of the following statement is a false statement by finding a counterexample
for each.
For all numbers x:
𝑥
1. = 1
2.
𝑥
𝑥+3
3
= 𝑥+1
3. √𝑥 2 + 16 = 𝑥 + 4
D. Use Inductive Reasoning to Solve an Application
Scientists often use inductive reasoning. For instance, Galileo Galilei (1564-1642)
used inductive reasoning to discover that the time required for a pendulum to complete the
swing, called the period of the pendulum, depends on the length of the pendulum. Galileo did
not have a clock, so he measured the periods of pendulum in “heartbeats.” The following table
shows some results obtained for pendulums of various lengths. For the sake of convenience,
a length of 10 inches has been designated as 1 unit.
Length of Pendulum
Period of Pendulum(in
The period of a
(in
units)
heartbeats)
pendulum is the
1
1
time it takes for
4
2
the pendulum to
swing from left to
9
3
right and back to
16
4
its
original
25
5
position.
36
6
Use the data in the above table and inductive reasoning to answer each of the following
questions.
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a. If a pendulum has a length of 49 units, what is its period?
b. If the length of a pendulum is quadrupled, what happens to its period
E. Use Deductive Reasoning to Establish a Conjecture
1. Consider the following procedure: Pick a number. Multiply the number by 8, add 6 to
the product, divide the sum by 2, and subtract 3.
2. Consider the following procedure: Pick a number. Add 3 to the number and multiply
the sum by 2. Subtract 6 from the product then divide the result by 2.
F. Each of four neighbors, Sean, Maria, Sarah, and Brian, has a different occupation (editor,
banker, chef, and dentist). From the following clues, determine the occupation of each
neighbor.
1. Maria gets home from work after the banker but before the dentist.
2. Sarah, who is the last to get home from work, is not the editor.
3. The dentist and Sarah leave for work at the same time.
4. The banker lives next door to Brian.
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4.2 Problem Solving with Patterns
Terms of a Sequence
An ordered list of numbers such as 5, 14, 27, 44, 65, … is called a sequence. The numbers
in a sequence that are separated by commas are the terms of the sequence. In the above sequence,
5 is the first term, 14 is the second term, 27 is the third term, 44 is the fourth term, and 65 is the
fifth term. The three dots “…” indicate that the sequence continues beyond 65, which is the last
written term. It is customary to use the subscript notation an to designate the nth term of a
sequence. That is,
a1 represents the first term of a sequence.
a2 represents the second term of a sequence.
a3 represents the third term of a sequence.
.
.
.
an represents the nth term of a sequence.
In the sequence 2, 6, 12, 20, 30, …, 𝑛2 + 𝑛,…
a1 = 2, a2 = 6, a3 = 12, a4 = 20, a5 = 30, and an = 𝑛2 + 𝑛
When we examine a sequence, it is natural to ask:


What is the next term?
What formula or rule can be used to generate the terms?
To answer these questions, we often construct a difference table, which shows the
differences between successive terms of the sequence. The following table is a difference table for
the sequence. The following table is a difference table for the sequence 2, 5, 8, 11, 14, …
Sequence:
2
First differences:
5
3
8
3
11
3
…
14
3
…
(1)
Each of the numbers in row (1) of the table is the difference between the two closest
numbers just above it (upper right number minus upper left number). The differences in row (1)
are called the first differences of the sequence. In this case, the first differences are all the same.
Thus, if we use the above difference table to predict the next number in the sequence, we predict
that the next term is 17 since 14 + 3 = 17. This prediction might be wrong; however, the pattern
shown by the first differences seems to indicate that each successive term is 3 larger than the
preceding term.
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The following table is a difference table for the sequences 5, 14, 27, 44, 65, …
Sequence:
5
First differences:
14
9
Second differences:
27
13
4
44
17
4
…
65
…
21
(1)
…
4
(2)
In this table, the first differences are not all the same. In such a situation it is often helpful
to compute the successive differences of the first differences. These are known in row (2). These
differences of the first differences are called the second differences. The differences of the
second differences are called the third differences.
To predict the next term of a sequence, we often look for a pattern in a row of differences.
For instance, in the following table, the second differences shown below are all the same constant,
namely 4. If the pattern continues, then a 4 would also be the next second difference, and we can
extend he table to the right as shown.
Sequence:
5
First differences:
14
9
Second differences:
27
13
4
44
17
4
…
65
…
21
4
(1)
4
(2)
Now we work upward. That is, we add 4 to the first difference 21 to produce the next first
difference, 25. We then add this difference to the fifth term, 65, to predict that 90 is the next term
in the sequence. This process can be repeated to predict additional terms of the sequence.
Sequence:
5
First differences:
14
9
Second differences:
27
13
4
44
17
4
65
21
4
90
25
4
(1)
(2)
Example 1: Predict the Next Term of a Sequence
Use a difference table to predict the next term in the sequence.
2, 7, 24, 59, 118, 207, …
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Solution:
Construct a difference table as shown below.
Sequence:
2
First differences:
7
5
Second differences:
Third differences:
24
17
12
59
35
18
6
118
59
24
6
207
89
30
6
332
125
36
6
(1)
(2)
(3)
The third differences, shown in row (3), are all the same constant, 6. Extending row (3) so
that it is includes an additional 6 enables us to predict that the next second difference will be 36.
Adding 36 to the first difference, 89, gives us the next first difference, 125. Adding 125 to the sixth
term, 207, yields 332. Using the method of extending the difference table, we predict that 332 is
the next term in the sequence.
Fibonacci Sequence
Fibonacci’s rabbit problem in chapter 1 is not a realistic model of
population growth of rabbits but is a very good example of a mathematical
problem solved using patterns. It is interesting to note that this famous rabbit
problem paved the way to the discovery of a phenomenal sequence of
numbers known as the Fibonacci sequence.
A sequence is an ordered list of numbers, separated by commas, are
called
the
terms of the sequence. From our discussion in section 1.2, we
Leonardo Pisano
knew that the first six terms of the Fibonacci sequence are 1, 1, 2, 3, 5, 8. If
we use the mathematician notation Fn to denote the nth term of the Fibonacci sequence, then,
For the first month, n =1, F1 = 1. For the second month, n = 2, F2 = 1.
For the third month, n =3, F3 = 2. For the fourth month, n = 4, F4 = 3.
For the fifth month, n =5, F3 = 2. For the fourth month, n = 4, F4 = 3.
The Fibonacci sequence then is the ordered list of numbers 1, 1, 2, 3, 5, 8, …, Fn, …where
the three dots indicate that the sequence continues beyond 8 and Fn.
How do we determine Fn, the nth term? Observe that,
F2 = F1
F3 = F2 + F1
F4 = F3 + F2
F5 = F4 + F 3
F6 = F5 + F 4
From these patterns, we conjecture that Fn = Fn – 1 + Fn – 2, for n ≥ 3. Fibonacci discovered
that a Fibonacci number can be found by adding its previous two Fibonacci numbers.
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The Fibonacci Numbers
F1 = 1, F2 = 1, and Fn = Fn – 1 + Fn – 2, for n ≥ 3.
Example 2: Finding a Fibonacci Number.
Use the definition of Fibonacci numbers to find the eight and tenth Fibonacci numbers.
Solution:
The eight Fibonacci number is the sum of the two previous Fibonacci numbers. Thus,
F8 = F7 + F 6
= (F6 + F5) + F6
= (8 + 5) + 8
= 13 + 8
= 21
The tenth Fibonacci number is the sum of the two previous Fibonacci numbers in an
ordered sequence. Thus,
F10 = F9 + F8
= (F8 + F7) + F8
= (21 + 13) + 21
= 34 + 21
= 55
It is easy to find the nth Fibonacci number Fn if the two previous
numbers, Fn-1 and Fn-2 are known. Suppose we want to find F20. Using the
definition, it is tedious and time consuming to compute F19 and F18 to
determine F20. Fortunately, Jacques Binet in 1543 was able to find a formula
for the nth Fibonacci number:
Binet’s Formula
Fn =
Jacques Binet
1
√5
1+√5
[(
2
𝑛
1−√5
𝑛
) −( 2 ) ]
Example 3: Use Binet’s formula and a calculator to find the 20th and 50th Fibonacci number.
Solution:
F20 =
1
√5
1+√5
[(
= 6,765
2
20
)
1−√5
−( 2
20
) ]
F50 =
1
√5
1+√5
[(
2
50
)
1−√5
50
−( 2 ) ]
= 12, 586, 269, 020
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Pascal’s Triangle
Another famous mathematician who loves patterns is Blaise Pascal (1623 – 1662). For
hundreds of years, many mathematicians were intrigued with the Pascal’s triangle. The figure
below illustrates the first seven rows of the Pascal’s triangle. As you can see, each row starts and
ends with the number 1. Any other number x is the sum of the two numbers in the previous row
closest to that number x. For instance, the number 15 in row 6 is the sum of numbers 5 and 10
closest to it in the previous row,
row 0
row 1
row 2
row 3
row 4
row 5
row 6
Blaise Pascal
In algebra, expanding (𝑥 + 𝑦)3 = 𝑥 3 + 3𝑥 2 𝑦 + 3𝑥𝑦 2 + 𝑦 3 is just a simple special product
process. But expanding (𝑥 + 𝑦)6 can be tedious. Amazingly, note that the numerical coefficients
of the expansion of (𝑥 + 𝑦)3 = 𝑥 3 + 3𝑥 2 𝑦 + 3𝑥𝑦 2 + 𝑦 3 are the entries of row 3 of the Pascal’s
triangle, i.e., 1, 3, 3, 1. Moreover, take note that the exponents of x in the expansion starts with 3
and decreasing in the succeeding terms while the exponents of y starts with 0 and increasing in the
remaining terms. Now, we expand (𝑥 + 𝑦)6 using the entries in row 6 (1, 6, 15, 20, 15, 6, 1) of the
Pascal’s triangle. The result is given below.
(𝑥 + 𝑦)6 = 𝑥 6 + 6𝑥 5 𝑦 + 15𝑥 4 𝑦 2 + 20𝑥 3 𝑦 3 + 15𝑥 2 𝑦 4 + 6𝑥𝑦 5 + 𝑦 6
Can you try expanding (𝑥 − 𝑦)7 ?
Suppose you add the horizontal entries in the rows of the Pascal’s triangle except row 0.
What pattern do you observe in these sums? Can you predict the sum of the sum of the entries in
row 10?
Row
1
2
3
4
5
6
10
Sum
2
4
8
16
32
64
?
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Another amazing discovery in Pascal’s triangle is that when you get the sum of the numbers
using lines as shown in the next figure, the Fibonacci sequence appears. The first seven Fibonacci
numbers 1, 1, 2, 3, 5, 8, 13 show up.
Website Application
Another equally famous problem involving patterns is the Tower of Hanoi, invented by
Edouard Lucas in 1883. The Tower of Hanoi is a puzzle consisting of three pegs and a number of
disks of distinct diameters piled as shown in the figure below.
The puzzle requires that all the disks be moved from the first peg to the third peg such that
the largest disk is on the bottom, the next largest disk is placed on top of the largest disk and so on
and that only one disk be moved at a time. All pegs may be used.
Determine the minimum number of moves required to transfer the disks from the first peg
to the third peg for each of the following situations. Visit the website
https://www.mathisfun.com/games/towerofhanoi.html for a nice simulation of the puzzle.
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EXERCISES 4.2
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
A. Use a difference table to predict the next term in the sequence.
1. 1, 14, 51, 124, 245, 426, ?
2. – 2, 2, 12, 28, 50, 78, ?
3. – 4, – 1, 14, 47, 104, 191, 314, ?
4. 5, 6, 3, – 4, – 15, – 30, – 49, ?
5. 2, 0, – 18, – 64, – 150, – 288, –490, ?
B. Use the given nth term formula to compute the first six terms of the sequence.
1. 𝑎𝑛 = 2−𝑛
2. 𝑎𝑛 = (−1)𝑛+1 𝑛2
3. 𝑎𝑛 =
𝑛2 −1
𝑛
𝑛
4. 𝑎𝑛 = 𝑛+1
5. 𝑎𝑛 = (−1)(𝑛2 − 𝑛 + 7)
C. Expand the following algebraic expressions using Pascal’s triangle.
1. (x + y)5
2. (x – 2y)4
3. (x + y)8
4. (3x + 2y)4
5. (2x2 – y3)5
D. Determine the minimum number of moves required to transfer all of the disks to another
peg for each of the following situations.
1. You start with four disks.
2. You start with five disks.
3. You start with six disks.
4. You start with seven disks.
5. You start with eight disks.
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4.3 Polya’s Problem-Solving Strategy
One of the recent mathematician who outlined a strategy for
solving problems form virtually any discipline is George Polya (18871985). In his book, How to Solve It, he writes, “A great discovery solves
a great problem but there is a grain of discovery in the solution of any
problem. Your problem may be modest; but if it challenges your curiosity
and brings into play your incentive faculties, and if you solve it by your
own means, you may experience the tension and enjoy the triumph of
discovery.” Because of his ideas, he is considered the father of problemsolving among mathematicians. The following four-step strategy is
named after him:
George Polya
Polya’s Four-Step Problem-Solving Strategy
1.
2.
3.
4.
Understand the problem
Devise a plan
Carry out the plan
Review the solution
Understand the Problem
This part of problem-solving is sometimes, if not always, neglected. In order to solve a
problem, one must first know what is being asked, and what information or data can be extracted
from what is given. Furthermore, one must see to it that he or she can state the problem in his or
her own words.
Devise a Plan
For this step, one must think of strategies to solve the problem. Some of these strategies
include organizing the given information using a list, table or chart; drawing a diagram; working
out the problem backwards; looking for a pattern; trying to solve a similar but simpler problem;
writing an equation; or simply guessing at a possible solution and then later checking if the result
is valid.
Carry Out the Plan
Carrying out a plan to solve the problem is basically implementing the strategy chosen in
the second step until the problem is solved or until a new course of action is suggested. One may
get ideas from others in deciding the best strategy to make sure that the best solution is employed.
Review the Solution
Questions like “Is your answer reasonable?” is important in checking the veracity of the
answer to the problem. For example, if one is looking for the dimensions of a rectangular box of
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least cost and his or her answer yields a negative length, he or she can automatically say that there
must be something wrong with the solution because there is no such box with negative dimensions.
Example 1: Apply Polya’s strategy in solving the following problem.
The GSW basketball team won three out of their last six games. In how many different
orders could they have attained three wins and three losses in six games?
Solution:
Understand the Problem. There are many different ways. GSW may have won three straight
wins and three losses (WWWLLL), or maybe they lost in the first three games and won in the last
three games (LLLWWW). Likewise, there are other several orders.
Devise a Plan. One can organize a list of all possibilities making sure that no entry will be
duplicated.
Carry Out the Plan. Three Ws must be presented in every entry without duplication The strategy
is to start the list with three consecutive wins. Next in the list are all the entries starting with two
consecutive wins, then next in the list are all the entries starting with a single win. Following this
pattern, consider starting with three consecutive losses and so on. Here are the different orders:
1.
2.
3.
4.
5.
6.
7.
8.
WWWLLL
WWLWLL
WWLLWL
WWLLLW
WLLLWW
WLLWLW
WLWWLL
WLWLWL
9. LLLWWW
10. LLWLWW
11. LLWWLW
12. LLWWWL
13. LWWWLL
14. LWWLWL
15. LWLLWW
16. LWLWLW
Review the Solution. The list is organized and has no duplicates, so there are sixteen (16) different
orders in which a basketball team can win exactly three out of six games.
Example 2: Solving a tour problem.
An agency charged Php 15,000.00 for a 3-day and 2-night tour in Macau and Php 20,000.00
for the same tour with a side trip in Hong Kong. Ten persons joined the trip, which enable them to
collect Php 170,000.00. How many tourists made a side trip to Hong Kong?
Solution:
Understand the Problem. There are two types of tourists in the situation given. Some purely
stayed in Macau while others made a side trip to Hong Kong. From the total collection, how much
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was the amount collected from those who made side trips to Hing Kong. It is needed to know how
many were bound in Macau and who made a side trip to Hong Kong.
Devise a Plan. Use x and y to represent the two types of tourists. Define these variables.
Let
x = number of tourists bound in Macau alone
y = number of tourists bound in Macau but who made a side trip to Hong Kong.
Hence, we have the following algebraic equations:
15,000x = amount collected from the tourists bound in Macau alone
20,000y = amount collected from the tourists bound in Macau but who made a side
trip to Hong Kong.
Carry Our the Plan. Write the equations and solve using the elimination method to the system of
equations.
Equations:
x + y = 10
15,000x + 20,000y = 170,000
(1)
(2)
To find the number of tourists bound in Macau but who made a side trip in Hong Kong,
we solve for y.
To do this, we use elimination by substitution.
a. Solve for y in (1)
x + y = 10
y = 10 – x
(3)
b. Substitute y = 10 – x in equation (2)
15,000x + 20,000(10 – x) = 170,000
15,000x + 200,000 – 20,000x = 170,000
- 5,000x = 170,000 – 200,000
- 5,000x = - 30,000
x=
−30,000
−5,000
x=6
Substituting x = 6 in equation (3), y = 10 – x = 10 – 6 = 4.
Therefore, four tourists made a side trip to Hong Kong.
Review the Solution. Since there are a total of 10 tourists, six of them only stayed in Macau while
four made a side trip to Hong Kong. Now, 15,000(6,000) + 20,000(4) = 170,000. This satisfies the
condition that the total amount collected for the whole trip is Php 170,000.00.
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Example 3
In consecutive turns of a Monopoly game, Stacy first paid £800 for a hotel. She then lost
half her money when she landed on Boardwalk. Next, she collected £200 for passing GO. She then
lost half for remaining money when she landed on Illinois Avenue. Stacy now has £2,500. How
much did she have just before she purchased the hotel?
Solution:
Understand the Problem. We need to determine the number of euro that Stacy had just prior to
her £800 hotel purchase.
Devise a Plan. We could guess and check, but we might need to make several guesses before we
found the correct solution. An algebraic method might work, but setting up the necessary equation
could be a challenge. Since we know the result, let’s try the method of working backwards.
Carry Out the Plan. Stacy must have had £5,000 just before she landed on Illinois Avenue;
£4,800 just before she passed GO; and £9,600 prior to landing on Boardwalk. This means she had
£10,400 just before she purchased the hotel.
Review the Solution. To check our solution, we start with £10,400 and proceed through each of
the transactions. £10,400 less £800 is £9,600. Half of £9,600 is £4,800. £4,800 increased by £200
is £5,000. Half of £5,000 is £2,500.
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EXERCISES 4.3
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
A. Apply the Polya’s Problem Solving Strategy by identifying your own problem and life.
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
B. Apply Polya’s Problem Solving Strategy (Guess and Check)
1. A baseball team won two out of their last four games. In how many different orders
could they have two wins and two losses in four games?
2. Determine the digit 100 places to the right of the decimal point in the decimal
7
representation .
27
3. The product of the ages, in years, of three teenagers is 4590. None of the teens are the
same age. What are the ages of the teenagers?
4. A hat and a jacket together cost Pnp 100.00. The jacket costs Php 90.00 more than the
hat. What are the cost of the hat and the cost of the jacket?
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Chapter 5: THE MATHEMATICS OF FINANCE
 Learning Objectives
At the end of this chapter, the student is expected to:
 Differentiate simple and compound interest;
 Define credit cards, stocks, bonds, and mutual funds;
 Solve problems involving simple and compound interest; and
 Display perseverance and patience in dealing with problems associated with
mathematics of finance.
Introduction
Everybody uses money. Sometimes you work for your money and other times your money
works for you. For example, unless you are attending college on a full scholarship, it is very likely
that you and your family have either saved money or borrowed money, or both, to pay for your
education. When we borrow money, we normally have to pay interest for that privilege. When we
save money, for a future purchase or retirement, we are lending money to a financial institution
and we expect to earn interest on our investment. We will develop the mathematics in this chapter
to understand better the principles of borrowing and saving. These ideas will then be used to
compare different financial opportunities and make informed decisions
5.1Simple Interest and Compound Interest
In the business world, an investor who places capital in a productive enterprise expects not
only the eventual return of his capital but also additional payment. An individual who lends his
capital expects the debtor to pay back not only the money originally borrowed but also an
additional amount. This additional payment or amount is called interest. The interest is the
compensation that a borrower of capital pays to a lender for its use. It can be viewed as a form of
rent that the borrower pays to the lender to compensate for the loss of opportunity to use the capital
to other productive financial transaction.
Therefore, interest is the fee paid for borrowed money. We receive interest when we let
others use our money (for example, by depositing money in a savings account or making a loan).
We pay interest when we use other people’s money (such as when borrow from a bank or a friend).
5.1.1 Simple Interest
It is an interest that is calculated on the balance owned but not on previous interest or in
other words if interest is computed on the original principal during the whole life of the investment,
the interest due at the end of the time is called simple interest. It is computed entirely on the original
principal (P), multiplied by the rate of interest (r), and the time (t). This leads to simple interest
formula.
Simple Interest Formula:
𝐼 = 𝑃𝑟𝑡
𝐼
𝑡 = 𝑃𝑟
𝐼
𝑃 = 𝑟𝑡
𝐼
𝑟 = 𝑃𝑡
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where
I = amount of interest
P = principal
r = rate per period of time
t = time between the date of loan is made and the date it matures
Definitions of Basic Terms





Principal (P) – It is the original amount borrowed.
Rate of Interest (r) – It is the percent that the borrower pays for the use of the money
commonly expressed as annual interest rate.
Time (t) – It is the length of time usually expressed in years.
Maturity Date or Due Date – It is the date on which the loan is to be repaid.
Maturity Value (M) – It is the total amount the borrower would need to pay back.
Examples:
1. Natasha invests P250,000 in a building society account. At the end of the year her account
is credited with 2% interest. How much interest had her P250,000 earned in the year?
Solution:
P = P250,000 r = 2% or 0.02
t = 1 year
𝐼 = 𝑃𝑟𝑡
= (250,000)(0.02)(1)
= 𝑃5,000
2. A business borrowed 10 million pesos from the bank. If he agrees to pay an 8% annual rate
of interest, calculate the amount of interest in (a) 5 years, (b) 10 years, and (c) 15 years.
Solution:
P = P10,000,000
r = 8%
a. For t = 5 years
𝐼 = 𝑃𝑟𝑡
= (10,000,000)(0.08)(5)
= 𝑃4,000,000.00
b. For t = 10 years
𝐼 = 𝑃𝑟𝑡
= (10,000,000)(0.08)(10)
= 𝑃8,000,000.00
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c. For t = 15 years
𝐼 = 𝑃𝑟𝑡
= (10,000,000)(0.08)(15)
= 𝑃12,000,000.00
Two methods for converting time from days to years
a. Exact Method
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑎𝑦𝑠
365
b. Ordinary Method (mostly used by businessmen)
𝑡=
𝑡=
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑎𝑦𝑠
360
3. Compute the simple interest of a loan amounting to P50,000.00 payable in 6 months if the
interest rate is 3.5%.
Solution:
P = P50,000.00
r = 3.5%
t = 6 months
𝐼 = 𝑃𝑟𝑡
6
)
12
= 𝑃 875.00 𝑖𝑛 𝑠𝑖𝑥 𝑚𝑜𝑛𝑡ℎ𝑠
= (50,000)(0.035)(
4. Mr. Flores plans to buy a Sala set from Department Store which cost P12,000.00. The loan
charges P1, 800.00 interest in 6 months. Find the simple interest rate.
Solution:
P = P 12,000.00
I = P 1,800.00
t = 6 months
𝑟=
𝐼
𝑃𝑡
=
1,800.00
= 0.3(100%) = 30%
6
(12,000)(12)
5. Ryan borrowed P750,000 from a bank to buy a car at 10% interest rate and earned P30,000
interest while clearing the loan, find the time for which the loan was given.
Solution:
P = P 750,000
r = 10%
I = P 30,000
𝑡=
=
𝐼
𝑃𝑟
30,000
( 750,000)(0.01)
= 4 𝑦𝑒𝑎𝑟𝑠
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Year Table
Day – of – the – Year – Table
Day of
Jan Feb Mar Apr May Jun Jul Aug Sept Oct Nov Dec
the
Month
1
1
32
60
91
121 152 182 213 244 274 305 335
2
2
33
61
92
122 153 183 214 245 275 306 336
3
3
34
62
93
123 154 184 215 246 276 307 337
4
4
35
63
94
124 155 185 216 247 277 308 338
5
5
36
64
95
125 156 186 217 248 278 309 339
6
6
37
65
96
126 157 187 218 249 279 310 340
7
7
38
66
97
127 158 188 219 250 280 311 341
8
8
39
67
97
128 159 189 220 251 281 312 342
9
9
40
68
99
129 160 190 221 252 282 313 343
10
10
41
69
100 130 161 191 222 253 283 314 344
11
11
42
70
101 131 162 192 223 254 284 315 345
12
12
43
71
102 132 163 193 224 255 285 316 346
13
13
44
72
103 133 164 194 225 256 286 317 347
14
14
45
73
104 134 165 195 226 257 287 318 348
15
15
46
74
105 135 166 196 227 258 288 319 349
16
16
47
75
106 136 167 197 228 259 289 320 350
17
17
48
76
107 137 168 197 229 260 290 321 351
18
18
49
77
108 138 169 199 230 261 291 322 352
19
19
50
78
109 139 170 200 231 262 292 323 353
20
20
51
79
110 140 171 201 232 263 293 324 354
21
21
52
80
111 141 172 202 233 264 294 325 355
22
22
53
81
112 142 173 203 234 265 295 326 356
23
23
54
82
113 143 174 204 235 266 296 327 357
24
24
55
83
114 144 175 205 236 267 297 328 358
25
25
56
84
115 145 176 206 237 268 298 329 359
26
26
57
85
116 146 177 207 2382 269 299 330 360
27
27
58
86
117 147 178 208 239 270 300 331 361
28
28
59
87
118 148 179 209 240 271 301 332 362
29
29
--88
119 149 180 210 241 272 302 333 363
30
30
89
120 150 181 211 242 273 303 334 364
31
31
90
151
212 243
304
365
According to Richard Aufmann, the day of the year table can be used to determine the
number of days from one date to another date. For instance, because June 30 is day 181 on the
table and November 11 is day 315, meaning there are 315 – 181 = 134 days from June 30 to
November 11.
The table can also be used to determine the due date of a loan. For instance, an 85 – day
loan made on march 15, which is day 74 is due on day 74 + 85 = day 159 which is June 8.
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Example:
1. Calculate the simple interest due on a P20,600.00 loan made on February 8 and repaid on
December 8 of the same year. The interest rate is 7%.
Solution:
February 8 is day 39 and December 8 is day 342.
342 – 39 = 303 days (term of the loan)
𝐼 = 𝑃𝑟𝑡
= (20,600)(0.07)(
= 𝑃1,213.68
303
)
360
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EXERCISES 5.1.1
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. Mrs. Rodriguez wants to purchase a washing machine listed at P25,000.00 cash and
P25,950.00 if paid at an instalment basis of 4 months. What is the rate of interest?
2. An interest of P850.00 was earned in 5 months on an investment at 10%. How much was
invested?
3. What principal will accumulate to P215,000.00 in 3 years at 12% simple interest?
4. A bank issued a 6-year loan of P500,000.00 with a simple interest of 7% to an employee.
Determine the interest which the employee must pay.
5. A cash of P250,000 is deposited to an account paying at 5% simple interest. How much is
the account after five years?
6. Find the interest on a loan of P65,000.00 at 12% interest which will be paid after 6 months.
7. A P10,000.00 savings account earned P1,400.00 interest in 3 years. What was the rate of
interest given?
8. Find the number of days from March 15 to September 15 of the same year and calculate
the simple interest due on a P35,800.00 loan made with an interest rate of 1.5%.
9. Calculate the simple interest due on a P25,400.00 loan made on June 30 and repaid on
February 25 of the following year with 1.65% given interest rate.
10. Find the due date on a 60-day loan made on November 11.
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5.1.2 Maturity Value
It is the total amount the borrower would need to pay back and is usually denoted by M.
To find the maturity value, we simply add interest to the principal.
Maturity Value
M=P+I
M = Maturity Value P = Principal I = amount of interest
Examples:
1. Calculate the maturity value of a P10,000.00 loan with 8% interest rate (a) in 5 years and
(b) in 8 months.
Solution:
P = P10,000.00
r = 8%
a. In 5 years
Find I:
𝐼 = 𝑃𝑟𝑡
= (10,000)(0.08)(5)
= 𝑃4,000.00
Find M:
𝑀=𝑃+𝐼
= 10,000 + 4,000
= 𝑃14,000.00
b. In 8 months
Find I:
𝐼 = 𝑃𝑟𝑡
= (10,000)(0.08)(
= 𝑃533.33
Find M:
8
)
12
𝑀=𝑃+𝐼
= 10,000 + 533.33
= 𝑃10,533.33
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We can also make use of other formula in computing for the maturity value. This can be done by
substituting Prt for I (interest).
M = P + I = P + Prt = P(1 + rt)
2. Calculate the maturity value of a simple interest, 9 months loan of P15,300. The interest
rate is 8%.
Solution:
P = P15,300.00
r = 8%
t = 9 months
𝑀 = 𝑃(1 + 𝑟𝑡)
9
)]
12
= 15,300(1 + 0.0675)
= 15,300 [1 + (0.08) (
= 15,300(1.0675)
= 𝑃16,332.74
3. The maturity value of a 4-month loan of P5,000.00 is P5075.00. What is the simple interest
rate?
Solution:
P = P5,000.00
M = P5,075.00
t = 4 months
Find I:
𝐼 = 𝑀−𝑃
= 5,075 − 5,000
= 75
Find r:
𝑟=
=
𝐼
𝑃𝑡
75
4
(5,000)(12)
75
1,666.67
= 0.045(100%) = 4.5%
=
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EXERCISES 5.1.2
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. Calculate the maturity value of a simple interest, a 10-month loan of P20,000.00 if the
interest rate is 3.75%.
2. A credit union has issued a 6-month loan of P10,500.00 at a simple interest rate of 2.5%.
What amount will be repaid at the end of six months.
3. An employee applied a P50,000.00 loan from the bank. If she agrees to pay the loan in 6
months with a simple interest rate of 1.25% per month. How much should he repay the
bank?
4. P45,000.00 is borrowed for 90 days at a 5% interest rate. Calculate the maturity value by
the exact method and by the ordinary method.
5. Joshua borrowed P4,895.00 from his employer. He promised to repay him in 60 days with
an interest of 10%. How much will he pay using the exact interest?
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5.1.3 Compound Interest
Compounding is the concept that any amount earned on an investment can be reinvested
to create additional earnings that would not be realized based on the original principal, or original
balance, alone The interest on the original balance alone would be called simple interest. The
additional earnings plus simple interest would be equal to the total amount earned from compound
interest.
In other words, an investment earns compound interest when the interest from each time
period is added to the principal. And the earns interest in the following time periods. As the
principal grows, the rate at which you earn interest grows as well, because you are earing “interest
on interest”. Compounding makes a significant difference in the final value of an investment.
Compounding increases the amount you earn when investing, but increase the costs when you
borrow money.
The compound interest formula calculates the amount of interest earned on an account or
investment where the amount earned is reinvested. By reinvesting the amount earned, an
investment will earn money based on the effect of compounding.
Examples:
1. Jonathan deposits P5,000.00 in a savings account earning 2% interest compounded
annually.
Solution:
Compounded annually means that the interest will be calculated once a year.
𝐼 = 𝑃𝑟𝑡 = (5,000)(0.02)(1) = 𝑃100.00
At the end of one year, his money on bank will be
𝑀 = 𝑃 + 𝐼 = 5,000 + 100 = 𝑃5100.00
During its second year,
𝐼 = 𝑃𝑟𝑡 = (5,100)(0.02)(1) = 𝑃102.00
At the end of the second year, the total amount in the account is
𝑀 = 𝑃 + 𝐼 = 5,100 + 102 = 𝑃5,202.00
the interest earned during the third year is calculated using the amount in the
account at the end of the second year (P5,202.00)
𝐼 = 𝑃𝑟𝑡 = (5,202)(0.02)(1) = 𝑃104.04
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Notice that the interest earned every year increases. This is what compound interest is all
about. However, compound interest is not only limited to annually, we also have semiannually or
twice a year, quarterly or four times a year, monthly or even daily. We call this frequency as
compounding period.
For instance, in our example number 1, if the interest is compounded semiannually,
meaning the first interest payment occurs after 6 months and the earned interest is added to the
account.
Interest earned after six months:
6
) = 𝑃50.00
12
𝑀 = 𝑃 + 𝐼 = 5,000 + 50 = 𝑃5,050.00
𝐼 = 𝑃𝑟𝑡 = (5,000)(0.02) (
Interest earned after second six months:
6
) = 𝑃50.50
12
𝑀 = 𝑃 + 𝐼 = 5,050 + 50.50 = 𝑃5,100.50
𝐼 = 𝑃𝑟𝑡 = (5,050)(0.02) (
The total amount in the account at the end of the first year is P5,100.50 which is called
compound amount.
Maturity value formula of 𝑀 = 𝑃(1 + 𝑟𝑡) can also be used to calculate M at the end of
six months.
2. Mr. Agoncillo deposited P16,400.00 in an account earning 3% interest, compounded
quarterly. How much is in the account at the end of 1 year.
Solution:
a. First Quarter
𝑀 = 𝑃(1 + 𝑟𝑡)
= 16,400 [1 + (0.03) (
3
)]
12
= 𝑃16,523.00 (end of the 1st quarter)
b. Second Quarter
𝑀 = 𝑃(1 + 𝑟𝑡)
3
)]
12
= 𝑃16,646.92 (end of the 2nd quarter)
= 16,523 [1 + (0.03) (
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c. Third Quarter
𝑀 = 𝑃(1 + 𝑟𝑡)
= 16,646.92 [1 + (0.03) (
3
)]
12
= 𝑃16,771.77 (end of the 3rd quarter)
d. Fourth Quarter
𝑀 = 𝑃(1 + 𝑟𝑡)
= 16,771.77 [1 + (0.03) (
3
)]
12
= 𝑃16,897.56 (end of the 4th quarter)
The total amount in the account at the end of 1 year is P16,897.56 known as the compound
amount.
Compound Amount Formula
𝑟 𝑛𝑡
𝑀 = 𝑃 (1 + )
𝑛
where:
M = compound amount
P = amount of money deposited
r = interest rate
n = number of compounding periods per year
t = the number of years
Examples:
1. Mr. Misa deposited P15,000.00 in an account earning 5% interest, compounded quarterly,
for a period of 2 years.
Solution:
P = P15,000.00
r = 5% or 0.05 n = 4
t=2
𝑟 𝑛𝑡
𝑀 = 𝑃 (1 + )
𝑛
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= 15,000 (1 +
0.05 (4)(2)
)
4
= 𝑃16,567.29 (compound amount after 2 years)
2. Calculate the future value of P7,500 earning 9% interest, compounded daily, for 3 years.
Solution:
P = P7,500.00 r = 9% or 0.09 n = 360
t=3
𝑟 𝑛𝑡
𝑀 = 𝑃 (1 + )
𝑛
0.09 (360)(3)
)
= 7,500 (1 +
360
= 𝑃 9,824.40 (the future value after 3 years)
3. How much interest is earned in 4 years on P8,000.00 deposited in an account paying 6%
interest, compounded semiannually.
Solution:
P = P8,000.00 r = 6% or 0.06 n = 2
t=4
Find M:
𝑟 𝑛𝑡
𝑀 = 𝑃 (1 + )
𝑛
0.06 (2)(4)
)
= 8,000 (1 +
2
Find I:
= 𝑃10,134.16 (compound amount)
𝐼 = 𝑀−𝑃
= 10,134.16 − 8,000
= 𝑃2,134.16 (earned interest in 4 years)
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EXERCISES 5.1.3
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. Find the compound amount of P35,000.00 compounded semiannually for 3 years at 15%
interest rate.
2. Tom deposits P2,000.00 into an account with an interest rate of 2.5% that is compounded
quarterly. Rounding to the nearest peso, what is the balance in Tom’s account after 5 years.
3. An engineer deposited P18,000.00 in a savings account at 8% interest rate. If the interest
is compounded monthly, what will be the amount of the deposit at the end of three years.
4. Accumulate P5,600.00 for 3 years at 5.5% compounded semiannually.
5. How much money should be invested in an account that earns 6% interest, compounded
semiannually in order to have P25,500.00 in 33 years. Use the formula below to find the
present value which was derived from the compound amount formula for P.
𝑀
𝑃=
𝑟 𝑟𝑡
(1 + 𝑛)
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5.2 Credit Cards and Consumer Loans
Credit Card
A credit card is a payment card issued to users (cardholders) to enable the cardholder to
pay a merchant for good and services based on the cardholder’s promise to the card issuer to pay
them for the amounts so paid plus the other agreed charges. The card issuer (usually a bank) creates
a revolving account and grants a line of credit to the cardholder, from which the cardholder can
borrow money for payment to a merchant or as a cash advance. In other words, credit cards
combine payment services with extensions of credit. Complex fee structures in the credit card
industry may limit costumers’ ability to comparison shop, helping to ensure that the industry is not
price-competitive and helping to maximize industry profits. Due to concerns about this, many
legislatures have regulated credit card fees.
Credit cards are best suited for financing extending over a shorter time period. Remember
that it does not give you more money, rather it enables you to have higher purchasing power in
your everyday life. And so it is important to be aware of the price of having a credit card. Similar
to all the services and products you use, you should be aware of the terms and prices. Remember
that it is expensive to postpone payments. Also keep in mind that the sooner you pay, the least
interest you pay. As long as you make your payments on time, there are no accruing interests.
Many credit cards charge annual fees but also come with interest-free grace periods,
balance transfers and rewards.
Usually credit card companies issue monthly bills. If the bill is paid in full before its due
date no charges is added otherwise interest charges will start to accrue. The most common method
of determining finance charges is the average daily balance method
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑑𝑎𝑖𝑙𝑦 𝑏𝑎𝑙𝑎𝑛𝑐𝑒 =
Examples:
𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑤𝑒𝑑 𝑒𝑎𝑐ℎ 𝑑𝑎𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑛𝑡ℎ
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑎𝑦𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑏𝑖𝑙𝑙𝑖𝑛𝑔 𝑝𝑒𝑟𝑖𝑜𝑑
1. An unpaid bill for P2,500.00 had a due date of January 15. A purchase of P1,650.00 was
made on January 18 and P560.00 was charge on January 27. A payment of P2,000.00 was
made on January 20. The next billing date is February 15. The interest on the average daily
balance is 1.25% per month. Find the finance charge on the February 15 bill.
Solution:
Prepare first a table showing this information
Date
Jan 15 –
Jan 17
Jan 18 –
Jan 19
Jan 20 –
Jan 26
Payments or
Purchases
Balance
each day
2,500
No. of days unit
balance changes
3
Unpaid balance
times no. of days
7,500
1,650
4,150
2
8,300
-2,000.00
2,150
7
15,050
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Jan 27 –
Feb 14
Total
560
2,750
19
51,490
P82,340.00
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑑𝑎𝑖𝑙𝑦 𝑏𝑎𝑙𝑎𝑛𝑐𝑒 =
𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑤𝑒𝑑 𝑒𝑎𝑐ℎ 𝑑𝑎𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑛𝑡ℎ
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑎𝑦𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑏𝑖𝑙𝑙𝑖𝑛𝑔 𝑝𝑒𝑟𝑖𝑜𝑑
82,340.00
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑑𝑎𝑖𝑙𝑦 𝑏𝑎𝑙𝑎𝑛𝑐𝑒 =
31
= P2,656.13
Use the day of the year table to identify the number of days in the billing period.
Finding the finance charge on the February 15 bill
𝐼 = 𝑃𝑟𝑡 = (2,656.13)(0.0125)(1) = 𝑃33.20
2. An unpaid bill P4,585.00 had a due date of March 2. A purchase of P15,000.00 was made
on March 8 and another was on March amounting to P3,200.00. An P875.00 was charge
on March 21. A payment of P10,000.00 was made on March 15. The interest on the average
daily balance is 2.3% per month. Find the finance charge on the April 2 bill.
Solution:
Prepare a table that shows the given data from the cited problem.
Date
March 2 –
March 7
March 8 –
March 9
March 10 –
March 14
March 15 –
March 20
March 21 –
April 2
Total
Payments or
Purchases
Balance
each day
4,585
No. of days unit
balance changes
6
Unpaid balance
times no. of days
27,510
15,000
19,585
2
39,170
3,200
22,785
5
113,925
-10,000
12,785
6
76,710
875
13,660
13
177,580
P434,895.00
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑑𝑎𝑖𝑙𝑦 𝑏𝑎𝑙𝑎𝑛𝑐𝑒 =
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑑𝑎𝑖𝑙𝑦 𝑏𝑎𝑙𝑎𝑛𝑐𝑒 =
𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑤𝑒𝑑 𝑒𝑎𝑐ℎ 𝑑𝑎𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑛𝑡ℎ
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑎𝑦𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑏𝑖𝑙𝑙𝑖𝑛𝑔 𝑝𝑒𝑟𝑖𝑜𝑑
434,895.00
31
= P14,028.87
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Use the day of the year table to identify the number of days in the billing period.
Finding the finance charge on the April 2 bill
𝐼 = 𝑃𝑟𝑡 = (14,028.87)(0.0235)(1) = 𝑃329.68
Annual Percentage Rate (APR)
A credit card’s interest rate is the price you pay for borrowing money. For credit cards, the
interest rates are typically stated as a yearly rate. This is called the annual percentage rate (APR).
On most cards, you can avoid paying interest on purchases if you pay your balance in full each
month by the due date.
Annual percentage rate is the annualized interest rate on a loan or investment which doesn’t
account for the effect of compounding. It is the annualized form of the periodic rate which when
applied to a loan or investment balance gives the interest expense or income for the period. In most
cases, it is the interest rate quoted by banks and other financial intermediaries on various products
like loans, mortgages, credit cards, deposits, etc. It is also called the nominal annual interest rate
or simple interest rate.
This APR was covered by the Republic Act No. 3765 otherwise known as the “Truth in
Lending Act”. It is an act requiring the disclosure of finance charges in connection with the
extension of credit.
The policy behind the law is to protect the people from lack of awareness of the true cost
of credit by assuring full disclosure of such cost with a view of preventing the uninformed use of
credit to the detriment of the national economy.
The law covers any creditor, which is defined as any person engaged in the business of
extending credit (including any person who as a regular business practice make loans or sells or
rents property or services on a time, credit, or instalment basis, either as principal or as agent) who
requires as an incident to the extension of credit, the payment of a finance charge.
A finance charge includes interest, fees, service charges, discounts, and such other charges
incident to the extension of credit as may be prescribed by the Monetary Board of the Bangko
Sentral ng Pilipinas through regulations.
This formula can be used to estimate the annual percentage rate (APR) on a simple interest
rate instalment loan.
2𝑛𝑟
𝐴𝑃𝑅 =
𝑛+1
where:
n = number of payments
r = simple interest rate
Examples:
1. An investors borrowed P35,000.00 from a bank that advertises a 12% simple interest rate.
He agrees to a 5 monthly instalment.
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a. Calculate its monthly payment
Solution:
P = P35,000.00
r = 12% or 0.12
𝐼 = 𝑃𝑟𝑡 = (35,000)(0.12) (
t = 5 months or 5/12
5
) = 𝑃1,750.00
12
𝑀 = 𝑃 + 𝐼 = 35,000 + 1750 = 𝑃36,750.00
𝑴𝒐𝒏𝒕𝒉𝒍𝒚 𝑷𝒂𝒚𝒎𝒆𝒏𝒕 =
36,750
= 𝑃7,350.00
5
Solving for the interest on the first month
1
𝐼 = 𝑃𝑟𝑡 = (35,000)(0.12) ( ) = 𝑃350.00
12
Solving for the interest on the second month
1
𝐼 = 𝑃𝑟𝑡 = (28,000)(0.12) ( ) = 𝑃280.00
12
Solving for the interest on the third month
1
𝐼 = 𝑃𝑟𝑡 = (20,930)(0.12) ( ) = 𝑃209.30
12
Principal – 7,000 = 28,000
7,350 – 280 = 7,070
28,000 – 7,070 = 20,930
7,350 – 209.30 = 7,140.70
Solving for the interest on the fourth month
1
𝐼 = 𝑃𝑟𝑡 = (13,789.30)(0.12) ( ) = 𝑃137.89 20,930 – 7,140.70 = 13,789.30
12
7,350 – 137.89 = 7,212.11
Solving for the interest on the fifth month
1
𝐼 = 𝑃𝑟𝑡 = (6,577.19)(0.12) ( ) = 𝑃65.77
13,789 – 7,212.11 = 6,577.19
12
First Payment: P7,000.00; Second Payment: P7,070.00; Third Payment: P7,140.70; Fourth
Payment: P7,212.11; and Fifth Payment: P6,577.19
These shows that each month the amount you owe is decreasing and not by a constant
amount.
Republic Act No. 3765 tells us that the interest rate for a loan be calculated only on the
amount owed at a particular time, not on the original amount borrowed.
b. Compute for the APR
n=5
2𝑛𝑟
𝐴𝑃𝑅 = 𝑛+𝑟 =
r = 12% or 0.12
2(5)(0.12)
5+1
= 0.2 or 20%
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The annual percentage are on the loan is approximately 20%. Recall that the simple interest
rate was 12% much less than the actual rate. The Truth in Lending act provides the consumer with
a standard interest rate, APR, so that it iss possible to compare loans. The 12% simple interest loan
described in problem no.1 is equivalent to an APR loan of about 20%.
2. A manager bought a brand new car amounting to P750,000.00 He gave a downpayment of
25% and the balance was agreed to be paid in 18 equal monthly instalments. The finance
charge on the balance was given at 12% simple interest.
a. Calculate for the finance charge.
b. Calculate the annual percentage rate in tow decimal places.
Solution:
a. Solving for the finance charge
Downpayment = 25% of 750,000
= 0.25(750,000)
= P187,500.00
Amount Finance = 750,000 – 187,500
= P562,500.00
Interest Owed = finance rate x amount financed
= 0.12(562,500)
= P67,500.00 (is the finance charge)
b. Solving for the annual percentage rate (APR)
n = 18
r = 0.12
2𝑛𝑟
𝐴𝑃𝑅 = 𝑛+𝑟 =
2(18)(0.12)
18+1
= 0.227369 or 22.74%
Consumer Loans
A consumer loan is when a person borrows money from a lender, either unsecured or
secured. There are several types of consumer loans and some of the most popular ones include
mortgages, refinances, home equity lines of credit, credit cards, auto loans, student loans, and
personal loans.
A consumer loan is a good alternative to a credit card if you want predictability with your
monthly expenses. A consume loan provides a set plan for your monthly down payments which
gives many a sense of security. You can arrive back from a vacation paid with a consumer loans
and not expect any surprises. You will simply start paying back a pre decided amount each month.
It is also called as consumer credit or consumer lending.
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The payment amount for these loans is given by the following formula
Payment Formula for an APR Loan
𝑃𝑀𝑇 = 𝐴 [
where:
𝑟
𝑛
𝑟 −𝑛𝑡
1 − (1 + 𝑛)
]
PMT = is the payment
A = is the loan amount
r = is the annual interest rate
n = is the amount of payments per year
t = is the number of years
Example:
1. A certain computer company is offering an 8% annual interest rate for 2 years on all their
computer gadget products. Joshua Emmanuel, a computer technician, decided to buy one
set of computer unit for P45,000.00. Find his monthly payment.
Solution:
r = 8% or 0.08 n =12
t = 2 years
𝑃𝑀𝑇 = 𝐴 [
𝑟
𝑛
𝑟 −𝑛𝑡
1 − (1 + 𝑛)
]
0.08
12
]
= 45,000 [
0.08 −(12)(2)
1 − (1 +
)
12
= 𝑃2,035.23 (monthly payment)
Calculate APR on Payday Loans
To calculate the APR on a short-term payday loan:
1.
2.
3.
4.
Divide the finance charge by the loan amount.
Multiply the result by 365.
Divide the result by the term of the loan.
Multiply the result by 100.
Example:
1. You get a payday loan for P500.00, and you pay a fee of P50.00. The loan must be repaid
within 14 days. What is the APR?
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Solution:
50
) 365
500
50
) 365
(
] 100%
𝐼 = [ 500
14
𝐼=(
𝐼=(
𝐼=(
36.50
) 100%
14
36.50
) 100%
14
𝐼 = 260.71%
Calculating Loan Payoffs
Loan payoffs is a complete repayment of a loan (principal plus interest), full discharge of
an obligation, or the return from a deal, decision or investment.
Your payoff amount is how much you will actually have to pay to satisfy the terms of your
mortgage loan an completely pay off your debt. Your payoff amount also includes the payment of
any interest you owe through the day you intend to pay off your loan.
APR Loan Payoff Formula
The payoff amount for a loan based on APR is given by
where:
𝑟 −𝑈
1 − (1 + 𝑛)
]
𝐴 = 𝑃𝑀𝑇 [
𝑟
𝑛
A = is the loan payoff
PMT = is the payment
r = is the annual interest rate
n = is the number of payments per year
U = is the number of remaining (or unpaid) balance
Example:
1. A lady wants to pay off the loan in 32 months. Her monthly obligation is P850.00 on a 3year loan with an annual percentage rate of 7.5%. Find the payoff amount.
Solution:
PMT = P850.00
r = 7.5% or 0.075
n = 12
U = 4 months
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𝑟 −𝑈
1 − (1 + 𝑛)
]
𝐴 = 𝑃𝑀𝑇 [
𝑟
𝑛
0.75 −4
1 − (1 + 12 )
]
= 850 [
0.75
12
= 𝑃3,347.53 (is the loan payoff)
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EXERCISES 5.2
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. A bill for P65,200.00 was due on August 2. Purchases of P3,800.00 were made on August
8 and P1,800.00 was charged on August 23. A payment of P2,500.00 was made on August
16. The interest on the average daily balance is 2.1% per month. Find the finance charge
on the September 2 bill.
2. Mrs. Guanzon bought a gold necklace amounting to P15,000.00. A 10% was given as a
required down payment and the balance is to be paid in 12 equal monthly instalments. The
finance charge on the balance is 5% simple interest.
a. Solve for the finance charge.
b. Calculate the annual percentage rate in a tenths place.
3. A doctor purchased a second-hand vehicle from his bestfriend’s show room for
P175,000.00. He was given a 20% annual interest rate for 2 years. Find his monthly
payment.
4. An Engineer borrowed P2,500.00 for 21 days and pays a fee of P50.00. What is the APR?
5. Mr. Soriano applied a loan for 16 months. His monthly payment is P750.00 on a 2-year
loan at an annual percentage rate of 10%. Find the payoff amount.
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5.3 Stocks, Bonds, and Mutual Bonds
Stocks
A stock is ownership in a company. When you buy a stock, (you are called stockholders)
you buy a piece of the company. So if the company does well, you do well. Congruently, if the
company tanks, your stock tanks. Just like bonds, there are many types of stocks because there are
many different types of companies out there. Large company stocks (large cap), mid cap stocks,
small cap stocks, international stocks, emerging stocks, tech stocks, etc. Historically, stocks have
an annual average return. However, remember that with more return comes more risk. So when
investing in stocks, keep in mind that you have to be able to handle the extra risk or volatility to
reap the potential reward in the long run.
A company may distribute profits to the owners (stockholders) in the form of dividends.
Most dividends are paid in the form of cash—for example, a company might declare a quarterly
dividend of P0.50 per share. However, though it’s less common, companies also have the option
of declaring stock dividends. When paying a stock dividend, a company issues additional shares
of stock proportional to existing investors’ holdings.
Calculate Dividends Paid to a Stockholder
Example:
1. A stock pays an annual dividend of P0.75 per share. Calculate the dividend paid to a
shareholder who has 350 shares of the company’s stock.
Solution:
(0.75 per share) (350 shares) = P262.50 (the shareholder receives P262.50 in dividends)
Before the stock dividends are handed out, they’re known as “stock dividends
distributable” and are listed in the stockholders’ equity section of the company’s balance sheet.
The first step in calculating stock dividends distributable is to divide that percentage by
100 to convert it inti a decimal. In our example, 10% would become 0.10. Next, multiply the
company’s total outstanding shares by this decimal. You can find the number of outstanding shares
in most stock quotes.
Finally, multiply this amount by the par value of the stock, which can usually be found in
the stockholders’ equity section of the balance sheet. This is typically a small amount, such as
P0.01, and it has no relation to the actual share price of the stock. Once you multiply these figures
by one another, the result is the amount the company would list as stock dividends distributable.
𝑆𝑡𝑜𝑐𝑘 𝐷𝑖𝑣𝑖𝑑𝑒𝑛𝑑 𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑎𝑏𝑙𝑒 =
𝑆𝑡𝑜𝑐𝑘 𝐷𝑖𝑣𝑖𝑑𝑒𝑛𝑑 %
× 𝑠ℎ𝑎𝑟𝑒𝑠 𝑜𝑢𝑡𝑠𝑡𝑎𝑛𝑑𝑖𝑛𝑔 × 𝑝𝑎𝑟 𝑣𝑎𝑙𝑢𝑒
100
Examples:
1. A company declares a stock dividend of 0.05 shares per outstanding share, and there are
100 million total shares outstanding before the stock dividend is paid. A quick look at the
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balance sheet tells us that the stock’s par value is P0.01 per share, so the stock dividend
distributable that the company will list on its balance sheet can be calculated as follows:
Solution:
𝑆𝑡𝑜𝑐𝑘 𝐷𝑖𝑣𝑖𝑑𝑒𝑛𝑑 𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑎𝑏𝑙𝑒 =
𝑆𝑡𝑜𝑐𝑘 𝐷𝑖𝑣𝑖𝑑𝑒𝑛𝑑 %
× 𝑠ℎ𝑎𝑟𝑒𝑠 𝑜𝑢𝑡𝑠𝑡𝑎𝑛𝑑𝑖𝑛𝑔 × 𝑝𝑎𝑟 𝑣𝑎𝑙𝑢𝑒
100
= 0.05 × 100,000,000 × 0.01
= 𝑃50,000.00
2. Joshua invested P120,000 in stocks and bonds. If he made a 20% profit on his stocks and
a 5% profit on his bonds, and the combined profit was P10,500.00, how much did Joshua
invest in stocks.
Solution:
Let x = amount invested in stock at 20%
120,000 – x = amount invested in bonds at 5%
Equation:
[(x)(20%)] + [(120,000 – x)(5%)] = 10,500
[(x)(0.20)] + [(120,000 – x)(0.05)] = 10,500
(0.20x + 6,000 – 0.05x) = 10,500
(20x + 600,000 – 5x) = 1,050,000
15x = 1,050,000 – 600,000
15x = 450,000
x = 30,000 (amount invested in stocks)
120,000 – x = 120,000 – 30,000
= 90,000 (amount invested in bonds)
Checking:
[(30,000)(20%)] + [(120,000 – 30,000)(5%)] = 10,500
[(30,000)(0.20)] + [(90,000)(0.05)] = 10,500
(6,000 + 4,500) = 10,500
10,500 = 10,500
Bonds
The best way to describe a bond is to think of it like a loan. You loan your money to the
government or a company, and in return they pay you interest for the term of that loan. Typically,
bonds are considered conservative types of investments because you can choose the length and
term of the bond and know exactly how much money will you get back at the end of the term or
“maturity”. There are many types of bonds: government bonds, corporate bonds, short-term bonds,
long-term bonds, municipal and inflation protection bonds, etc. Generally, bonds are less risky
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than stocks and the main way you lose money on a bond is if the company or government issuing
the bond defaults on their obligations. Historically, bonds have an annual average total return of
6.3%.
Bonds are subject to market risk and interest rate risk if sold prior to maturity. Bond values
will decline as interest rates rise and bonds are subject to availability and change in price.
Examples:
1. Harold invested P30,000.00 in various stocks and bonds. He earned 6% on his bonds and
12% on his stocks. If Harold’s total profit on both types of investments was P2,460.00,
how much of the P30,000.00 did he invest in bonds?
Solution:
Let x = is the amount invested at 6% on his bonds
30,000 – x = is the amount invested at 12% on his stocks
Equation:
(interest earned at 6%) + (interest earned at 12%) = 2,460
[(x)(0.06)] + [(30,000 – x)(0.12)] = 2,460
0.06x + 3,600 – 0.12x = 2460
6x + 360,000 – 12x = 246,000
-6x = 246,000 – 360,000
-6x = -114,000
x = 19,000 (amount invested in bonds)
30,000 – x = 30,000 – 19,000
= 11,000 (amount invested in stocks)
Checking:
[(19,000)(0.06)] + [(30,000 – 19,000)(0.12)] = 2,460
1,140 + (11,000)(0.12) = 2,460
1,140 + 1,320 = 2,460
2,460 = 2,460
Mutual Funds
Mutual funds represent another way to invest in stocks, bonds, or cash alternatives. You
can think of a mutual fund like a basket of stocks or bonds. A mutual fund investor is buying part
ownership of the mutual fund company and its assets. Basically, your money is pooled, along with
the money of other investors, into a find, which then invests in certain securities according to a
stated investment strategy. The fund is managed by a fund manager who reports to board directors.
By investing in the fund, you own a piece of the pie (total portfolio), which could include anywhere
from a few dozens to hundreds of securities. This provides you with both a convenient way to
obtain professional money management and instant diversification that would be more difficult
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and expensive to achieve on your own. Every mutual fund publishes a prospectus. Before investing
in a mutual fund, get a copy and carefully review the information it contains, such as the fund’s
investment objective, risks, fees, and expenses. Carefully consider those factors as well as others
before investing.
Mutual fund units, or shares, can typically be purchased or redeemed as needed at the
fund’s current net asset value (NAV) per share, which is sometimes expressed as NAVPS. A fund’s
NAV is derived by dividing the total value of the securities in the portfolio by the total amount of
shares outstanding.
The Net Asset Value of a Mutual Fund Formula is
𝐴−𝐿
𝑁𝐴𝑉 =
𝑁
where:
A = is the total fund assets
L = is the total fund liabilities
N = is the number of shares outstanding
Example:
1. A mutual fund has P600,000,000.00 worth of stock, P5,000,000.00 worth of bonds, and
P1,000,000.00 in cash. The fund’s total liabilities amount to P2,000,000.00. There are
25,000,000 shares outstanding. You invest P15,000.00 in this fund.
a. Solve for the Net Asset Value.
b. How many shares will you buy?
Solution:
a. Find the NAV:
𝐴−𝐿
𝑁
(600,000,000 + 5,000,000 + 1,000,000) − 2,000,000
=
25,000,000
𝑁𝐴𝑉 =
= 𝑃24.16
b. Find the number of shares:
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠ℎ𝑎𝑟𝑒𝑠 =
𝑎𝑚𝑜𝑢𝑛𝑡 𝑖𝑛𝑣𝑒𝑠𝑡𝑒𝑑
𝑐𝑜𝑠𝑡 𝑝𝑒𝑟 𝑠ℎ𝑎𝑟𝑒
15,000
24.16
= 621 shares
=
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Treasury Bills
Treasury bills or popularly known as T-Bills, are peso-denominated short-term fixed
income securities issued by the Republic of the Philippines through Bureau of Treasury. With a
minimum of P200,000.00, you can already enjoy high yields.
T-Bills are issued at a discount to the maturity value. Rather than paying a coupon rate of
interest, the appreciation between issuance price and maturity price provides the investment return.
For instance, a 26-week T-bill is prices at P9,800.00 on issuance to pay P10,000 in six months. No
interest payments are made.
Investors buying treasury bills on auction day, in the days when paper bills were still issued.
You can purchase treasury bills at a bank, though a dealer or broker, or online from a website like
Treasury Direct. The bills are issued through an auction bidding process, which occurs weekly.
Treasury bills among the safest investments in the market. They are backed by the full faith
and credit of the Philippine government, and they come in maturities ranging from four weeks to
one year. When buying Treasury bills, you will find that quotes are typically given in terms of
their discount, so you will need to calculate the actual price.
Keep in mind that the Treasury does not make separate interest payments on Treasury bills.
Instead, the discounted price accounts for the interest that you will earn.
Example:
1. A certain Electric Company invest in a P60,000.00 Philippine Treasury bill at 4.46%
interest for 30 days. The bank through which the bill is purchased charges a service fee of
P20.00. What is the cost of the treasury bill?
Solution:
Principal = P60,000.00
r = 4.46% or 0.0446
t = 30 days or 30/360
𝐼 = 𝑃𝑟𝑡
= (60,000)(0.0446)(
= 𝑃223.00
30
)
360
Cost = (face value – interest) + service fee
= (60,000 – 223) + 20
= P59,797.00
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EXERCISES 5.3
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. Bob earned P420.00 on his investment in bonds and stocks. If his bonds return 2% and his
stock returned 6% and his total investment was P10,000.00, how much did he invest in
bonds and stocks?
2. A stock pays an annual dividend of P0.85 per share. Calculate the dividends paid to a
shareholder who has 650 shares of the company’s stock.
3. Mr. Mendoza invested some of his P18,000.00 in bonds and made a 5% profit and the rest
in bonds that made a 12% profit. If the profit on the 12% bond was P885.00 more than the
profit on the 5% bonds, how much did Mr. Mendoza invest in the 5% bonds.
4. A manager invested a P20,000.00 in bonds that made an 8% profit and the rest in bonds
that made a 7% profit. If the profit on the 8% bonds was P700.00 more than the profit on
the 7% bonds, how much did he invest in the 7% bonds?
5. Mr. Cruz has P36,000.00 to invest, some in bonds and the rest in stocks. He has decided
that the money invested in bonds must be at least twice as much as that in stocks. But the
money invested in bonds must be greater than P20,000.00. If the bonds earn 5% and the
stocks earn 7%, how much money should be invested in each to maximize profit?
6. A mutual fund has P659 million worth of stock, P550,000.00 in cash, and P2,500,000.00
in other assets. The fund’s total liabilities amount to P2,500,000.00. There are 20 million
shares outstanding. You invest P12,000.00 in this fund.
a. Solve for the NAV.
b. How many shares will you buy?
7. A P30,000.00 Philippine Treasury bill, purchased at 1.6% interest, matures in 85 days. The
purchaser is charged a service fee of P30.00. What is the cost of the treasury bill?
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Chapter 6: THE STATISTICAL TOOLS
 Learning Objectives
At the end of this chapter, the student is expected to:
 apply a variety of statistical tolls to process and manage numerical data;
 use the methods of linear regression and correlations to predict the value of a
variable given certain conditions; and
 recognize the importance of testing of hypotheses in making decisions.
Introduction
Statistics involves the collection, organization, summarization, presentation, and
interpretation of data. It has two branches: descriptive statistics and inferential statistics.
Descriptive statistics is the term given to the analysis of data that helps describe, show or
summarize data in a meaningful way. When using descriptive statistics, it is useful to summarize
a group of data using a combination of tabulated description (i.e., tables), graphical description
(i.e., graphs and charts) and statistical commentary (i.e., a discussion of the results). The branch
that allows to make predictions (“inferences”) from the data is called inferential statistics. With
inferential statistics, it takes data from samples and make generalizations about a population.
For instance, you might stand in a mall and ask a sample of 100 people if they like shopping
at SM. You could make a bar chart of yes or no answers (that would be a descriptive statistics) or
you could use your research (and inferential statistics) to reason that around 75-80% of the
population (all shoppers in all malls) like shopping at SM.
Testing the significance of the difference between two means, two standard deviations, two
proportions, or two percentages, is an important area of inferential statistics. Comparison between
two or more variables often arises in research or experiments and to be able to make valid
conclusions regarding the results of the study, one has to apply an appropriate test statistic. This
chapter deals with the discussion of the different test statistics that are commonly used in research
studies under inferential statistics.
6.1 Hypothesis Testing
In Statistics, decision-making starts with a concern about a population regarding its
characteristics denoted by parameter values. We might be interested in the population parameter
like the mean or the proportion. For instance, you are deciding to put up a business selling cars.
Your first course before spending money in business is to know which car sells the most these
days. Before you open a business of selling Toyota, Mitsubishi, Hyundai, Honda, Nissan, or
Suzuki, you need to gather information which among these get the most number of sales. How
many existing distributors of these cars are out there? Do you want to compete? To answer these
questions, you need to gather data. What type of data? And where will you get them? You simply
need to do a survey. These concerns can be addressed in a procedure in Statistics called hypothesis
testing.
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Hypothesis
A hypothesis is a conjecture or statement which aims to explain certain phenomena in the
real world. Many hypotheses, statistical or not, are products of man’s curiosity. To seek for the
answers to his questions, he tries to find and present evidences, then tests the resulting hypothesis
using statistical tools and analysis. In statistical analysis, the truth of which will be either accepted
or rejected within a certain critical interval.
The hypothesis that is subjected to testing to determine whether its truth can be accepted
or rejected is the null hypothesis by Ho. This hypothesis states that there is no significant
relationship or no significant difference between two or more variables, or that one variable does
not affect another variable. In statistical research, the hypotheses should be written in null form.
For example, suppose you want to know whether method A is not more effective than method B
in teaching high school mathematics. The null hypothesis for this study will be: “There is no
significant difference between the effectiveness of method A and method B.”
Another type of hypothesis is the alternative hypothesis, denoted by Ha. This is the
hypothesis that challenges the null hypothesis. The alternative hypothesis for the example above
can be: “There is a significant difference between the effectiveness of method A and method B.”
or “Method A is more effective than method B,” or Method A is less effective than method B,”
depending on whether the type of test is either one-tailed or two-tailed. These will be discussed in
the succeeding lessons.
Significance Level
To test the null hypothesis of no significance in the difference between the two methods in
the above example, one must set the level of significance first. This is the probability of having a
Type I error and is denoted by the symbol 𝛼. A Type I error is the probability of accepting the
alternative hypothesis when, in fact, the null hypothesis is true. The probability of accepting the
null hypothesis when, in fact, it is false is called a Type II error and it is denoted by the symbol 𝛽.
The most common level of significance is 5%.
Table 1. Four Possible Outcomes in Decision-Making
Decisions about the Ho
Ho is true.
Reject
Do not Reject Ho
(or Accept Ho)
Type I error
Correct Decision
Reality
Ho is false. Correct Decision
Type II error
If the null hypothesis is true and accepted, or if it is false and rejected, the decision is
correct. If the null hypothesis is true and reject, the decision is incorrect and this is a Type I error.
If the null hypothesis is false and accepted, the decision is incorrect and this is a Type II error. For
instance, Sarah insists that she is 31 years old when, in fact, she is 35 years old. What error is Sarah
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committing? Mary is rejecting the truth. She is committing a Type I error. Another example, a man
plans to go hunting the Philippine monkey-eating eagle believing that it is a proof of his mettle.
What type of error is this? Hunting the Philippine eagle is prohibited by law. Thus, it is not a good
sport. It is a Type II error. Since hunting the Philippine monkey-eating eagle is against the law, the
man may find himself in jail if he goes out of his way hunting endangered species.
In decisions that we make, we form conclusions and these conclusions are the bases of our
actions. But this is not always the case in Statistics because we make decisions based on sample
information. The best that we can do is to control the probability with which an error occurs. This
is the reason why we are assigning small probability values to each of them.
One-Tailed and Two-Tailed Tests
A test is called a one-tailed test if the rejection region lies on one extreme side of the
distribution and two-tailed if the rejection region is located on both ends of the distribution.
0.025 or
2.5% .
0.025 or
2.5% .
0.05
5% .
0.05
5% .
or
(B)
(A)
(C)
Figure 1. Two-tailed (A) and One-tailed (A & B) tests
In figure 1.A (two-tailed), the rejection region is the areas to the extreme left and right of
the curve marked by the two vertical lines. In figure 1.B&C (both one-tailed), the rejection region
is the area to the left (left tail) and to the right (right tail) of the vertical line under the bell curve,
respectively.
Steps in Testing Hypothesis
Below are the steps when testing the truth of a hypothesis.
1.
2.
3.
4.
5.
6.
7.
Formulate the null hypothesis. Denote it as Ho and the alternative hypothesis as Ha.
Set the desired level of significance (𝛼).
Determine the appropriate test statistic to be used in testing the null hypothesis.
Compute for the value of the statistic to be used.
Compute for the degrees of freedom.
Find the tabular value using the table of values for different tests from the appendix tables.
State the Decision Rule: If the computed value is less than the tabular value, accept the null
hypothesis. If the computed value is greater than the tabular value, reject the null
hypothesis.
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8. Compare the computed value to the tabular value. Make a conclusion using the result of
the comparison.
Degree of Freedom (df)
The degree of freedom gives the number of pieces of independent information available
for computing variability. For any statistical tool used in testing hypothesis, the number of degrees
of freedom required will vary depending on the size of the distribution. For a single group of
population, the number of degrees of freedom is N – 1, where N is the population. For two groups,
the formula for df is: N1 + N2 – 2 for t-test and N – 2 for Pearson r. These test statistics will be
discussed later in this chapter.
6.1.1 Tests Concerning Means
6.1.1.1 z-test on the Comparison between the Population Mean and the Sample Mean
If the population mean (𝜇) and the population standard deviation (𝜎) are known,
and 𝜇 will be compared to a sample mean (𝑥̅ ), use the formula below.
𝑧=
(𝑥̅ −𝜇)
𝜎
∙ √𝑛, where n is the number of sample.
The tabular values of 𝑧 can be obtained from the following table:
Table 2. Summary Table of Critical Values
Test Type
One-tailed Test
Level of Significance
0.10
0.05
0.025
0.01
±1.28
Two-tailed Test ±1.645
±1.645 ±1.96 ±2.33
±1.96
±2.33 ±2.58
Decision Rule:
Example 1
Reject Ho if |𝑧| ≥ |𝑧𝑡𝑎𝑏𝑢𝑙𝑎𝑟 |.
A company, which makes a battery-operated toy car, claims that its products have a mean
life span of 5 years with a standard deviation of 2 years. Test the null hypothesis that 𝜇 = 5 years
against the alternative hypothesis that years if a random sample of 40 toy cars was tested and found
to have a mean life span of only 3 years. Use a 5% level of significance.
Solution:
1. Ho : The mean lifespan of battery-operated toy cars is 5 years. (𝜇 = 5)
Ha : The mean lifespan of battery-operated toy cars is 5 years. (𝜇 ≠ 5)
2. 𝛼 = 0.05, two-tailed
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3. Use z-test as test statistic.
4. Computation:
Given 𝑥̅ = 3, 𝜇 = 5, 𝑛 = 40, 𝜎 = 2
(𝑥̅ − 𝜇)
𝑧=
∙ √𝑛
𝜎
(3 − 5)
=
∙ √40
2
= −6.32
5. Critical Value: 𝑧 < −1.96 𝑎𝑛𝑑 𝑧 > 1.96
6. Decision Rule: Reject Ho if |𝑧| ≥ |𝑧𝑡𝑎𝑏𝑢𝑙𝑎𝑟 |.
7. Since the computed|𝑧|, which is 6.32, is greater than |𝑧𝑡𝑎𝑏𝑢𝑙𝑎𝑟 |, which is 1.96, therefore,
reject Ho. Hence, there is a significant difference between the population and sample mean
lifespan of battery-operated toy cars.
Example 2
A manufacturer of bicycle tires has developed a new design which he claims has an average
lifespan of 5 years with a standard deviation of 1.2 years. A dealer of the product claims that the
average lifespan of 150 samples of the tires is only 3.5 years. Test the difference of the population
and sample means at 5% level of significance.
Solution:
1. Ho : There is no significant difference between the population and sample mean of
bicycle tires’ lifespan.
(𝑥̅ = 𝜇)
Ha : There is a significant difference between the population and sample mean of bicycle
tires’ lifespan. (𝑥̅ < 𝜇)
2. 𝛼 = 0.05, one-tailed, left tail
3. Use z-test as test statistic.
4. Computation:
Given 𝑥̅ = 3.5, 𝜇 = 5, 𝑛 = 150, 𝜎 = 1.2
𝑧=
(𝑥̅ − 𝜇)
∙ √𝑛
𝜎
=
(3.5−5)
1.2
= −15.31
∙ √150
5. Critical Value: 𝑧 < −1.645
6. Decision Rule: Reject Ho if |𝑧| ≥ |𝑧𝑡𝑎𝑏𝑢𝑙𝑎𝑟 |.
7. Since the computed|𝑧|, which is 15.31, is greater than |𝑧𝑡𝑎𝑏𝑢𝑙𝑎𝑟 |, which is 1.645, therefore,
reject Ho. Hence, there is a significant difference between the population and sample mean
of bicycle tires’ lifespan.
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EXERCISES 6.1.1.1
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. A researcher used a developed problem solving test to randomly select 50 Grade 6 pupils.
In this sample, 𝑥̅ = 80 and s = 10. The 𝜇 and the standard deviation of the population used
in the standardization of the test were 75 and 15, respectively. Use the 95% confidence
level to answer the following questions:
a. Does the sample mean differ significantly from the population mean?
b. Can it be said that the sample mean is above average?
2. The owner of a factory that sells a particular bottled fruit juice claims that the average
capacity of their product is 250 ml. To test the claim, a consumer group gets a sample of
100 such bottles, calculates the capacity of each bottle, and then finds the mean capacity to
be 248 ml. The standard deviation is 5 ml. Is the claim true?
3. In a plant nursery, the owner thinks that the lengths of seedlings in a box sprayed with a
new kind of fertilizer has an average height of 26 cm after three days and a standard
deviation of 10 cm. One researcher randomly selected 80 such seedlings and calculated the
mean height to be 20 cm and the standard deviation was 10 cm. Will you conduct a onetailed test or two-tailed test? Proceed with the test using 𝛼 = 0.05.
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6.1.1.2 t-test on the Comparison between the Population Mean and the Sample Mean
The t-test can be used to compare the means when the population mean (𝜇) is known
but the population standard deviation (𝜎) is unknown.
When the population standard deviation is unknown but the sample standard
deviation can be computed, the t-test can also be used instead of the z-test. The formula
is given below:
𝑡=
(𝑥̅ − 𝜇)
∙ √𝑛
𝑠
The denominator of the formula, s, divided by the √𝑛 for t is called the standard
error of the statistic. It is the standard deviation of the sampling distribution of a
statistic for random samples n.
Decision Rule:
Example 1
Reject Ho if |𝑡| ≥ |𝑡𝑡𝑎𝑏𝑢𝑙𝑎𝑟 |.
The average length of time for people to vote using the old procedure during a presidential
election period in precinct A is 55 minutes. Using computerization as a new election method, a
random sample of 20 registrants was used and found to have a mean length of voting time of 30
minutes with a standard deviation of 1.5 minutes. Test the significance of the difference between
the population mean and the sample mean.
Solution:
1. Ho : There is no significant difference between the population and sample mean of length
of time for people to vote using the old and new procedure.
(𝑥̅ = 𝜇)
Ha : There is a significant difference between the population and sample mean length of
time for people to vote using the old and new procedure.
(𝑥̅ < 𝜇)
2. 𝛼 = 0.05, one-tailed, left tail
3. Use t-test as test statistic.
4. Computation:
Given 𝑥̅ = 30, 𝜇 = 55, 𝑛 = 20, 𝑠 = 1.5
𝑡=
(𝑥̅ − 𝜇)
∙ √𝑛
𝑠
=
(30−55)
1.5
= −74.54
∙ √20
5. df = n – 1 = 20 – 1 = 19
6. Tabular Value: t = 1.729 (from Appendix)
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7. Decision Rule: Reject Ho if |𝑡| ≥ |𝑡𝑡𝑎𝑏𝑢𝑙𝑎𝑟 |.
8. Since the computed|𝑡|, which is 74.54, is greater than |𝑧𝑡𝑎𝑏𝑢𝑙𝑎𝑟 |, which is 1.729, therefore,
reject Ho. Hence, there is a significant difference between the population and sample mean
length of time for people to vote using the old and new procedure. It implies that using
computerization method in election gives short period of time to vote compare to the old
procedure.
Example 2
An experiment study was conducted by a researcher to determine if a new time slot has an
effect on the performance of pupils in Mathematics. Fifteen randomly selected learners
participated in the study. Toward the end of the investigations, a standardized assessment was
conducted. The sample mean was 85 and the standard deviation of 3. In the standardization of the
test, the mean was 75 and the standard deviation was 10. Based on the evidence at hand, is the new
time slot effective? Use 5% level of significance.
Solution:
1. Ho: There is no significant difference between the population and sample mean of
performance in Mathematics in a new time slot. (𝑥̅ = 𝜇)
Ha: There is a significant difference between the population and sample mean of
performance in Mathematics in a new time slot. (𝑥̅ > 𝜇)
2. 𝛼 = 0.05, one-tailed, right tail
3. Use t-test as test statistic.
4. Computation:
Given 𝑥̅ = 85, 𝜇 = 75, 𝑛 = 15, 𝑠 = 3
(𝑥̅ − 𝜇)
𝑡=
∙ √𝑛
𝑠
=
5.
6.
7.
8.
(85−75)
= 12.91
3
∙ √15
df = n – 1 = 15 – 1 = 14
Tabular Value: t = 1.761 (from Appendix)
Decision Rule: Reject Ho if |𝑡| ≥ |𝑡𝑡𝑎𝑏𝑢𝑙𝑎𝑟 |.
Since the computed|𝑡|, which is 12.91, is greater than |𝑧𝑡𝑎𝑏𝑢𝑙𝑎𝑟 |, which is 1.761, therefore,
reject Ho. Hence, there is a significant difference between the population and sample mean
of performance in Mathematics in a new time slot. It implies that there is an effect of
students’ performance in Mathematics when it changed the time slot.
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EXERCISES 6.1.1.2
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. Drinking water has become an important concern among people. The quality of drinking
water must be monitored as often as possible during the day for possible contamination.
Another variable of concern is the pH below 7.0 is acidic while a pH above 7.0 is alkaline.
A pH of 7.0 is neutral. A water-treatment plant has a target pH of 8.0. based on 16 random
water samples, the mean and the standard deviation were found to be 7.6 and 0.4,
respectively. Does the sample mean provide enough evidence that it differs significantly
from the target mean? Use 𝛼 = 0.05, two – tailed test.
2. The following sample of eight measurements was randomly selected from a normally
distributed population: 12, 10, 9, 8, 15, 10, 11, and 13. Test for significant difference
between the sample mean and the population mean of 10. Use 𝛼 = 0.05.
3. An experiment study was conducted by a researcher to determine if a new time slot has an
effect on the performance of pupils in Mathematics. Fifteen randomly selected learners
participated in the study. Toward the end of the investigation, a standardized assessment
was conducted. The sample mean was 85 and the standard deviation was 3. In the
standardization of the test, the mean was 75 and the standard deviation was 10. Based on
the evidence at hand, is the new time slot effective? Use 𝛼 = 0.05.
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6.1.1.3 t-test Concerning Means of Independent Samples
When two samples are drawn from normally distributed population with the
assumption that their variances are equal, the t-test with the given formula should be
used.
̅̅̅1 − ̅̅̅
𝑥2
𝑥
𝑡=
(𝑛1 − 1)𝑠1 2 + (𝑛2 − 1)𝑠2 2 𝑛1 + 𝑛2
√[
][
]
𝑛1 + 𝑛2 − 2
𝑛1 𝑛2
where
𝑥1 ̅̅̅
̅̅̅,
𝑥2 = means
𝑛1 , 𝑛2 = sample sizes
𝑠1 , 𝑠2 = variances
Decision Rule:
Example 1
Reject Ho if |𝑡| ≥ |𝑡𝑡𝑎𝑏𝑢𝑙𝑎𝑟 |.
A course in Physics was taught to 10 students using the traditional method. Another group
of students went through the same course using another method. At the end of the semester, the
same test was administered to each group. The 10 students under method A got an average of 82
with a standard deviation of 5, while the 11 students under method B got an average of 78 with a
standard deviation of 6. Test the null hypothesis of no significant difference in the performance of
the two groups of students at 5% level of significance.
Solution:
1. Ho: There is no significant difference between the average scores of the two groups of
students.
(𝑥
̅̅̅1 = ̅̅̅)
𝑥2
Ha: There is a significant difference between the average scores of the two groups of
students.
(𝑥
̅̅̅1 > ̅̅̅)
𝑥2
2. 𝛼 = 0.05, one-tailed, right tail
3. Use the t-test as test statistic.
4. Computation:
𝑥2 = 78, 𝑛1 = 10, 𝑛2 = 11, 𝑠1 = 5, 𝑠2 = 6
Given ̅̅̅
𝑥1 = 82, ̅̅̅
̅̅̅
𝑥1 − ̅̅̅
𝑥2
𝑡=
(𝑛1 − 1)𝑠1 2 + (𝑛2 − 1)𝑠2 2 𝑛1 + 𝑛2
√[
][
]
𝑛1 + 𝑛2 − 2
𝑛1 𝑛2
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=
√[
=
=
5.
6.
7.
8.
82 − 78
(10 − 1)(5)2 + (11 − 1)(6)2 10 + 11
][
]
10 + 11 − 2
(10)(11)
√[
4
(9)(25) + (10)(36) 21
][
]
19
110
4
= 1.65
2.4245
df = 𝑛1 + 𝑛2 − 2 = 10 + 11 – 2 = 19
Tabular Value: t = 1.729 (from Appendix)
Decision Rule: Reject Ho if |𝑡| ≥ |𝑡𝑡𝑎𝑏𝑢𝑙𝑎𝑟 |.
Since the computed|𝑡|, which is 1.645, is less than |𝑧𝑡𝑎𝑏𝑢𝑙𝑎𝑟 |, which is 1.729, therefore,
accept Ho. Hence, there is no significant difference between the average scores of the two
groups of students. It implies that there is no significant difference in using method A and
method B in their students’ performance in Physics.
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EXERCISES 6.1.1.3
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. An investigator thinks that people under the age of forty have vocabularies that are different
than those of people over sixty years of age. The investigator administers a vocabulary test
to a group of 31 younger subjects and to a group of 31 older subjects. Higher scores reflect
better performance. The mean score for younger subjects was 14.0 and the standard
deviation of younger subject's scores was 5.0. The mean score for older subjects was 20.0
and the standard deviation of older subject's scores was 6.0. Does this experiment provide
evidence for the investigator's theory?
2. An investigator predicts that dog owners in the country spend more time walking their dogs
than do dog owners in the city. The investigator gets a sample of 21 country owners and
23 city owners. The mean number of hours per week that city owners spend walking their
dogs is 10.0. The standard deviation of hours spent walking the dog by city owners is 3.0.
The mean number of hours’ country owners spent walking their dogs per week was 15.0.
The standard deviation of the number of hours spent walking the dog by owners in the
country was 4.0. Do dog owners in the country spend more time walking their dogs than
do dog owners in the city?
3. An investigator theorizes that people who participate in a regular program of exercise will
have levels of systolic blood pressure that are significantly different from that of people
who do not participate in a regular program of exercise. To test this idea, the investigator
randomly assigns 21 subjects to an exercise program for 10 weeks and 21 subjects to a nonexercise comparison group. After ten weeks the mean systolic blood pressure of subjects
in the exercise group is 137 and the standard deviation of blood pressure values in the
exercise group is 10. After ten weeks, the mean systolic blood pressure of subjects in the
non-exercise group is 127 and the standard deviation on subjects in the non-exercise group
is 9.0. Please test the investigator's theory using an alpha level of 0.05.
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6.1.1.4 t-test on the Significance of the Difference Between Two Correlated Means
When comparing two correlated means, the t-test is the appropriate statistic. A
typical example is when comparing the results of the pre-test and post-test
administered to group of individuals. The two tests must be the same and the given
formula should be used.
∑𝑑
𝑡=
2
2
√(𝑛 ∑ 𝑑 ) − (∑ 𝑑 )
𝑛−1
where d = difference between the pre-test and post-test scores
n = number of samples
Example 1
To determine whether the students’ performance in College Algebra improved after
enrolling in the subject for one term, a 60-item pre-test and post-test were administered to them on
the first and the last days of classes, respectively. The same test was given as pre-test and posttest.
The results are as follows:
Student Pre-Test Score Post-Test Score
A
34
45
B
23
32
C
40
46
D
31
57
E
24
39
F
45
48
G
27
27
H
32
33
I
12
18
J
45
45
d
-11
-9
-6
-26
-15
-3
0
-1
-6
0
𝒅𝟐
121
81
36
676
225
9
0
1
36
0
∑ 𝑑 = −77 ∑ 𝑑 2 = 1,185
Solution:
1. Ho: There is no significant difference between the pre-test and post-test of the students’
performance in College Algebra. (𝜇1 = 𝜇2 )
Ha: There is a significant difference between the pre-test and post-test of the students’
performance in College Algebra. (𝜇1 < 𝜇2 )
2. 𝛼 = 0.05, one-tailed, left tail
3. Use the t-test as test statistic.
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4. Computation:
𝑡=
=
∑𝑑
−77
2
(𝑛 ∑ 𝑑 ) −−(∑(−77)
𝑑)2 2
√10(1,185)
𝑛10−−
11
=
=
−77
√5,921
9
−77
25.65
= −3.002
5.
6.
7.
8.
df = n – 1 = 10 – 1 = 9
Tabular Value: t = 2.821(from Appendix)
Decision Rule: Reject Ho if |𝑡| ≥ |𝑡𝑡𝑎𝑏𝑢𝑙𝑎𝑟 |.
Since the computed|𝑡|, which is 3.002, is greater than |𝑧𝑡𝑎𝑏𝑢𝑙𝑎𝑟 |, which is 2.821, therefore,
reject Ho. Hence, there is a significant difference between the pre-test and post-test of the
students’ performance in College Algebra. It implies that the performance of the students
in Algebra is significantly improved.
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EXERCISES 6.1.1.4
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. Suppose we were interested in determining whether two types of music, A and B, differ
with respect to their effects on sensory-motor coordination. We test some subjects in the
presence of Type-A music and other subjects in the presence of Type-B music. With the
design for correlated samples, we test all subjects in both conditions and focus on the
difference between the two measures for each subject. To obviate the potential effects of
practice and test sequence in this case, we would also want to arrange that half the subjects
are tested first in the Type-A condition, then later in the Type-B condition, and vice versa
for the other half.
Student
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
A
10.2
8.4
17.8
25.2
23.8
25.7
16.2
21.5
21.1
16.9
24.6
20.4
25.8
17.1
14.4
2. Consider an experimenter interested in whether
the time it takes to respond to a visual signal is
different from the time it takes to respond to an
auditory signal. Ten subjects are tested with both
the visual signal and with the auditory signal. (To
avoid confounding with practice effects, half are
in the auditory condition first and the other half
are in the visual task first). The reaction times (in
milliseconds) of the ten subjects in the two
conditions are shown on the right side.
B
13.2
7.4
16.6
27.0
27.5
26.6
18.0
23.4
23.4
21.1
23.8
20.2
29.1
17.7
19.2
Subject
1
2
3
4
5
6
7
8
9
10
Visual
420
235
280
360
305
215
200
460
345
375
Auditory
380
230
300
260
295
190
200
410
330
380
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6.1.1.5 z-test on the Significance of the Difference Between Two Independent
Proportions
There are certain situations when the data to be analyzed involve population
proportions or percentages. For instance, a shoe company may want to know the
proportions of defective shoes to be delivered in other countries. To determine if there
is a significant difference between proportions of two variables, the z-test will be used.
𝑧=
𝑝1 − 𝑝2
𝑝1 𝑞1 𝑝2 𝑞2
+
√ 𝑛
𝑛2
1
where 𝑝1 = proportion of first sample
𝑝2 = proportion of second sample
𝑞1 = 1 - 𝑝1
𝑞2 = 1 - 𝑝2
𝑛1 = number of cases in the first sample
Example 1
𝑛2 = number of cases in the second sample
A sample survey of a presidential candidate in the Philippines shows that 120 of 200 male
voters dislike candidate X and 175 of 250 female voters dislike the same candidate. Determine
175
120
and 250, is significant or not at 1%
whether the difference between the two sample proportions,
200
level of significance.
Solution:
1. Ho: There is no significant difference between the proportion of the male votes and the
proportion of female votes. (𝑝1 = 𝑝2 )
Ha: There is a significant difference between the proportion of the male votes and the
proportion of female votes. (𝑝1 ≠ 𝑝2 )
2. 𝛼 = 0.01, two-tailed
3. Use the z-test as test statistic.
4. Computation:
120
175
Given 𝑝1 = 200, 𝑝2 = 250
𝑧=
𝑝1 − 𝑝2
𝑝1 𝑞1 𝑝2 𝑞2
+
√ 𝑛
𝑛2
1
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120 175
−
200 250
=
120
120
175
175
√(200) (1 − 200) (250) (1 − 250)
+
200
250
=
=
=
−0.1
120 80
175 75
√(200) (200) (250) (250)
+
200
250
−0.1
√0.24 + 0.21
200 250
−0.1
0.045
= −2.22
5. Tabular Value: z = 2.58
6. Decision Rule: Reject Ho if |𝑧| ≥ |𝑧𝑡𝑎𝑏𝑢𝑙𝑎𝑟 |.
7. Since the computed|𝑧|, which is 2.22, is less than |𝑧𝑡𝑎𝑏𝑢𝑙𝑎𝑟 |, which is 2.58, therefore, accept
Ho. Hence, there is no significant difference between the proportion of the male votes and the
proportion of female votes in their dislike for candidate X.
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EXERCISES 6.1.1.5
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. Two types of medication for hives are being tested to determine if there is a difference in
the proportions of adult patient reactions. Twenty out of a random sample of 200 adults
given medication A still had hives 30 minutes after taking the medication. Twelve out of
another random sample of 200 adults given medication B still had hives 30 minutes after
taking the medication. Test at a 5% level of significance.
2. A research study was conducted about gender differences in “sexting.” The researcher
believed that the proportion of girls involved in “sexting” is less than the proportion of
boys involved. The data collected in the spring of 2010 among a random sample of middle
and high school students in a large school district in the southern United States is
summarized in the table. Is the proportion of girls sending sexts less than the proportion of
boys “sexting?” Test at a 5 % level of significance.
Males
Females
Sent “sexts”
183
156
Total number surveyed
2231
2169
3. Researchers conducted a study of smartphone use among adults. A cell phone company
claimed that iPhone smartphones are more popular with whites (non-Hispanic) than with
African-Americans. The results of the survey indicate that of the 232 African-American
cell phone owners randomly sampled, 5% have ab iPhone. Of the 1,343 white cell phone
owners randomly sampled, 10% own an iPhone. Test at the 5% level of significance. Is the
proportion of white iPhone owners greater than the proportion of African-American iPhone
owners?
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6.1.2 Significance of the Difference Between Variances
6.1.2.1 Analysis of Variance
When the variances of two or more independent samples differ, the appropriate test
statistic to determine the significance of such difference is the analysis of variance
(ANOVA), which makes use of the F ratio or variance ratio. The various groups being
compared are assumed to belong to a population with a normal distribution, each group
randomly selected and independent from the other groups. The variables from each
group also have standard deviations that are approximately equal.
Steps in Solving the Analysis of Variance
1. State the null hypothesis.
2. Set the level of significance.
3. Accomplish the ANOVA table.
The ANOVA Table
Source of Sum of
df
Variance Square
Mean
Square
Between
SSB
dfB = k – 1
Within
SSW
Total
TSS
dfW = N – k 𝑀𝑆𝑊 =
where 𝑆𝑆𝐵 =
∑(∑ 𝑋𝐴𝑖 )2
𝑛𝐴𝑖
2
𝑇𝑆𝑆 = ∑ 𝑋𝑖 −
dfT = N – 1
−
𝑀𝑆𝐵 =
F
𝑆𝑆𝐵
𝑑𝑓𝐵
𝑆𝑆𝑊
𝑑𝑓𝑊
𝐹
=
𝑀𝑆𝐵
𝑀𝑆𝑊
(∑ 𝑋𝑖 )2
𝑁
(∑ 𝑋𝑖 )2
𝑁
𝑆𝑆𝑊 = 𝑇𝑆𝑆 − 𝑆𝑆𝐵
𝑁 = sample size
𝑘 = number of columns
𝑋 = observed value
𝑛 = number of rows
𝐴 = given factor or category
𝑖 = individual observation of cell
4. Find the tabular value of F at the given level of significance (from Appendix)
5. State the Decision Rule: If the computed value is less than the tabular value, accept the
null hypothesis. If the computed value is greater than the tabular value, reject the null
hypothesis.
6. Interpret the result.
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Example 1
Determine who among the three salesmen will most likely be promoted based on their
monthly sales in pesos. Use 5% level of significance.
Sales of Three Candidates for Promotion (A, B, C)
A
B
C
12,000
15,500
12,800
10,000
12,500
16,000
10,900
12,000
15,000
18,000
13,000
12,700
16,000
14,000
15,000
14,400
15,000
13,000
14,400
12,300
12,000
15,500
15,000
16,000
18,800
19,000
16,000
Solution:
1. Ho: There is no significant difference between the mean sales of the three candidates for
promotion.
Ha: There is a significant difference between the mean sales of the three candidates for
promotion.
2. 𝛼 = 0.05
3. Accomplish the ANOVA Table.
A
12,000
10,000
10,900
18,000
16,000
14,400
14,400
15,500
18,800
B
15,500
12,500
12,000
13,000
14,000
15,000
12,300
15,000
19,000
C
12,800
16,000
15,000
12,700
15,000
13,000
12,000
16,000
16,000
∑ 𝐴 =130,000
∑ 𝐵 =128,300
∑ 𝐶 =128,500
3.1 Sum of Squares
A2
144,000,000
100,000,000
118,810,000
324,000,000
256,000,000
207,360,000
207,360,000
240,250,000
324,000,000
∑ 𝐴2 =1,921,780,000
B2
240,250,000
156,250,000
144,000,000
169,000,000
196,000,000
225,000,000
151,290,000
225,000,000
361,000,000
∑ 𝐵 2 =1,867,790,000
C2
163,840,000
256,000,000
225,000,000
161,290,000
225,000,000
169,000,000
144,000,000
256,000,000
256,000,000
∑ 𝐶 2 =1,856,130,000
Find SSB:
𝑆𝑆𝐵 =
∑(∑ 𝑋𝐴𝑖 )2 (∑ 𝑋𝑖 )2
−
𝑛𝐴𝑖
𝑁
2
2
(∑ 𝐴) + (∑ 𝐵) + (∑ 𝐶)2 (∑ 𝐴 + ∑ 𝐵 + ∑ 𝐶 )2
=
−
𝑛𝐴𝑖
𝑁
=
(130,000)2 + (128,300)2 + (128,500)2 (130,000 + 128,300 + 128,500)2
−
9
27
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= 5,541,460,000 −
(386,800)2
27
= 5,541,460,000 − 5,541,268,148.15
𝑆𝑆𝐵 = 191,851.85
Find TSS:
𝑇𝑆𝑆 = ∑ 𝑋𝑖 2 −
2
(∑ 𝑋𝑖 )2
𝑁
2
2
= ∑𝐴 + ∑𝐵 + ∑𝐶 −
(∑ 𝐴 + ∑ 𝐵 + ∑ 𝐶 )2
𝑁
= 1,921,780,000 + 1,867,790,000 + 1,856,130,000 − 5,541,268,148.15
= 5,645,700,000 − 5,541,268,148.15
= 104,431,851.85
Find SSW:
𝑆𝑆𝑊 = 𝑇𝑆𝑆 − 𝑆𝑆𝐵
= 104,431,851.85 − 191,851.85
= 104,240,000
1.2. degrees of freedom
dfB = 𝑘 – 1 = 3 – 1 = 2
dfW = 𝑁 – 𝑘 = 27 – 3 = 24
dfT = 𝑁 – 1 = 27 – 1 = 26
1.3. Mean of Squares
𝑀𝑆𝐵 =
𝑀𝑆𝑊 =
𝑆𝑆𝐵 191,851.85
=
= 95,925.93
𝑑𝑓𝐵
2
𝑆𝑆𝑊 104,240,000
=
= 4,343,333.33
𝑑𝑓𝑊
24
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1.4. F – Value
𝐹=
𝑀𝑆𝐵
95,925.93
=
= 0.0221
𝑀𝑆𝑊 4,343,333.33
The ANOVA Table
Sum of
Source of Square
Variance
Between
Within
Total
191,851.85
104,240,000
104,431,851.85
df
Mean
Square
F
2
95,925.93
0.0221
24
26
4,343,333.33
2. Tabular Value: F = 3.40
3. Decision Rule: If the computed value is less than the tabular value, accept the null
hypothesis. If the computed value is greater than the tabular value, reject the null
hypothesis.
4. Since the computed F-value, which is 0.0221, is less than the tabular value, which is 3.40,
so the null hypothesis is accepted. Hence, there is no significant difference between the
mean sales of the three candidates for promotion. It implies that the three salesmen have
almost equal chances of promotion.
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EXERCISES 6.2
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. Zelazo et al. (1972) investigated the variability in age at first walking in infants. Study
infants were grouped into four groups, according to reinforcement of walking and
placement: (1) active (2) passive (3) no exercise; and (4) 8-week control. Sample sizes
were 6 per group, for a total of n=24. For each infant, study data included group assignment
and age at first walking, in months.
The following are the data and consist of recorded values of age (months) by group:
Active Group
9.00
9.50
9.75
10.00
13.00
9.50
Passive Group
11.00
10.00
10.00
11.75
10.50
15.00
No-Exercise Group
11.50
12.00
9.00
11.50
13.25
13.00
8-Week Control
13.25
11.50
12.00
13.50
11.50
12.35
2. Four brands of flashlight batteries are to be compared by testing each brand in five
flashlights. Twenty flashlights are randomly selected and divided randomly into four
groups of five flashlights each. Then each group of flashlights uses a different brand of
battery. The lifetimes of the batteries, to the nearest hour, are as follows:
Brand A
Brand B
Brand C
Brand D
42
28
24
20
30
36
36
32
39
21
28
38
28
32
28
28
29
27
33
25
Preliminary data analyses indicate that the independent samples come from normal
populations with equal standard deviations. At the 5% significance level, does there appear
to be a difference in mean lifetime among the four brands of batteries?
3. The times required by three workers to perform an assembly-line task were recorded on
five randomly selected occasions. Here are the times, to the nearest minute.
Hank
8
10
9
11
10
Joseph
8
9
9
8
10
Susan
10
9
10
11
9
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6.2 Correlation and Regression Analysis
Look at these pictures. What do they show?
When we say “healthy students are better students,” are we saying that the academic
performance of a student depends on his health?
If you know the monthly net profit of a company for a period of time, can you predict its
net profit for the coming months?
When one applies a job, what requirements are needed for submission? Why are they
required? Can a hiring officer predict the kind of worker an applicant will be based on the
submitted requirements?
These are some of the real-life situations that are require decision-making that will be
discussed in this topic. We will learn how to determine whether there is a relationship between
two variables using correlation analysis. We will also learn how to predict the value of one variable
in terms of the other variable using regression analysis.
Correlation Analysis
Why do most students who excel in English do not do well in Mathematics? Have you ever
wondered whys some of your friends who are good in Mathematics do not have high grades in
English? Did it occur to you to find out if there exists a relationship between academic performance
in English and achievement in Mathematics? The statistical procedure that is used to determine
whether a relationship between two variables is called correlation analysis.
Correlation
Correlation analysis measures the association or the strength of the relationship
between two variables say, x and y.
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The relationship or correlation between two variables may be described in terms of direction and
strength.
The direction of correlation may be positive, negative, or zero.



Two variables are positively correlated if the values of the two variables both
increase or both decrease.
Two variables are negatively correlated if the values of one variable increase
while the values of the other decrease.
Two values are not correlated or they have zero correlation if one variable neither
increases nod decreases while the other increases.
The strength of correlation may be perfect, very high, moderately high, moderately
low, very low, and zero. The discussion of the strength is found in the succeeding box.
Example 1
Suppose a ten-item test in English and a ten-item test in Mathematics were administered
to ten students. The scores of the students are tabulated below. It must be determined if the
scores in Mathematics quiz (here labelled variable x) and the English quiz (labelled variable y)
are correlated or not.
Student
1
2
3
4
5
6
7
8
9
10
Mathematics
Score
(x)
4
5
9
2
8
1
2
7
6
4
English Score
(y)
5
4
8
3
9
2
1
6
7
5
The scatter graph of the data above is given below. Note that x-axis represents the scores
in Mathematics and y-axis shows the scores in English. Each point in the graph below is an ordered
pair (x, y) corresponding to the score obtained by a student in the two subjects.
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10
9
8
7
6
5
4
3
2
1
0
0
2
4
6
Mathematics Score
8
10
The graph above indicates a direct correlation between variables x and y which appears to
be increasing.
Example 2
Suppose the scores of the students in those two subjects happen to be as follows:
Mathematics Score
(x)
9
3
4
7
6
1
2
5
10
2
Student
1
2
3
4
5
6
7
8
9
10
English Score
(y)
3
6
7
4
2
9
8
4
2
10
The scatter graph of the data above looks like this:
12
English Score
10
8
6
4
2
0
0
2
4
6
Mathematics Score
8
10
12
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This time the trend of the data is decreasing, hence, the variables are negatively correlated.
Example 3
Suppose the same students have the following scores.
Mathematics
Score
(x)
9
2
6
3
4
5
3
6
8
2
Student
1
2
3
4
5
6
7
8
9
10
English Score
(y)
8
9
3
7
7
5
6
7
4
2
English Score
The scatter graph of the data above looks like this:
10
9
8
7
6
5
4
3
2
1
0
0
1
2
3
4
5
6
Mathematics Score
7
8
9
10
The scatter of the data is neither increasing nor decreasing. It represents a zero correlation.
While a scatter plot may be a convenient way of inspecting correlation between two
variables, it does not offer a measure of the strength of the correlation. Fortunately, Karl Pearson
(1857-1936) developed and perfected a formula that can give a numerical value to measure the
strength of correlation. This formula does not only show how greatly two data sets are correlated
but also reveals if the correlation is direct or inverse, or if the data sets are not correlated. The
formula named after him is called the Pearson Product-Moment Correlation Coefficient.
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6.2.1 Pearson Product-Moment Correlation Coefficient
The most common statistical tool in measuring the linear relationship between two random
variables, x and y, is the linear correlation coefficient commonly called the Pearson ProductMoment Correlation Coefficient or Pearson r for short. It became the basis of different theories in
the fields of heredity, psychology, anthropometry, and statistics. It can be used to determine the
linearity of the relationships between two variables. The Pearson r formula is given by,
𝑟=
𝑛 ∑ 𝑥𝑦 − ∑ 𝑥 ∑ 𝑦
√[𝑛 ∑ 𝑥 2 − (∑ 𝑥 )2 ][𝑛 ∑ 𝑦 2 − (∑ 𝑦)2 ]
Note that the results of r should be interpreted only after its value has been found to be
significant. We will use the measuring devise to determine the strength of the computed r, as shown
below.
Pearson r
±1.0
±0.75 𝑡𝑜 ± 0.99
±0.50 𝑡𝑜 ± 0.74
Qualitative Description
Perfect Correlation/Relationship
Very High Correlation/Relationship
Moderately High Correlation/Relationship
±0.25 𝑡𝑜 ± 0.49
Moderately Low Correlation/Relationship
0
Zero or No Correlation/Relationship
±0.01 𝑡𝑜 ± 0.24
Very Low Correlation/Relationship
Consider the data in Example 1 of this section. Organize the data as shown in the table
below.
Mathematics Score English Score
x2
y2
xy
(x)
(y)
4
5
16
25
20
5
4
25
16
20
9
8
81
64
72
2
3
4
9
6
8
9
64
81
72
1
2
1
4
2
2
1
4
1
2
7
6
49
36
42
6
7
36
49
42
4
5
16
25
20
∑ 𝑥 = 48
∑ 𝑦 = 50
∑ 𝑥 2 = 296 ∑ 𝑦 2 = 310 ∑ 𝑥𝑦 = 298
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Solution:
𝑟=
=
=
=
𝑛 ∑ 𝑥𝑦 − ∑ 𝑥 ∑ 𝑦
√[𝑛 ∑ 𝑥 2 − (∑ 𝑥)2 ][𝑛 ∑ 𝑦 2 − (∑ 𝑦)2 ]
(10)(298) − (48)(50)
√[(10)(296)− (48)2 ][(10)(310) − (50)2 ]
580
√(656)(600)
580
62.73755
= 0.92
This result is in conformity with the scatter plot in Example 1 of this section. The computed
r is almost 1, hence, it has a very high positive correlation. This the reason why the scatter plot in
Example 2 in this section is increasing from left to right.
Using the data in Example 2 of this section, we have the following computations.
Mathematics Score English Score
x2
y2
xy
(x)
(y)
9
3
81
9
27
3
6
9
36
18
4
7
16
49
28
7
4
49
16
28
6
2
36
4
12
1
9
1
81
9
2
8
4
64
16
5
4
25
16
20
10
2
100
4
20
2
10
4
100
20
∑ 𝑥 = 49
∑ 𝑦 = 55
∑ 𝑥 2 = 325 ∑ 𝑦 2 = 379 ∑ 𝑥𝑦 = 198
Solution:
𝑟=
=
𝑛 ∑ 𝑥𝑦 − ∑ 𝑥 ∑ 𝑦
√[𝑛 ∑ 𝑥 2 − (∑ 𝑥)2 ][𝑛 ∑ 𝑦 2 − (∑ 𝑦)2 ]
(10)(198) − (49)(55)
√[(10)(325)− (49)2 ][(10)(379) − (55)2 ]
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=
=
−715
√(849)(765)
−715
805.906322
= −0.89
The computed r is – 0.89, hence, it has a very high correlation. This is the reason why the
scatter plot in Example 2 of this section is decreasing from left to right.
We now compute the r of the data on two non-correlated variables in Example 3 of this
section.
Mathematics Score English Score
x2
y2
xy
(x)
(y)
9
8
81
64
72
2
9
4
81
18
6
3
36
9
18
3
7
9
49
21
4
7
16
49
28
5
5
25
25
25
3
6
9
36
18
6
7
36
49
42
8
4
32
4
32
2
2
4
4
4
∑ 𝑥 = 48
∑ 𝑦 = 58
∑ 𝑥 2 = 252 ∑ 𝑦 2 = 370 ∑ 𝑥𝑦 = 278
Solution:
𝑟=
=
=
=
𝑛 ∑ 𝑥𝑦 − ∑ 𝑥 ∑ 𝑦
√[𝑛 ∑ 𝑥 2 − (∑ 𝑥)2 ][𝑛 ∑ 𝑦 2 − (∑ 𝑦)2 ]
(10)(278) − (48)(58)
√[(10)(252)− (48)2 ][(10)(370) − (58)2 ]
−4
√(216)(336)
−4
269.39933
= −0.01
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Since the computed r is almost zero, then it has little or zero linear correlation. This
conforms with the scatter plot in Example 3 in this section. The graph is neither increasing nor
decreasing and therefore the two sets of data are not correlated.
Example 4
Test the hypothesis that there is no significant relationship between mental ability and
English proficiency at 5% level of significance.
Mental Ability (x)
50
54
50
51
49
46
48
47
44
44
46
45
48
53
54
33
34
English Proficiency (y)
200
198
200
203
186
205
185
197
183
171
179
185
184
190
191
170
168
Solution:
1. Ho: There is no significant relationship between mental ability and English proficiency.
Ha: There is a significant relationship between mental ability and English proficiency.
2. 𝛼 = 5% 𝑜𝑟 0.05
3. Pearson r will be used to test the hypothesis.
4. Computation
Mental
Ability (x)
50
54
50
51
49
46
48
47
44
English
Proficiency
(y)
200
198
200
203
186
205
185
197
183
x2
y2
xy
2,500
2,916
2,500
2,601
2,401
2,116
2,304
2,209
1,936
40,000
39,204
40,000
41,209
34,596
42,025
34,225
38,809
33,489
10,000
10,692
10,000
10,353
9,114
9,430
8,880
9,259
8,052
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44
46
45
48
53
54
33
34
∑ 𝑥 = 796
MATHEMATICS IN THE MODERN WORLD
171
179
185
184
190
191
170
168
∑ 𝑦 = 3,195
𝑟=
=
=
=
1,936
2,116
2,025
2,304
2,809
2,916
1,089
1,156
2
∑ 𝑥 = 37,834
29,241
32,041
34,225
33,856
36,100
36,481
28,900
28,224
∑ 𝑦2 =
602,625
7,524
8,234
8,325
8,832
10,070
10,314
5,610
5,712
∑ 𝑥𝑦 =
150,401
𝑛 ∑ 𝑥𝑦 − ∑ 𝑥 ∑ 𝑦
√[𝑛 ∑ 𝑥 2 − (∑ 𝑥)2 ][𝑛 ∑ 𝑦 2 − (∑ 𝑦)2 ]
(17)(150,401) − (796)(3,195)
√[(17)(37,834)− (796)2 ][(17)(602,625) − (3,195)2 ]
13,597
√(9,562)(36,600)
13,597
18,707.46375
= 0.73
5. df = N – 2 = 17 – 2 = 15
6. Tabular Value: r = 0.482 (from Appendix)
7. Decision Rule: If the computed value is less than the tabular value, accept the null
hypothesis. If the computed value is greater than the tabular value, reject the null hypothesis.
8. Since the computed r (0.73) is greater than the tabular value (0.482), so reject the null
hypothesis. Hence, there is a significant relationship between mental ability and English
proficiency. It shows that there is a moderately high positive relationship between the two
variables.
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EXERCISES 6.2.1
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. Below are the data for six participants giving their number of years in college (X) and their
subsequent yearly income (Y). Income here is in thousands of pesos, but this fact does not
require any changes in our computations. Test whether there is a relationship with Alpha
= .05.
No. of Years
of College
(x)
Income
(y)
0
1
3
4
4
6
15
15
20
25
30
35
2. Yvonne is a good student, but at times she doesn’t get enough sleep. She hypothesizes that
when she gets more sleep she does better on tests. To test her hypothesis, she tracked how
she did on a number of tests, based on how many hours of sleep she got on the night
previous. She inputs the following data into her excel file to compute the correlation
coefficient equation.
Hours of Sleep
(x)
Test Score
(y)
8
8
6
5
7
6
81
80
75
65
91
80
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6.2.2 Regression Analysis
Regression analysis is used when predicting the behavior of a variable. The regression
equation explains the amount of variations observable in the independent variable x. It is actually
an equation of a straight line in the form:
𝑦 = 𝑏𝑥 + 𝑎
where y = criterion measure
x = predictor
a = ordinate or the point where regression line crosses the yaxis
b = the slope of the line.
To get the regression equation, the values of a and b are computed using the formula below.
∑ 𝑦 ∑ 𝑥 2 − ∑ 𝑥 ∑ 𝑥𝑦
𝑎=
𝑛 ∑ 𝑥 2 − (∑ 𝑥 )2
and
where n = number of pairs
Example 1
𝑏=
𝑛 ∑ 𝑥𝑦 − ∑ 𝑥 ∑ 𝑦
∑ 𝑥 2 − (∑ 𝑥 )2
The data in the table represent the membership at a university Mathematics club during the
past 5 years. Find the regression equation to predict the membership 5 years from now.
Number of Years (x)
1
2
3
4
5
Membership (y)
25
30
32
45
50
Number of
Years (x)
1
2
3
4
5
∑ 𝑥 = 15
x2
xy
1
4
9
16
25
2
∑ 𝑥 = 55
25
60
96
180
250
∑ 𝑥𝑦 = 611
Solution:
Membershi
p (y)
25
30
32
45
50
∑ 𝑦 = 182
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Find a:
Find b:
𝑎=
∑ 𝑦 ∑ 𝑥 2 − ∑ 𝑥 ∑ 𝑥𝑦
𝑛 ∑ 𝑥 2 − (∑ 𝑥)2
𝑏=
5(611) − 15(182)
5(55) − (15)2
325
=
50
182(55) − 15(611)
5(55) − (15)2
845
=
50
=
𝑛 ∑ 𝑥𝑦 − ∑ 𝑥 ∑ 𝑦
∑ 𝑥 2 − (∑ 𝑥)2
=
= 6.5
= 16.9
Substitute the values of a and b in the equation y = bx + a.
y = 6.5x + 16.9
Since you need to predict the membership five years from now, or at year 10, substitute 10 for x
in the equation.
y = 6.5(10) + 16.9
= 81.9
≈ 82
Thus, five years from now, the Mathematics club would have 82 members.
Example 2
The following data pertains to the heights of father and their eldest sons in inches. If there
is a significant relationship between two variables, predict the height of the son if the height of his
father is 78 inches. Use 5% level of significance.
Height of the Father
(x)
71
69
69
65
66
63
68
70
60
58
Height of the Son
(y)
71
69
71
68
68
66
70
72
65
60
Solution:
1. Ho: There is no significant relationship between heights of father and their eldest sons.
Ha: There is a significant relationship between heights of father and their eldest sons.
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2. 𝛼 = 5% 𝑜𝑟 0.05
3. Pearson r will be used to test the hypothesis.
4. Computation
Height of the
Father
(x)
71
69
69
65
66
63
68
70
60
58
∑ 𝑥 = 659
Height of the
Son
(y)
71
69
71
68
68
66
70
72
65
60
∑ 𝑦 = 680
𝑟=
x2
y2
xy
5041
4761
4761
4225
4356
3969
4624
4900
3600
3364
2
∑ 𝑥 = 43,601
5041
4761
5041
4624
4624
4356
4900
5184
4225
3600
2
∑ 𝑦 = 46,356
5041
4761
4899
4420
4488
4158
4760
5040
3900
3480
∑ 𝑥𝑦 = 44,947
𝑛 ∑ 𝑥𝑦 − ∑ 𝑥 ∑ 𝑦
√[𝑛 ∑ 𝑥 2 − (∑ 𝑥)2 ][𝑛 ∑ 𝑦 2 − (∑ 𝑦)2 ]
=
(10)(44,947) − (659)(680)
√[(10)(43,601)− (659)2 ][(10)(46,356) − (680)2 ]
= 0.95
5. df = N – 2 = 10 – 2 = 8
6. Tabular Value: r = 0.632 (from Appendix)
7. Decision Rule: If the computed value is less than the tabular value, accept the null
hypothesis. If the computed value is greater than the tabular value, reject the null hypothesis.
8. Since the computed r (0.95) is greater than the tabular value (0.632), so reject the null
hypothesis. Hence, there is a significant relationship between heights of father and their
eldest sons. It shows that there is a very high positive relationship between the two variables.
We can now proceed to regression analysis since there was a significant relationship
between heights of father and their eldest sons.
Find a:
Find b:
𝑎=
∑ 𝑦 ∑ 𝑥 2 − ∑ 𝑥 ∑ 𝑥𝑦
𝑛 ∑ 𝑥 2 − (∑ 𝑥)2
𝑏=
𝑛 ∑ 𝑥𝑦 − ∑ 𝑥 ∑ 𝑦
∑ 𝑥 2 − (∑ 𝑥)2
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=
680(43,601) − 659(44,947)
(10)(43,601)− (659)2
= 16.55
=
10(44,947) − 659(680)
(10)(43,601)− (659)2
= 0.78
Substitute the values of a and b in the equation y = bx + a.
y = 0.78x + 16.55
Since you need to predict the height of the son if the height of the father is 78 inches, substitute
78 for x in the equation.
y = 0.78(78) + 16.55
= 77.39
≈ 77 inches
Thus, the predicted height of the son whose father’s height is 78 inches is 77 inches.
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EXERCISES 6.2.2
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. The time x in years that an employee spent at a company and the employee’s hourly pay,
y, for 5 employees are listed in the table below. Find the equation of regression line and
predict the hourly pay rate of an employee who has worked for 20 years.
No. in Years Hourly Pay
(x)
(y)
5
25
3
20
4
21
10
35
15
38
2. The table below shows the number of absences, x, in a Calculus course and the final exam
grade, y, for 7 students. Find the equation of regression line and predict the test score for a
student with 5 absences.
Number of
Absences
(x)
1
0
2
6
4
3
3
Test Score
(y)
95
90
90
55
70
80
85
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6.2.3 Spearman’s Rank Correlation Coefficient Spearman rho (𝝆)
Beauty contests are very popular not only among Filipinos but also to many people around
the world. Normally, when the names of the five finalists are announced, people place their own
bets on who will be the queen and the runners-up. Very often, they are happy about the results.
This happens when their ranks agree with the ranks assigned by the board of judges. There might
be some slight differences between the ranks assigned by the people and those by the board of
judges but if overall, there is a positive correlation (or agreement) between these ranks, then
everyone will be happy about the results.
In this next statistical measure, we shall be concerned with correlation between ranks. Like
in simple correlation, we have cases of positive correlation, zero correlation, or negative
correlation. A positive rank correlation indicates that those categories that are given high ranks by
one judge (or rater) are also the categories that are assigned high ranks by the other rater. Or those
with low ranks in one have also low ranks in the other. A negative rank correlation is the reverse.
It means that those categories who were assigned high ranks by the first rater is given low ranks
by the second rater, or vice versa.
The most common method used in rank correlation is the statistics developed by Spearman
where the coefficient used is symbolized by 𝜌 (rho, Greek letter for r). To compute 𝜌, we use the
formula:
6 ∑ 𝑑2
𝜌 = 1−
𝑛(𝑛2 − 1)
where d = difference between ranks
n = number of categories given ranks.
In interpreting the computer 𝜌, we use the same qualitative interpretation as the one we use
in interpreting Pearson r.
Example 1
In a contest for Mr. Campus Personality, two judges gave their ratings to 8 candidates.
Transform the ratings to ranks and compute the coefficient of rank correlation. Interpret the
result.
Candidate
1
2
3
4
5
6
7
8
Judge 1
98
97
95
90
89
88
85
85
Judge 2
94
97
98
95
92
90
89
85
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Solution:
Candidat
e
1
2
3
4
5
6
7
8
Judge 1
(x)
98
97
95
90
89
88
85
85
Judge 2
(y)
94
97
98
95
92
90
89
85
𝜌=1−
Rx
Ry
d
d2
1
2
3
4
5
6
7.5
7.5
4
2
1
3
5
6
7
8
-3
0
2
1
0
0
0.5
-0.5
9
0
4
1
0
0
0.25
0.25
6 ∑ 𝑑2
6(14.5)
=1−
= 0.83
2
𝑛(𝑛 − 1)
8(82 − 1)
∑ 𝑑 2 = 14.5
Interpretation: The computed 𝜌 (0.83) indicates a “very high positive correlation” between the
ranks. This means that those candidates who received high ranks from the first judge are also the
candidates who received the same high ranks from the second judge. Similarly, those candidates
who were ranked low by the first judge were also ranked low by the other judge. This means that
the rankings of the two judges have a very high degree of agreement. It also implies that as to the
selection of Mr. Campus Personality, the two judges have more or less the same taste.
Example 2
Ten instructors were rated by third- and fourth-year students on their “master of subject
matter” and the results were tabulated. What is the Spearman rho value for the data? At 5% level
of significance, determine if there is a significant relationship in the scores obtained by the
teachers.
Instructor
1
2
3
4
5
6
7
8
9
10
3rd Year (x)
44
45
38
32
46
47
37
35
27
40
4th Year (y)
46
43
40
30
39
37
44
46
48
50
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Solution:
1. Ho: There is no significant relationship between the ratings given to the ten instructors by
third- and fourth-years students.
Ha: There is a significant relationship between the ratings given to the ten instructors by
third- and fourth-years students.
2. 𝛼 = 5% 𝑜𝑟 0.05
3. Spearman rho (𝜌)will be used to test the hypothesis.
4. Computation
Instructo
r
1
2
3
4
5
6
7
8
9
10
3rd Year
(x)
44
45
38
32
46
47
37
35
27
40
4th Year
(y)
46
43
40
30
39
37
44
46
48
50
Rx
Ry
d
d2
4
3
6
9
2
1
7
8
10
5
3.5
6
7
10
8
9
5
3.5
2
1
0.5
-3
-1
-1
-6
-8
2
4.5
8
4
0.25
9
1
1
36
64
4
20.25
64
16
∑ 𝑑2
= 215.5
𝜌 = 1−
= 1−
6 ∑ 𝑑2
𝑛(𝑛2 − 1)
6(215.5)
10(102 − 1)
= 1 − 1.31
= −0.31
5. df = N – 2 = 10 – 2 = 8
6. Tabular Value: 𝜌 = 0.643 (from Appendix)
7. Decision Rule: If the computed value is less than the tabular value, accept the null
hypothesis. If the computed value is greater than the tabular value, reject the null hypothesis.
8. Since the absolute value of the computed 𝜌 (0.31) is less than the tabular value (0.643), so
the null hypothesis accepted. Hence, there is no significant relationship between the ratings
given to the ten instructors by third- and fourth-years students. It implies that the ratings of
the third- and fourth-years students are not the same.
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EXERCISES 6.2.3
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. The scores for nine students in history and algebra are as follows:
History Algebra
34
31
25
32
16
46
9
23
40
9
7
48
28
31
9
4
Compute the Spearman rank correlation.
2. The left side of Figure 1 displays the association between the IQ of each adolescent in a
sample with the number of hours they listen to rock music per month. Determine the
strength of the correlation between IQ and rock music using both the Pearson’s correlation
coefficient and Spearman’s rank correlation. Compare the results.
IQ Rock
99
2
120
0
98
25
102
45
123
14
105
20
85
15
110
19
117
22
90
4
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Chapter 7: THE MATHEMATICS OF GRAPHS
 Learning Objectives
At the end of this chapter, the student is expected to:
 differentiate Eulerian from Hamiltonian graphs;
 apply Euler and Hamiltonian paths to solve problems; and
 solve problems using graphs.
Introduction
Graph theory is a branch of Mathematics that was developed after Leonhard Euler (1707 –
1783), a Swiss mathematician, solved an eighteen century problem involving the seven bridges of
Konigsberg in Old Prussia. The city of Konigsberg (now Kaliningrad, Russia) has four districts
divided by the Pregel River. Seven bridges connected these districts as shown in the figure below.
In Euler’s time, people were puzzled if there is a travel route that would only cross each of the
seven bridges exactly one. Euler proved in 1736 that it is impossible to take a stroll that would
lead them across each bridge and return to the starting point without traversing the same bridge
twice. Problems involving connections such as the seven bridges of Konigsberg is the subject
matter of this chapter.
At present, graph theory finds many applications in the social sciences (social network
sites), computer science (networks of communication), chemistry (chemical structure).
Communication arts (networks of communication), and operations research (network analysis).
7.1 Graph
A graph is a collection of points called vertices or nodes and line segment or curves called
edges that connect the vertices.
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The position of the vertices, the lengths of the edges, and the shape of the edges do not
matter in a graph. Sometimes the edges are given orientations and are presented by arrows or are
given values (weights). But it is the number or vertices and which of them ae joined by edges that
matter most.
Graphs can be used to illustrate huge connections such as social networks in Facebook,
flight destinations of airlines, the simple community garbage collection route, or even the computer
system connectivity in a school.
Example: Constructing a Graph
Brunei
Singapore
Kuala
Lumpur
Ho Chi Minh
Bangkok
Macau
Hog Kong
Taipei
Seoul
Tokyo
Manila
The following table lists eleven cities connected by Cebu Pacific airline flights. The symbol
indicates that the cities have direct flights.
Manila
Tokyo
Seoul
Taipei
Hong Kong
Macau
Bangkok
Ho Chi Minh
Kuala Lumpur
Singapore
Brunei
Draw a graph that presents this information where each vertex represents a city and an edge
connects two vertices if the two cities have a direct flight.
Use your graph to determine which city has the most and least number of direct flights.
Solution:
a. Draw eleven vertices (in any configuration you wish) to represent the eleven cities, and
connect the vertices with edges according to the table.
Seoul
Hong Kong
Ho Chi Minh
Bangkok
Taipei
Macau
Tokyo
Kuala Lumpur
Singapore
Manila
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b. The Manila vertex has nine edges attached to it; hence, Manila has the most number of
direct flights. On the other hand, the Macau vertex is connected to only one node;
hence, Macau has the least number of direct flights. It is important to note also that the
vertex of Brunei is not connected to any node; hence, Brunei does not have a direct
flight to any of the ten cities.
Some Definitions









A loop is the edge connecting a vertex to itself.
If two vertices are connected by more than one edge, these edges are called multiple edges.
A graph with no loops and no multiple edges is called a simple graph.
A path is an altering sequence of vertices and edges. It can be seen as a trip from one vertex
to another using the edges of the graph.
A graph is connected if there is a path connecting all the vertices.
If a path begins and end s with the same vertex, it is called a closed path or a circuit or
cycle.
Two vertices are adjacent if there is an edge joining them.
If every pair of vertices of graph are adjacent, the graph is complete. A complete graph
with n vertices is denoted by Kn.
The degree of a vertex is the number of edges attached to it.
Examples of Graph:
Null or Disconnected Graph. The graph below is a null or disconnected graph since it has
four vertices but no edges. The degree of each vertex is 0.
Graph with a Loop. The loop connects vertex A to itself. The degree of a loop is 2.
B
A
Graph with Multiple Edges. Both graphs G1 and G2 on the next page are connected and
have multiple edges connecting vertices A and B. The degrees of vertices of A and B in G 1 are
both equal to 3 while that of G2 are both equal to 4.
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G1:
MATHEMATICS IN THE MODERN WORLD
A
B
G2:
A
B
Complete Graph. A complete graph is a connected graph in which every edge is drawn
between vertices. It should not contain multiple edges.
K1: One Vertex:
K2: Two Vertices:
K3: Three Vertices:
K4: Four Vertices:
K5: Five Vertices:
Let e be the number of edges in a complete graph. From the results above, we find that for:
K1: e = 0, degree of the vertex is 0.
K2: e = 1, degree of the vertex is 1.
K3: e = 3, degree of the vertex is 2.
K4: e = 6, degree of the vertex is 3.
K5: e = 10, degree of the vertex is 4.
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What about Kn, a complete graph with n vertices? How many edges could we possibly
get and what is the degree of each vertex in the graph? The number of edges is equal to:
𝑒𝑛 =
𝑛(𝑛 − 1)
2
for n ≥ 3 while the degree of each vertex is obviously equal to n – 1.
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MATHEMATICS IN THE MODERN WORLD
EXERCISES 7.1
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
Aileen
Bien
Charles
David
Erica
Fred
Gladys
X
X
X
X
Gladys
X
X
X
X
X
X
X
X
X
X
X
Fred
X
X
X
Erica
David
Charles
Bien
Aileen
1. An “X” in the table below indicates that the corresponding people are connected on
Facebook. Draw a graph in which each vertex represents a name and an edge connects
two vertices if the two friends are connected on Facebook.
X
X
X
X
X
X
2. Draw a graph that represents the information given in the table below involving teachers
and subjects that are assigned to them in a semester.
Mathematics
College and
Calculus I
Number
in the Modern Advanced
with Analytic Theory
World
Algebra
Geometry
X
X
X
Leroy
X
Joan
X
X
Mark Gil
X
X
Kiervin
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7.2Euler Paths and Circuits
Euler Circuits
An Euler circuit is a closed path that uses every edge, but never
uses the same edge twice. The path may cross through vertices more
than once.
In the Konigsberg bridges problem, finding a path crosses each
bridge exactly once and returning to the starting point is the same
as finding an Euler circuit in the graph below.
Leonhard Euler
Euler proved that the graph does not have an Euler circuit because for an Euler circuit to
exist, the degree of each vertex in the graph must be even. Apparently, all the vertices in the
Konigsberg bridges problem have odd degrees; hence not Eulerian. Consequently, he formulated
the following theorem:
Eulerian Graph Theorem
A connected graph is Eulerian if and only if every vertex of the graph is of even degree.
Note that the Eulerian Graph Theorem only guarantees that if the degrees of all the vertices
in a graph are even, an Euler circuit exists, but it does not tell us how to find one.
Example 1: Determine whether the following graph is Eulerian. If it is, find an Eulerian circuit. If
A
it is not, explain why.
E
B
D
C
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Solution:
The degree of each of the vertices is 4 (even); hence, the graph is Eulerian. The path A –
D – B – E – C – A – E – D – C – B – A starts at vertex A and ends at vertex A; hence it is circuit.
Moreover, it uses all the edges all the edges only once; hence it is an Euler circuit.
Euler Path
An Euler path is a path that uses every edge in the graph exactly once but it does not
start and end at the same vertex.
Example 2: Determine whether the following graph is Eulerian. If it is, find a Eulerian circuit. If
it is not, can you find an Euler path?
C
B
L
K
A
D
J
G
I
H
E
F
Solution:
Using the Eulerian Graph Theorem, this graph is not Eulerian since vertices A and J both
have odd degrees. But the path A – B – C – D – E – F – G – H – I – J – D – G – A – L – K – J uses
every edge without duplication, hence the graph contains an Euler path. Furthermore, it can be
noted that the path starts at A but ends at J, the vertices having odd degrees.
Euler Path Theorem
A connected graph contains an Euler path if and only if the graph has two vertices of
odd degrees with all other vertices of even degrees. Furthermore, very Euler path must start at
one of the vertices of odd degrees and end at the other.
Example 3: An Appointment of Euler Path Theorem
Below is the map of all the trails in a national park. A biker would like to traverse all the
trails exactly once.
a. Is it possible for the biker to plan a trip that traverses all the trails exactly once?
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b. Is it possible for him to traverse all the trails and return to the starting point without
repeating any trail in the trip?
B
A
C
D
E
G
F
Solution:
a. By the Euler Path Theorem, the map shows an Euler path since the graph has two vertices
of odd degree with all other vertices of even degree. By trial and error, the path A – B – E
– F – D – B – C – F – G – C – A – G uses every edge without duplication, hence an Euler
path. Thus, it is possible for the biker to plan a trip that traverses all the trails exactly once.
The trip starts at point A, a vertex with an odd degree and ends at point G, the other vertex
with an odd degree.
b. Using the Eulerian Graph Theorem, this graph is not Eulerian since vertices A and G both
have odd degrees. Thus, it is not possible for the biker to traverse all the trails and return
to the starting point without repeating any trail in the trip.
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EXERCISES 7.2
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. Determine whether the graph is Eulerian. If it is, find an Eulerian circuit. If it is not, explain
why? If the graph does not have an Euler circuit, does it have an Euler path? If so, find one
If not, explain why.
a. A
C
B
A
c.
B
F
C
E
E
G
F
D
D
A
A
b.
d.
B
E
B
C
D
E
C
D
2. For each of the networks below, determine whether it has an Euler path. If it does, find
one.
a.
c.
b.
d.
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7.3Hamiltonian Paths and Circuits
Hamiltonian
A Hamiltonian path is a path that visits each vertex
of the graph exactly once.
A Hamiltonian circuit is a path that uses each vertex
of a graph exactly once and returns to the starting vertex. A
graph that contains a Hamiltonian circuit is called
Hamiltonian.
Sir William Rowan Hamilton
In Euler circuits, closed paths use every edge exactly once, possibly visiting a vertex more
than once. On the contrary, in Hamiltonian circuits, paths visit each vertex exactly once, possibly
not passing through some of the edges. But unlike the Euler circuit, where the Eulerian Graph
Theorem is used to determine whether it contains an Euler circuit or not, there is no straightforward
criterion to determine whether or not a Hamiltonian circuit exists in a graph. Fortunately, the
following theorem can help:
Dirac’s Theorem
Consider a connected graph with at least three vertices and no multiple edges. Let n be
𝑛
the number of vertices in the graph. If every vertex has degree of at least 2 , then the graph
must be Hamiltonian.
Example 1:
Determine whether the graph below is Hamiltonian or not. If it is, find a Hamiltonian
circuit. If it is not, explain why.
A
C
B
E
G
F
D
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Solution:
𝑛
There are seven vertices, hence 2 = 3.5. Since vertex A is a degree 2, less than 3.5, Dirac’s
Theorem does not apply here. But it does not necessarily follow that the graph is not Hamiltonian.
In fact, it is. Consider the path A – B – C – E – D – F – G – A. This path visits each vertex only
once in the graph and returns to its starting point, therefore, it is Hamiltonian
Example 2: An Application of Hamiltonian Circuits
The graph below shows the available flights of a popular airline. An edge between two
vertices indicates that there is a direct flight between the two cities. Apply Dirac’s Theorem to
verify that the graph is Hamiltonian. Then find a Hamiltonian circuit.
Seoul
Tokyo
Hong Kong
Taipei
Macau
Bangkok
Ho Chi Minh
Manila
Kuala Lumpur
Singapore
Solution:
𝑛
There are ten vertices in the graph, so n = 10 and = 5. Now, vertex Manila has nine edges,
2
Tokyo has five, Seoul has six, Teipei has six, Hong Kong has seven, Macau has nine, Bangkok
has six, Ho Chi Minh has five, Kuala Lumpur has five, and Singapore has five. Using Dirac’s
𝑛
Theorem, if each vertex has a degree of at least 2 = 5, then the graph is Hamiltonian. This means
that the graph contains a circuit that visits each vertex and returns to its starting point without
visiting a vertex more than once. By trial and error, one Hamiltonian circuit is Manila – Tokyo –
Seoul – Hong Kong – Macau – Bangkok – Ho Chi Minh – Kuala Lumpur – Singapore – Manila.
In example 2, there is a number of different paths which are Hamiltonian. For example,
Manila – Tokyo – Seoul – Hong Kong – Macau – Bangkok – Ho Chi Minh – Kuala Lumpur –
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Singapore – Taipei – Manila is another Hamiltonian circuit that represents a sequence of flights
that visits each city and returns to the starting city without visiting any city twice. Although
generally the lengths of the edges do not matter in the graph, there is now concern over the route
that minimizes the distance travelled. In other words, there is a need to know which of these
Hamiltonian routes is the cheapest. Hence, it is but important that one focuses on the distances
between cities. These distances can be presented using weighted graphs.
Weighted Graphs
A weighted graph is a graph in which each edge is associated with a value, called
weight.
Example 3: An Application of Hamiltonian Circuits
Cebu
Cagayan
de Oro
Davao
Palawan
Ozamis
Manila
Cebu
Cagayan
de Oro
Davao
Palawan
Ozamis
Manila
The table below lists down the distance (miles) between the cities having direct routes as
well as the corresponding distances between them.
355
485
355
137
485
137
-
589
240
118
358
354
414
477
148
64
589
358
477
240
354
148
118
414
64
495
133
495
363
133
363
-
a. Draw a graph that represents this information where each vertex represents a city and an
edge connects two vertices if the two cities have a direct flight with their corresponding
weights.
b. Find two different routes that visit each of the places and return to its starting point without
visiting any city twice. Compare the total number of miles travelled by each of these routes.
Solution:
a. The graph along with the weights of the edges is shown on the next page.
b. One Hamiltonian circuit is Ozamis – Cagayan de Oro – Cebu – Palawan –
Manila – Davao – Ozamis. The total distance travelled is 64 + 137 + 354 + 358
+ 589 + 133 = 1,635 miles. Another node route is Ozamis – Davao – Cagayan
de Oro – Palawan – Cebu – Manila – Ozamis. This circuit has a total distance
of 133 + 118 + 414 + 354 + 355 + 477 = 1,851 miles. Obviously, the first node
route is shorter than the second.
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355
Manila
Cebu
148
477
354
455
358
589
Ozamis
Palawan
363
64
240
414
133
137
495
Davao
Cagayan de Oro
118
In Example 3, we computed two Hamiltonian routes. But these results do not guarantee
that one of them is the shortest distance travelled. From the solution in Example 3, is Ozamis –
Cagayan de Oro – Cebu – Palawan –Manila – Davao – Ozamis the shortest route? There is no
guarantee. If this is the case, how can the shortest route be determined after visiting all the cities
exactly once and going back to the origin city? One method is to down all the Hamiltonian circuits,
compute the total weight, and choose the smallest total weight. Unfortunately, this is tedious
especially when the number of possible circuits is too large. However, there are two algorithms,
the greedy algorithm and the edge-picking algorithm, that can help in finding a good solution.
Note that both of these algorithms apply only to complete graphs.
The Greedy Algorithm
1. Choose a vertex to start at, and then travel along the connected edge that has the smallest
weight. (if two or more edges have the same weight, pick any one.)
2. After arriving at the next vertex, travel along an edge of the smallest weight that
connects to a vertex not yet visited. Continue this process until you have visited all
vertices.
3. Return to the starting vertex.
Example 4:
Aaron, Belle, Carol, Donna, Eric, and Fe are best of friends. The figure below shows the
distances (in kilometers) from a friend’s place to another. If Aaron wants to visit each of his
friend’s houses exactly once, what is the shortest route that he must take?
Aaron
Carol
13
12
1
3
5
7
8
4
Fe
Eric
6
14
15
2
11
Belle
9
10
Donna
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Solution: Using the Greedy Algorithm
To find the least route that Aaron can take, one can find a Hamiltonian circuit using the
greedy algorithm. By trial and error, one Hamilton circuit is from Aaron’s house – Belle’s house
– Carol’s house – Donna’s house – Eric’s house – Fe’s house – Aaron’s house. The total weight
of the circuit is 1 + 2 + 3 + 9 + 6 + 12 = 33. But there are other Hamiltonian circuit from Aaron’s
house. Consider the Hamiltonian circuit from Aaron’s house – Eric’s house – Fe’s house – Carol’s
house – Donna’s house – Belle’s house and back to Aaron’s house. The total weight of this circuit
is 5 + 6 + 7 + 3 + 10 + 1 = 32. Ironically, this circuit has a weight lesser than the weight of the
circuit derived using the greedy algorithm. Thus, the greedy algorithm only attempts to give a
circuit of minimal total weight, although it does not always succeed.
The Edge-Picking Algorithm
1. Mark the edge of the smallest weight in the graph. (If two or more edges have the same
weight, pick any one.)
2. Mark the edge of the next smallest weight in the graph, as long as it does not complete
a circuit and does not add a third marked edge to a single vertex.
3. Continue this process until you can no longer mark any edges. Then mark the final edge
that completes the Hamiltonian circuit.
13
12
1
3
5
7
8
4
6
Fe
Eric
15
14
2
11
9
Solution: Using the Edge-Picking Algorithm
To find the route with the least distance that Aaron can take, one can find a Hamiltonian
circuit using the edge-picking algorithm. First, mark the line segment from Aaron’s house to
Belle’s house, of weight 1. Next, mark the segment from Belle’s to Carol’s house, of weight 2,
followed by Carol’s to Donna’s house, of weight 3, followed by Eric’s to Fe’s house, of weight 6.
Take note that we cannot mark the segment from Eric’s house to Aaron’s house because it can
complete a circuit. Also, we cannot mark the segment from Carol’s to Fe’s house because it can
make the third marked edge on a vertex. Finally, to complete the circuit, we mark the line segment
from Fe’s house back to Aaron’s. The final Hamiltonian circuit, of the total weight 1 + 2 + 3 + 6
+ 9 + 12 = 33, is Aaron’s house – Belle’s house – Carol’s house – Donna’s house –Eric’s house –
Fe’s house and back to Aaron’s house. A Hamiltonian circuit forms a complete loop so we can
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actually start from any of the vertices. It is important to note that we can reverse the direction in
which we follow the circuit.
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EXERCISES 7.3
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. A garbage collector would like to collect the garbage in all the streets of a subdivision
along a shortest possible path. Is this an Eulerian or Hamiltonian problem? Explain why?
2. A school bus driver would like to bring the kids back to their homes along a least expensive
route. Is this an Eulerian or Hamiltonian problem? Explain why?
3. Below is the map of streets in a subdivision. A garbage collector would like to collect the
garbage of residents along a shortest possible path.
a. Is it possible for the garbage collector to find the most efficient route to collect all the
garbage with no street to be traversed more than once?
b. Is it possible to plan a trip that traverses all the streets and returns to the starting point
without repeating any street in the trip?
4. Determine whether the graph is Hamiltonian. If it is, find a Hamiltonian circuit. If it is not,
explain why.
A
A
a.
b.
F
F
B
E
G
J
B
E
I
C
D
D
H
C
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7.4Graph Coloring
Graph coloring started in the mid-1800’s when Francis Guthrie tried to color the map of
England so that it would be easy to distinguish the countries sharing a common border. He made
sure that countries with the same border must have different colors. After many attempts, he found
out that a maximum of four colors we required to color the map.
In graph coloring, each vertex of a graph will be assigned one color in such a way that no
two adjacent vertices have the same color. The interesting idea here is to determine the minimum
number of distinct colors to be used so that each vertex of a graph is colored such that no two
adjacent vertices have the same color. A practical application of the graph coloring problem is in
scheduling meetings or events.
Planar Graph
A planar graph is a graph that can be drawn so that no edges intersect each other
(except at vertices)
A
A
Non - Planar
Planar
The Chromatic Number of a Graph
The minimum number of colors needed to color a graph so that no edge connects vertices
of the same color is called the chromatic number.
2-Colorable Graph Theorem
A graph is 2-colorable if and only if it has no circuits that consist of an odd number of
vertices.
Four-Color Theorem
The chromatic number of a planar graph is at most 4.
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Example 1:
Consider the complete graphs K4 and K5. Determine their chromatic number.
Solution:
First, assign vertex A with one color, say red, then vertex B with another color, say blue.
Since you cannot color two adjacent vertices using the same color, use green to color the vertex C,
and finally, yellow to color the vertex D. Thus, K4 is four-colorable. It is important to note that K4
is planar, hence the Four-Color Theorem is satisfied.
A
B
C
D
Previously, it is seen that K5 is not planar so the Four-Color Theorem does not hold here.
Now, assign each vertex of the graph with one color in such a way that no two adjacent vertices
have the same color as shown below. Thus, the chromatic number of K5 is 5. Can you find the
chromatic number of K8?
E
F
I
H
G
Example 2:
Six college accreditation committees need to hold meetings on the same day, but some
teachers belong to more than one committee. In order to avoid members missing meetings, the
meetings need to be scheduled at different time slots. An “X” in the table on the next page indicates
that the two corresponding committees share at least one member. Use graph coloring to determine
the minimum number of time slots necessary to ensure that all faculty members can attend all
meetings.
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Faculty Instruction
Faculty Development
Outreach Program
Physical Facility
Library Facility
Student Welfare
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
Student Welfare
(SW)
Library Facility
(LF)
Physical Facility
(PF)
Outreach
Program (OP)
Faculty
Development
(FD)
Committee
Faculty
Instruction (FI)
MATHEMATICS IN THE MODERN WORLD
X
X
X
X
-
Solution:
First, draw a graph representing the six committees using six vertices or nodes in any
configuration. An edge connects two committees that share at least one member. Then assign
each vertex of the graph with one color in such a way that no two adjacent vertices have the
same color.
FD
FI
OP
SW
LF
PF
Obviously, the graph is not 2-colorable because there are circuits of odd length, but the
graph is 3-colorable. Hence, the minimum number of time slots necessary to ensure that all faculty
members can attend all meeting is 3.
First time slot: Faculty Instruction, Student Welfare
Second slot: Faculty, Outreach Program
Third slot: Library Facility, Physical Facility
Example 3:
The fictional map on the next page shows the boundaries of barangays on a
rectangular town.
a. Represent the map of a graph
b. Find a coloring of the graph using the fewest possible number of colors
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c. Color the map according to the graph coloring theorem.
Solution:
First, represent each barangay using vertices A, B, C, D, E, F, G, H, I, and J.
G
H
A
I
J
F
B
D
E
C
Second, connect two vertices with an edge if the two barangays share the same boundary.
Third, color the vertices of the resulting graph so that no edge connects two vertices with
the same color. Coloring is not unique.
H
G
A
J
I
B
D
F
C
E
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Finally, color each barangay in the map according to the color of its assigned vertex in
the previous step.
H
A
G
J
I
F
B
C
D
E
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EXERCISES 7.4
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. Color the graph according to the graph coloring concepts discussed in section 7.4.
A
Determine its chromatic number.
F
B
E
C
D
2. The fictional map below shows the boundaries of countries on a rectangular continent.
a. Represent the map as a graph
b. Find a coloring of the graph using the fewest possible number of colors.
c. Color the map accordingly using one of the graph coloring theorems.
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Debate
Club
Student
Publication
Fitness
Club
X
X
X
X
X
X
X
Fitness
Club
X
Arts Club
X
Science
Club
Student
Council
X
X
X
X
Student
Publication
Student
Council
Science
Club
Arts Club
X
Debate
Club
Honor
Society
Math Club
Honor
Society
Club
Math Club
3. Eight senior high-school student clubs need to hold meetings on the first day of school.
However, some students belong to more than one of these clubs so clubs that share
members cannot meet at the same time. How many different time slots are required so that
all members can attend all meetings? An “X” in the table below indicates that the two
corresponding clubs share at least one member. Use graph coloring concepts to solve the
minimum number of time slots necessary to ensure that all members can attend all
meetings.
X
X
X
X
X
X
X
X
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7.5Trees
A tree is a mathematical structure which is a type of graph which has the following
properties:
1. undirected;
2. connected (each of the vertices is connected or linked to at least one other vertex); and
3. acyclic (there is only on route from any vertex to any other vertex or has no cycle).
An example of a tree is the Philippine Judiciary Organization Chart as shown in the figure
below.
Some common terminologies related to the graph of trees are illustrated in the
following diagram.
a
root
a
b'
b
internal vertices
b'
b
parent of e
c
d
e
d'
c'
c
d
d'
c'
terminal vertices
child of vertex d
e
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A full m-ary tree is a tree in which all external vertices are at the same time depth or has
exactly m children. Tree T1 in the following figure has exactly two children per vertex and hence
is called a full binary tree. Tree T2 is a full ternary tree because each vertex has three child vertices.
T2
T1
The height of a rooted tree is the number of edges along the longest path from
the root vertex to the farthest child vertex. Given a full m-ary tree T of height h, T has:
𝑣=
i.
𝑚ℎ+1 −1
𝑚−1
vertices;
𝑚ℎ −1
𝑖 = 𝑚−1 internal vertices; and
ii.
𝑡 = 𝑚ℎ terminal vertices.
iii.
The most common application of the tree is determining the number of matches that must
be played to determine the champion in a single elimination tournament.
Example 1:
A local basketball league implements a new ruling of single-elimination wherein a team is
eliminated after a single loss. If there are 16 teams to compete in the tournament, how many
matches must be played to determine the champion?
Solution:
The number of teams represent the terminal vertices so t =16. It is a binary tree since there
are two teams competing in each ball-game, thus m = 2.
Substituting the values of t and m to determine the number of internal vertices, we get:
𝑖=
𝑚ℎ − 1 16 − 1
=
= 15.
𝑚−1
2−1
Thus, there will be 15 matches before a champion will be declared.
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EXERCISES 7.5
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. The NBA League will implement a new game rule of single-elimination in which a team
is to be eliminated after a single loss. If there are 32 teams to compete in the tournament,
how many matches must be played to determine the champion?
2. The PBA League also plans to follow the new NBA ruling of single-elimination. If there
are 2 teams to compete in the tournament, how many matches must be played to determine
the champion? To help the participants understand the scheduling of the games, draw a full
binary tree and explain how your proposed scheduling will work until the champion is
proclaimed.
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Chapter 8: THE MATHEMATICS OF PATTERNS AND
SYMMETRIES
 Learning Objectives
At the end of this chapter, the student is expected to:
 draw the image of a polygon after the reflection and specified rotation;
 use geometric concepts, especially isometries in describing and creating designs;
and
 Apply concepts in geometry for the enrichment of Filipino culture and arts.
Introduction
The Fibonacci sequence of numbers {1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …} was already
discussed in Chapters 1 and 4. Each number in the sequence is the sum of two consecutive numbers
before it. There is an underlying pattern in the sequence since each number is generated by repeated
applicaton of an operation to get the succeeding numbers. This is an example of a numerical
pattern. What about logical patterns? Have you tried taking an IQ test? Patterns may be numerical,
logical, or geometric. This lesson focuses on geometric patterns and in particular, isometries. The
four types of transformations, symmetry, and pattern, tessellation, and fractal geometry will be
discussed in this chapter.
It is ideal to start with the concept of motif. Any artistic creation starts with a motif.
According to Grunbaum and Shephard (1987), a motif is “any non-empty plane set”. Any object
drawn in a plane is a motif. When you repeat the drawing of the fish in the plane not only once,
but several times, you have a pattern. A pattern can be described as “repetition of a ‘motif’ in the
plane” (Grunbaum and Shephard, 1987). An isometry is the rotation of a motif in a fixed angle
about a fixed point. Each rotation of a figure is an isometry. The image of the basic motif under
the additional number of rotations is a pattern (Renee Scott, 2008).
8.1Transformations and Isometries
Transformation
The four types of transformations in the plane are rotation, translation, reflection, and
dilation. Rotation turns a figure about a certain point in a plane. The figure below shows the
rotation of a polygon. The basic motif here is the polygon. Translation slides a figure in any
particular direction or distance. Reflection mirrors a figure over a line. Dilation shrinks or expands
a figure by some scaling factor. Each side of the polygon gets smaller or larger with the same
scale.
Translation
Reflection
Rotation
Dilation
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The motif of the next figure is a bicycle. You can see the bicycle translated, reflected,
rotated, and dilated.
The basic motif on the next figure is an equilateral triangle. A triangle is translated, changed
places from left to right or from down up. The triangle motif also undergoes rotation. Rotating the
equilateral triangle by 60° clockwise, the side that looks like the shorter diagonal of a parallelogram
becomes one side of the rotated triangle. Reflection can be observed along the vertical lines
dividing the panels. Each observed along the vertical lines dividing the panels. Each triangle is
mirrored to the other side of the vertical lines. There is no dilation here since all triangles are of
the same size.
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Translation and reflection can be combined to yield an effect shown below. This
transformation is known as glide reflection. It is a combination of a translation and a reflection.
Glide Reflection
Isometries
There are four transformations but only three of them are isometries. These isometric
transformations are reflection, rotation, and translation. The characteristics of an isometry is that
the original figure and the resulting figure after a transformation are congruent. Dilation is a result
of stretching or shrinking of an object. Hence, the mew figure is no longer congruent to the original
one. This makes dilation not an isometry.
Isometries are also formed from transformations consisting of any combinations of the
three operations. A combined translation and reflection is called glide reflection. Another isometry
is obtained after a reflection is followed by a rotation as shown in the figure below. Here is how
to do this transformation with a triangle. First, draw three circles centered at the rotation point.
Each circle passes through the vertices of the triangle. Rotate each of the three vertices by any
desired angle. Then connect the three rotated vertices which forms the rotated triangle.
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EXERCISES 8.1
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. Work on the following activity and answer the following questions below.
a. Draw the image of the given polygon under a reflection in a mirror line AB.
A
A
B
B
b. Draw the image of the given quadrilateral under the specified rotation:
b.1 Counter clockwise at 90° arount point O.
b.2 Clockwise at 90° arount point O.
B
O
A
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c. Create a tiling design by translating the polygon below.
d. Combine translation and reflection to create a glide reflection of the figure below.
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8.2 Symmetry
There are many objects in nature that are symmetrical. The letter M for instance is
symmetrical, whereas the letter G is not. Your face is symmetrical and, in fact, the human body
also symmetric. The picture of the cathedral below is symmetric. Why is that so? Imagine a vertical
line from the tip of the crucifix to the bottom of the church door. The distance of each point on the
right side of the façade to this imaginary vertical line is exactly the same as the distance of each
point on the left side. The left side and right side wings of the butterfly (see the figure below) is
also symmetrical. The leaves and the eagle as well are symmetries.
In a previous section, it was discussed that the combined isometric transformation of
translation followed by reflection yields a glide reflection. Recall the concept of composition of
functions in Algebra. The composition of a function f and a function 𝑔 is denoted by (𝑓 ∘ 𝑔)(𝑥 ) =
𝑓(𝑔(𝑥 )). Here, the variable x is first applied to the function f. This notion of composition in
Algebra is closely related to the transformation of the figure due to a glide reflection. First, the
figure was translated, and then reflected. This composition of isometries in the plane is called a
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symmetry. Mathematically, it simply means mapping the pattern in the plane back onto itself.
There are three broad types of symmetries. These are the rosette patterns, the frieze patterns,
and the wallpaper patterns. The rosette pattern, has only one reflections and rotation, and has no
translations or glide reflections. The frieze pattern has reflections and rotations. It has reflections
and rotations. It also contains translations and glide reflections but only along one line. The third
type of symmetry is the wallpaper pattern which has rotations, reflections, and glide reflections.
This symmetry group also has translations in two linearly independent directions.
Consider the figure below. One can perform seven rotations about its center point and seven
reflections along some lines passing through the center point. Each of these symmetric
transformations generates a new figure that overlaps with the original figure. It takes seven
rotations of an angle – 51.43° to get the figure back to its original position. For the reflections,
imagine a line between each pair of adjacent figures. These are seven lines for this figure which
determine the seven reflections. This group of seven rotations and seven reflections is called the
symmetric group D7. In general, any symmetric group involving reflections and rotations are called
dihedral group.
Rosette Goups
Consider another symmetry group consisting of 12 rotations. Examine the figure below. Its
center point is located on the center circle. Unlike the case above, any reflection cannot be done
on this object because it will not generate a figure that overlaps with the original figure. Hence,
the figure below has a symmetry of 12 rotations and which is called R12 illustrate the larger class
of rosette group of symmetry
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Frieze Groups
Now consider a symmetry without a center point, and translate the figure to the right. The
figure below has a motif consisting of four pairs of rectangles, each pair of the same size. This
motif completes the figure by moving it to the right or left at a fixed distance. This translation
symmetry belongs to the frieze group of symmetries and is called a frieze pattern. The distance of
translation is minimum. With this restriction there are only seven frieze groups. The other frieze
groups have a combination of translation with rotation and translation with reflection (Eck, n.d.).
The following are Conway’s seven frieze group patterns (“Frieze Patterns”, 2013)
1. Hop. This pattern only involves translation.
2. Step. The second frieze pattern is a combination of translation and reflection shown by
the following figure. Conway also called it glide reflection symmetry.
3. Sidle. The third consists of translation and vertical reflection symmetries.
4. Spinning Hop. The fourth contains translation and rotation (by a half-turn or rotation at
180° angle) symmetries.
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5. Spinning Sidle. The fifth contains translation, glide reflection and rotation (by a halfturn or rotation at 180° angle) symmetries.
6. Jump. The sixth contains translations and horizontal reflection symmetries.
7. Spinning Jump. Finally, the seventh frieze pattern contains all symmetries
(translation, horizontal and vertical reflection, and rotation).
Wallpaper Groups
If translation symmetry is added in a second, independent direction, one gets wallpaper
groups. It turns out that there are only 17 different wallpaper groups (again, considering only
discrete groups). Below are some examples. These were made using the groups p6m, pgg, and
p4m, respectively from the left to right. As always, you have to imagine the patterns extended
infinitely in all directions.
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If a wallpaper group has any rotational symmetry, then the smallest rotational symmetry
must be one of 180°, 120°, 90°, or 60° angle. A wallpaper group can also have reflection
symmetries and glide reflection symmetries. An “m” in the group name indicates a reflection
symmetry, while “g” indicates glide reflection symmetry.
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EXERCISES 8.1
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. The human face is an imperfect symmetry. How would you look if half of your face is the
reflection of the other? Paste it in ta short bond paper.
2. Mark all symmetries for each frieze group pattern. That is, identify and mark all
translations, rotations, reflections and glide-reflections if present.
a.
b.
c.
3. For each of the patterns (a) through (c) shown below, determine the wallpaper symmetry
group exhibited by the pattern. Show in detail how you determined the group.
a
b
c
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8.3Tessellations
A tessellation is a pattern covering a plane by fitting together replicas of the same basic
shape. The word tessellation comes from Latin word tessera, which means a square tablet or a die
used in gambling.
Snake Skin
Honeycomb
Tessellations have been created by nature and man either by accident or
design. Examples range from simple hexagonal pattern of the bees’
honeycomb, snake skin, or a tiled floor to intricate decorations used by the
Moors in 13th century Spain or the elaborate mathematical, but artistic, mosaics
created by Maurits Cornelis Escher in the 20 th century. Although Sumerians
used mosaics as early as 4000 B.C., Escher came to be known as the “Father
of Tessellations”. Some of Escher’s famous tessellations are Horsemen,
Lizard, and Snakes.
Horsemen
Lizard
Maurits Escher
Snakes
In geometry terminology, a tessellation is a pattern resulting from the arrangement of
regular polygons to cover a plane without any gap or overlap. The patterns are continuously
repeated (Scott, 2008).
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Regular Tessellation
A regular tessellation is a tessellation made up of congruent regular polygons. It has the
following properties:
1. The tessellation must tile a floor (that goes on forever) with no overlaps or gaps.
2. The tiles must be the same regular polygons.
3. All vertices must look the same.
Vertex
The three regular tessellations are shown below:
Hexagon
𝟔∙𝟔∙𝟔
Squares
𝟒∙𝟒∙𝟒∙𝟒
Triangles
𝟑∙𝟑∙𝟑∙𝟑∙𝟑∙𝟑
How to name a tessellation?
Step1.
Find the regular polygon with the least number of sides.
Step 2.
Find the longest consecutive run of this polygon, that is, two or more repetitions of
this polygon around the vertex.
Step 3.
Indicate the number of sides of this regular polygon.
Step 4.
Proceeding in a clockwise order, indicate the number of sides of each polygon as
you see them in the arrangement.
Examples:
𝟔 ∙ 𝟔𝟎° = 𝟑𝟔𝟎°
Triangular Tiling
𝟑∙𝟑∙𝟑∙𝟑∙𝟑∙𝟑
or 𝟑𝟔
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𝟗𝟎°
𝟗𝟎°
𝟗𝟎°
𝟗𝟎°
𝟒 ∙ 𝟗𝟎° = 𝟑𝟔𝟎°
𝟑 ∙ 𝟏𝟐𝟎° = 𝟑𝟔𝟎°
Square Tiling
𝟒 ∙ 𝟒 ∙ 𝟒 ∙ 𝟒 or 𝟒𝟒
Hexagonal Tiling
𝟔 ∙ 𝟔 ∙ 𝟔 or 𝟔𝟑
Can other regular polygons tessellate?
Pentagon?
Heptagon?
Octagon?
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Semi-Regular Tessellations
Semi-regular tessellations (or Archimedean tessellations) are regular tessellations of two
or more different polygons around a vertex and each vertex has the same arrangement of polygons.
Vertex
Trihexagonal Tiling
𝟑∙𝟔∙𝟑 ∙𝟔
Vertex
Truncated Square Tiling
𝟒∙𝟖∙𝟖
Vertex
Snub Square Tiling
𝟑∙𝟑∙𝟒∙𝟑∙𝟒
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Demi-Regular Tessellations
A demi-regular tessellation is an edge-to-edge tessellation, but the order or arrangement of
polygons at each vertex is not the same.
p4m, *442
p4g, 4*2
pgg, 2×
p6m, *632
(33.42; 32.4.3.4)1
(33.42; 32.4.3.4)2
(36; 32.62)
p6m, *632
p6m, *632
p6m, *632
p6m, *632
(36; 32.4.3.4)
(3.4.6.4; 32.4.3.4)
(3.4.6.4; 33.42)
(3.4.6.4; 3.42.6)
(3.12.12; 3.4.3.12)
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EXERCISES 8.3
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. Name the following semi-regular tessellations.
a.____________________
b.____________________
c.____________________
d.____________________
e.____________________
f.____________________
2. Name the following demi-regular tessellations.
a.____________________
b.____________________
c.____________________
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8.4Fractals
There are figures that are very interesting, mathematically speaking. These figures can be
found in nature. Take for instance the Barnsley Fern below.
The fern is created by the computer. See how each branch of the leaf is intricately designed.
Each branch becomes smaller and smaller but with a scaling factor. Each point of the fern has an
exact location in the xy plane determined by a function. The function which iterates a figure to
make it smaller and smaller or bigger and bigger using a scaling factor is called fractals.
What are fractals?
Fractals are mathematical constructs characterized by self-similarity. This means that as
one examines finer and finer details of the object, the magnified area is seen to be similar to the
original but is not identical to it. Two objects are self-similar if they can be turned into the same
shape by either stretching or shrinking (and sometimes rotating). The family of ducks below is
similar but not self-similar because the ducklings only look the same without regard to
measurement.
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On the other hand, the school of fish below is self-similar because a uniform stretching and
shrinking made them all the same. Here, the uniform stretching and shrinking is done by a scaling
factor. Self-similar objects do not have beginning or ending, and they form an endless sequence.
A fractal is “a geometric pattern that is repeated at ever smaller
scales to produce irregular shapes and surfaces that cannot be
represented by classical geometry”. It comes from the Latin adjective
“fractus” or verb “frangere” which means to break. Fractal geometry
is a discipline named and popularized by the mathematician Benoit
Mandelbrot (1924-2010). This category of geometry describes a set
of curves many of which were rarely seen before the advent of
computers.
Mandelbrot wrote The Fractal Geometry of nature (1997) and
he started: “Clouds are not spheres, and bark is not smooth, nor does
lightning travel in a straight line.” Some popular fractals below are
Sierpinski triangle, Pascal’s Triangle, Koch snowflakes, fractal trees,
and Barnsley ferns.
Benoit Mandelbrot
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In summary, a fractal is a geometric shape which:
1. is self-similar, and
2. has fractional (fractal) dimension.
Fractals are about four things: fractions, functions, graphs, and imaginary numbers. Fractal
geometry is a useful tool in quantifying the structure of a wide range of objects in nature, from
pure mathematics, through physics and chemistry, to biology and the medical sciences like on
pathology, neuropsychiatry, and cardiology. Some fractals that can be found in nature are the
following:
Fractals offer artists a way to create imaginary landscapes with the help of technological
tools. Many movie backgrounds are created using fractal graphics.
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Iteration
Iteration means repeating a process over and over again. In Mathematics, iteration means
repeating a function over and over. The Iteration Function System (IFS) is a method for generating
fractals involving a large number of calculations of a simple formula. Recursion is a special kind
of iteration. With recursion, there is given starting information and a rule for how to use it to get
new information. Then the rule is repeated using the new information as though it were the starting
information. What comes out of the rule goes back into the rule for the next iteration. A classic
example of a recursion is the Fibonacci sequence.
The scaling factor is a fraction, with a value of less than 0.1, used to specify the distance
from one plotted point to the next plotted point relative to the distance from the original plotted
point to one of the fixed points. The scaling factor ultimately governs how diffused or focused the
resulting fractal pattern will be. The fractal picture below illustrates the notion of iterating a
geometric construction.
Example 1: The Cantor Set
The cantor set is a fractal that can be formed using IFS. The Cantor set is formed by the
following algorithm.
Step 1:
Begin with the set [0,1].
Step 2:
Divide the existing segments into thirds.
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Step 3:
Remove the middle third.
Step 4:
Go to Step 2.
Dimensions
In Euclidean geometry, a one-dimensional line segment has only one length, a twodimensional triangle covers an area in a plane, and a three-dimensional pyramid occupies a volume
in space. A line segment is one-dimensional, a triangle or square is two-dimensional, and a pyramid
or cube is three-dimensional. Intuitively, dimension has something to with the number of distance
measurements needed to specify the size of an object in the Euclidean world.
For fractal objects such as, the Cantor set, the Sierpinski triangle, among others, the
dimension cannot be determined by simply counting the number of distance measurements. What
is the dimension of a fractal object that is fractured and scattered in space? Many resort to a
definition of dimension based on the concept of capacity, that is, how much space on object
actually takes up in reality. First, the capacity definition is applied to a line, triangle, and cube to
recover the Euclidean dimensions 1, 2, and 3, respectively. It is then found that the fractal
dimension d is not necessarily a whole integer but can be take on any value between the integers.
The formula for the dimension of a fractal is:
log 𝑛
𝑑=
1
log
𝑟
where:
r = ratio of the length of the new object to the length of the original object
n = the number of the new objects
Example 2: Sierpinski Triangle or Sierpinski Gasket
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One classic example of self-similarity is Sierpinski triangle or Sierpinski gasket. Let an
equilateral triangle be decomposed into three congruent figures, each of which is exactly half the
size of the original triangle. If any of the three smaller pieces is magnified by a factor of two, an
exact replica of the original triangle is obtained. That is, the original triangle consists of three selfsimilar copies of itslelf, each with the magnification factor of two.
The Sierpinski triangle can be constructed as follows:
Step 1: Begin with an equilateral triangle (although the actual shape does not really matter).
Step 2: Find the midpoint of each side.
Step 3: Connect the midpoints by a straight line.
Step 4: Observe that you created three or more triangles, one on top and two at the bottom.
The middle triangle is hollow.
Step 5: Repeat the process with all three triangles.
First Iteration
Observe the two triangles above. The first is the original equilateral triangle with sides
measuring one unit each. The recursive procedure is to replace the triangle with three smaller
triangle shares a vertex with the large triangle. Repeat the procedure.
Note that in the first iteration, the triangle has 3 “miniature” triangles. Each side is half the
length of a side of the original triangle. Each “miniature” triangle looks exactly like the original
triangle when magnified by a scaling factor or magnification.
Take the result and repeat (iterate).
Second Iteration
Third Iteration
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Notice that the lower left portion of the triangle is exactly the same as the entire triangle
when magnified by a factor of two. It is self-similar.
Fourth Iteration
Fifth Iteration
Thus, the Sierpinski triangle or gasket begins as an equilateral triangle, with each side
as one unit but as the recursive procedure (replacing the triangle with three smaller congruent
equilateral triangles such that each smaller triangle shares a vertex with the large triangle)
continues without end, the area of the triangle converges to zero. The fifth iteration in the figure
above shows a very large number of iterations. Could you believe that the Sierpinski triangle
does not have an area?
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EXERCISES 8.4
Name: ______________________________________________Score: ____________________
Section: _____________________________________________Date: _____________________
1. Create a fractal by starting with a line segment, dividing the segment into four equal
lengths, and replacing both the second and the third sections by two segments whose
lengths are one-fourth the length of the original segment. Repeat this process to four
iterations. Find the dimension of this fractal.
2. Create a fractal by starting with a square, dividing each line segment into three equal
lengths, and replacing the middle third of each side with three line segments whose lengths
are one-third the length of the original segment. This is the first iteration. Repeat this
process and draw the next iteration of this fractal. Find the dimension of this fractal.
3. Draw the first to fourth iterations of the Sierpinski triangle and complete the table below
by finding:
a. The number of new triangles drawn at each stage.
b. The length of each side of the triangle drawn at indicated iteration.
c. The area of each new triangle.
d. The total area of all triangles.
Iteration
Number of
Triangles
Length of
Sides
Area of Each
Triangles
Area of All
Triangles
First
Second
Third
Fourth
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Artmann, B. (2018). Euclidean Geometry. In Encyclopedia Britannica. Retrieved from
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Aufmann,
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(2018).
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Mathematical
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Bayer,
D.
(2003).
The
Seventeen
Wallpaper
Patterns.
http://www.math.columbia.edu/~buyer/symmetry/wallpaper
Retrieved
from
Connors, M. A. (2018). Exploring Fractals: From Cantor dust to the Fractal Skewed Web.
Department of Mathematics and Statistics, University of Massachusetts Amhest. Retrieved
from http://www.math.umass.edu/~mconnors/fractal.html
Coolmath.com. (2018). What are tessellations? Retrieved from http://www.coolmath.com/lessontesellations-1
Grunbaum, B. &Shephard, G.C. (1987). Tilings and Patterns. New York, NY: W. H. Freeman and
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