Real Numbers
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chapter
Real Numbers
Learning Outcomes
In this chapter you will learn:
ÂÂ How to define the real numbers
ÂÂ About factors, multiples and prime factors
ÂÂ How to write a whole number as a product of prime factors
ÂÂ The properties of integers
ÂÂ The properties of rational numbers
ÂÂ About rounding and significant figures
__
__
ÂÂ How to geometrically construct √
2 and √ 3
ÂÂ About orders of magnitude and scientific notation
1
1
Real Numbers
YOU SHOULD
REMEMBER...
–8 –7 –6 –5 –4 –3 –2 –1
0
1
2
3
4
5
6
7
8
This is not a very rigorous definition of the real numbers. However, it will
serve our purposes. The discovery of a proper rigorous definition of the real
numbers was one of the most important developments in the mathematics
of the nineteenth century. The main contributors to the field were a French
mathematician, Augustin-Louis Cauchy (1789–1857), and two German
mathematicians, Richard Dedekind (1831–1916) and Karl Weierstrass
(1815–1897).
KEY WORDS
Augustin-Louis
Cauchy (1789–1857)
Richard Dedekind
(1831–1916)
Karl Weierstrass
(1815–1897)
The natural numbers, the integers, the rational numbers and
the irrational numbers are all subsets of the real number system.
1.1 Factors, Multiples
and Prim e Factors
Natural Numbers
Rationals
The natural numbers are the ordinary counting numbers.
The set of natural numbers is an infinite set. This means that
the set is never-ending. The letter N is used to label the set
of natural numbers.
Reals
Irrationals
Integers
Naturals
Factors
For example, all the factors of 24 are {1, 2, 3, 4, 6, 8, 12, 24}.
As you can see, 1 is a factor of 24 and 24 is a factor of 24.
2
1
Multiples
Real Numbers
multiple
The multiples of 6 are {6, 12, 18, 24, 30, 36 ...}.
As you can see, the set of multiples is an infinite set, i.e. it goes on forever.
Prime Numbers
Prime numbers
7 is a prime number as it has two factors only: 1 and 7.
2 is the only even prime number. Its two factors are 1 and 2.
11 is the first two-digit prime. Its two factors are 1 and 11.
1 is not a prime as it has one factor only, itself.
0 is not a prime as it has an infinite number of factors, and it is not a natural number.
s
co mp os ite nu mb er
Euclid
The Fundamental Theorem of Arithmetic
The Fundamental Theorem of Arithmetic is an important result that shows that the primes are the
building blocks of the natural numbers.
3
Real Numbers
1
The highest common factor of 12 and 20 is 4, as 4 is the largest natural number that divides evenly
into both 12 and 20.
2
1.2
1, 1.
1.1,
ACTIVITies 1.
The lowest common multiple of 3 and 4 is 12, as 12 is the
smallest number that both 3 and 4 divide evenly into.
In Activities 1.1 and 1.2 you discovered how to find the HCF and LCM of two natural numbers,
by writing each number as a product of primes.
R
Wo rke d E x am p l e 1 . 1
Express 240 as a product of prime numbers.
Solution
2 240
Start with the lowest prime
that is a factor.
2 120
Wo rke d E x a m p le 1.2
R
Find (i) the HCF and (ii) the LCM of 512
and 280.
Solution
Express both numbers as a product of primes.
2 60
2 512
2 280
2 30
2 256
2 140
3 15
2 128
2 70
5 5
2 64
5 35
1
2 32
7 7
240 = 2 × 2 × 2 × 2 × 3 × 5
2 16
1
∴ 240 = 2
4 × 3 × 5
2 8
2 4
2 2
1
3 × 5 × 7
512 = 2
9and 280 = 2
4
(i) HCF (512, 280) = 2
3 = 8
(ii) LCM (512, 280) = 29 × 5 × 7 = 17,920
1
Wo rke d Ex am p l e 1 . 3
R
Real Numbers
Show that if p is a prime number and p divides evenly into r 2, then p divides evenly into r.
Solution
Let r = r 1a 1r 2a 2r 3a 3... r na n, where each rî is prime and each aî ∈ N.
Therefore, using the rules of indices:
r 2 = (r 1a 1r 2a 2r 3a 3... r na n)2
= r 12 a 1r 22 a 2r 32 a 3... r n2 a n
If p | r2 (p divides r2), then p | one of rî2aî.
Since each rî is prime, ⇒ p | one of rî.
Hence, p divides r 1a 1r 2a 2r 3a 3... r na n= r.
Wo rke d Ex am p l e 1 . 4
R
Periodical cicadas are insects with very long larval periods and brief adult lives.
For each species of periodical cicada with larval period of 17 years, there is a
similar species with larval period of 13 years.
If both the 17-year and 13-year species emerged in a particular location in 2011,
when will they next both emerge in that location?
Solution
Here we are looking for the LCM of 17 and 13. As both numbers are prime, the LCM of 13 and 17 is
17 × 13 = 221.
Therefore, both species will next appear together in that location in the year 2232.
Wo rke d Ex am p l e 1 . 5
R
Evaluate (i) 5! and (ii) 5 × 4!
Solution
(i)
5! = 5 × 4 × 3 × 2 × 1
∴ 5! = 120
(ii) 5 × 4! = 5 × (4 × 3 × 2 × 1)
= 120
5
1
Infinitude of Primes
Real Numbers
There are infinitely many prime numbers.
In the next activity you will discover a proof,
by contradiction, that this statement is correct.
3
1.3
ACTIVITy 1.
Exercise 1.1
1. Express the following numbers as a product of
prime factors:
(i) 160 (ii) 273 (vi) 1,870 (ix) 1,224
(iii) 128 (vii) 10,500 (x) 38,016
(iv) 368
(b) Hence, find the LCM and HCF for each set
of numbers.
(i) 102 and 170
(ii) 117 and 130 (vii) 69 and 123
(iii) 368 and 621 (viii) 20, 30 and 60
(iv) 58 and 174
(v) 60 and 765
(vi) 123 and 615
(ix) 8, 10 and 20
(x) 294, 252 and 210
3. Let n be a natural number.
What is the HCF of:
(i) n and 2n (ii) n and n2
334,6112,
4. If 3 divides
divide 334,611.
explain why 3 must also
5. r is an even natural number and s is an odd
natural number. The prime factorisations of r
and s are:
r = r 1a 1r 2a 2r 3a 3... r na n
s = s 1q 1s 2q 2s 3q 3... s mq m
6. Let u be any natural number with prime
factorisation,
u = u 1f 1u 2f 2u 3f 3... u nf n
(v) 1,155 (viii) 102
2. (a) Express each of the following pairs of
numbers as the product of prime factors.
6
Proof by contradiction
(i) Explain why one of the primes,
r 1, r 2, r 3, ... r nmust be 2.
(ii) Explain why none of the primes,
s 1, s 2, s 3, ... s mis 2.
(i) Explain why none of the primes,
u 1, u 2, u 3, ... u ndivide u + 1.
(ii) Hence, write down the HCF of u and
u + 1.
7. Let u be any natural number with prime
factorisation:
u = u 1f 1u 2f 2u 3f 3... u nf n
(i) If u is even, then find the HCF of u and
u + 2.
(ii) If u is odd, then what is the HCF of u and
u + 2.
8. Kate has two pieces of material. One piece
is 72 cm wide and the other piece is 90 cm
wide. She wants to cut both pieces into strips
of equal width that are as wide as possible.
How wide should she cut the strips?
9. Tom exercises every 14 days and Katie every
nine days. Tom and Katie both exercised
on March 12. On what date will they both
exercise together again?
10. Ms Hoover has 160 crayons and 30 colouring
books to give to her students. If each student
gets an equal number of crayons and an equal
number of colouring books what is the largest
number of students she can have in her class?
11. Bart is making a board game with dimensions of
16 cm by 25 cm. He wants to use square tiles.
What are the dimensions of the largest tile he
can use?
13. The Ulam numbers, u
n, n ∈ N, are defined as
follows:
(i) Find F(1) and F(2).
(ii) Explain why F(n) is always a natural
number.
(iii)Consider the sequence:
n, F(n), F(F(n), F(F(F(n))) ....
Construct the first 15 terms of the
sequence for each of the following values
of n:
6, 10, 15, 32 and 17.
1= 1,
u
u 2= 2
(iv) Based on your results make a conjecture
about the sequence.
and each successive natural number, m,
where m > 2, u mis an Ulam number if and
only if it can be written uniquely as the sum
of two distinct Ulam numbers.
(v) Test your conjecture for n = 39.
(i) Explain the words in bold.
(i) How many terms are in the sequence?
(ii) Why is 5 not an Ulam number?
(ii) Explain why none of the terms are prime.
(iii) Find the first 10 Ulam numbers.
(iii) Construct a sequence of 20
consecutive natural numbers none of
which are prime.
(iv) Is it possible to have a set of consecutive
natural numbers of any given size, that
does not contain any prime numbers?
Explain.
14. Let n be a natural number. Define the
function, F(n) as follows:
n
if n is even.
F(n) = __
2
3n + 1
F(n) = _______
if n is odd.
2
1
Real Numbers
12. Beginning on Monday of each week and
running until Friday, The Breakfast Show gives
away €100 to every 100th
caller who gets
through to the show. During the week before
a Saturday night concert, the show offers two
free tickets to the concert for every 70th
caller.
How many callers must get through before
one wins the tickets and the €100?
15. Consider the finite sequence:
10! + 2, 10! + 3, 10! + 4, ... 10! + 9, 10! + 10
1.2 Integers and R ational Num bers
The integers are made up of zero and all the positive and negative whole numbers. Mathematicians use
the letter Z to represent the set of integers.
Properties of Integers
a + b and a × b are integers whenever a and b are integers. (Closure property)
a + b = b + a and a × b = b × a (Commutative properties)
(a + b) + c = a + (b + c) and (a × b) × c = a × (b × c) (Associative properties)
a × (b + c) = (a × b) + (a × c) (Distributive property)
a + 0 = a and a × 1 = a for all integers a. (Identity elements)
For every integer a, there exists an integer –a, such that a + (–a) = 0.
We say –a is the additive inverse of a. (Additive inverse)
7
1
Rational Numbers
Real Numbers
The letter Q is used to represent the set of rational numbers.
Rational numbers are also called fractions.
All integers can be written in the form _ba
and hence are rational numbers.
For example, 3 = _31 .
The properties listed for the integers also carry over into the rationals.
Two fractions are equivalent if they have the same value. For example, _12 = _ 36 .
The reciprocal of a fraction is found by turning the fraction upside down (inverting).
__
11
12
For example, the reciprocal of __
is .
12
11
Every fraction, with the exception of zero, has a multiplicative inverse. The product of a number
and its multiplicative inverse is always 1. The multiplicative inverse of a fraction is its reciprocal.
For example, the multiplicative inverse of _ 34 is _ 43 .
Wo rke d Ex am p l e 1 . 6
R
Use the properties of the integers to prove that for any integer a × 0 = 0.
Solution
0 + 0 = 0 (Identity)
⇒ a × (0 + 0) = a × 0 (Multiplying both sides by a)
⇒ a × 0 + a × 0 = a × 0 (Distributive property)
a × 0 + a × 0 + (–a × 0) = a × 0 + (–a × 0) ( Adding –a × 0 to both sides)
a × 0 + [a × 0 + (–a × 0)]= a × 0 + (–a × 0) (Associative property)
a × 0 + 0 = 0 (Additive inverse)
∴ a × 0 = 0 (Identity)
R
Wo rke d Ex am p l e 1 . 7
R
Wo rke d E x a m p le 1.8
Use the properties of the integers to prove that
for all integers a and b, (a)(–b) = –ab.
Use the properties of the integers to prove that for
all integers a and b, (–a)(–b) = ab.
Solution
Solution
(a)(–b) + (a)(b) = a(–b + b) (Distributive
property)
(–a)(–b) + (–a)(b) =
–a(–b + b) (Distributive
property)
= a(0) (Additive inverse)
∴ (a)(–b) + (a)(b) = 0
⇒ (–a)(–b) + (–a)(b) = 0
Therefore, (a)(–b) is the additive inverse
of (a)(b).
Therefore, (–a)(–b) is the additive inverse of (–a)(b).
But (a)(b) = ab and its additive inverse is –ab.
∴ (a)(–b) = –ab
8
= –a(0) (Additive inverse)
But (–a)(b) = –ab and its additive inverse is ab.
∴ (–a)(–b) = ab
1
Wo rke d Ex am p l e 1 . 9
R
Real Numbers
(i) Find the sum of the squares of the four consecutive integers, –2, –1, 0 and 1.
(ii) Show that the sum of the squares of any four consecutive integers is always an even number.
Solution
(i) (–2)2 + (–1)2 + (0)2 + (1)2 = 4 + 1 + 0 + 1
=6
(ii) n, n + 1, n + 2, and n + 3, n ∈ Z, are four consecutive integers.
∴ (n)2 +(n + 1)2 +(n + 2)2 +(n + 3)2 = n2 + n2 + 2n +1 + n
2+ 4n + 4 + n
2+ 6n + 9
= 4n2+ 12n + 14
= 2(2n2 + 6n + 7) [an even number as 2 is a factor]
∴ The sum of any four consecutive integers is an even number.
Exercise 1.2
5. Use the properties of the integers to prove
that –5 × 8 = –(5 × 8).
1. Evaluate each of the following:
(i) –8 – 5 + 13
(iii) (–5)(–8)
(ii) (2)(–3)
(iv) 3(4)2 + 2(4) – 56
3(5 – 2)2 – 3(4 – 2)2 + 5(3)3
(v) ________________________
(5 – 2)2
2. Evaluate each of the following, leaving your
answers in their simplest form:
3
5
2 2 __
1 __ × 1 __ + __
×
4
3 3 6
14 __
___
________________
3
(vi)
(i) ×
1
15 8
7 __ – 4
2
8.4(19.6 – 12.2)2
1
(ii) 2 __ × 5
(vii) _______________
2
(14.4 – 12.2)3
3
__
2
(iii) 1 × 1 __
1 1
(iv) __ – __
4
3
3 13
__
– ___
5 20
______
(viii)
5
__
8
(ix) 3(2.5 – 1.2)2
2 8
5
2 __
__
(v) +
3 6
6. A box contains oranges and
grapes. An equal number of
the oranges and grapes are
rotten. _ 32 of all the oranges
are rotten and _ 43 of all the
grapes are rotten.
What fraction of the total number
of pieces of fruit in the box are rotten?
7. Alice and Bob share an allotment. The ratio of
the area of Alice’s portion to the area of Bob’s
portion is 3 : 2. They each grow vegetables
and fruit on the allotment.The entire allotment
is covered by vegetables and fruit in the ratio
7 : 3. On Alice’s portion of the allotment, the
ratio of vegetables to fruit is 4 : 1. What is the
ratio of vegetables to fruit in Bob’s portion?
8. Three cans of juice fill _ 23 of a 1 litre jug.
How many cans of juice are are needed to
completely fill eight 1 litre jugs?
3. Use the properties of the integers to prove
that 4 × –5 = –(4 × 5).
9. A rectangle has a length of _ 35 units and an area
of _13 units2. What is the width of the rectangle?
4. Use the properties of the integers to prove
that –3 × –6 = (3 × 6).
10. In the diagram, the number line is marked at consecutive integers, but the numbers themselves are
not shown. The four large red dots represent two numbers that are multiples of 3 and two numbers
that are multiples of 5. Which of the black points represents a number that is a multiple of 15?
Give an explanation for your choice.
+
A
+
B
+
+
P
+
J
+
Q
+
G
+
R
+
+
C
+
+
D
+
+
+
9
Real Numbers
1
11. The integers from 1 to 9 are listed on a whiteboard. If an additional m 8s and n 9s are added to the
list, then the average of all the numbers is 7.3. Find m + n.
12. Five square tiles are shown. Each tile has a side of integer length. The side lengths can be arranged
as consecutive integers. The sum of the areas of the five squares is 1,815.
(i) Show that the sum of the
squares of five consecutive
integers is divisible by 5.
(ii) Find the dimensions of
the largest square.
13. Seán has a pile of tiles, each measuring 1 cm by 1 cm. He tries to put these small tiles together to
form a larger square of length n cm, but finds that he has 92 tiles left over. If he had increased the
side length to (n + 2) cm, then he would have been 100 tiles short. How many tiles does Seán have?
14. A palindromic number is a positive integer that is the same when read forwards or backwards.
For example, 31213 and 1237321 are palindromic numbers.
(i) Find the total number of three digit palindromic numbers.
(ii) Determine the total number of palindromic numbers between 106 and 107 .
(iii) If the palindromic numbers in part (ii) are written in order, find the 2,125th
number on the list.
1.3 Irr ational Num bers
In the right-angled triangle shown, the value for x can be found
using the theorem of Pythagoras. Here is the solution:
x2=
12 +
1
12
x
x 2= 1 + 1
x 2= 2
1
__
x=√
2 (as x > 0)
__
Can √
2 be written as a fraction? This problem preoccupied the ancient
Greek mathematicians for many
__ years. Around 500 bc, Hippasus, a follower
of Pythagoras, proved that √
2 could not be written as a fraction. Pythagoras,
who did not believe in the existence of irrational numbers, was so enraged by
this proof that he had Hippasus thrown overboard from a ship and Hippasus
subsequently drowned.
__ Numbers that cannot be written as fractions are called
irrational numbers. √
2 was the first known irrational number.
Hippasus, a follower
of Pythagoras
__
While √
2 cannot be written__as a fraction, it is possible to
find an approximation
for √
2 . A calculator gives the
__
approximation √ 2 = 1.414213562, but this decimal
goes on forever with no pattern or repetition.
10
Real Number System
1
__
Proof That
3 Is Irrational
√
__
To prove: √ 3 is irrational.
__
Proof: Assume that √
3 is rational and can therefore be written in the form _ ba , a, b ∈ Z, b ≠ 0.
Also, assume that the fraction _ba is written in simplest terms, i.e. HCF(a, b) = 1.
__
√ 3 = _ ba ,
2
⇒ 3 = __
a2 (squaring both sides)
b
∴ a2= 3b2 (*)
As b2 is an integer, a2 has to be a multiple of 3, which means that 3 divides a 2.
If 3 divides a2 , then 3 divides a. (Worked Example 1.3)
∴ a = 3k, for some integer k. Substituting 3k for a in (*) gives,
Real Numbers
The proof of this result is another example of proof by contradiction.
(3k)2 = 3b2
9k2 = 3b2
⇒ b2 = 3k2
As k2 is an integer, b2 has to be a multiple of 3, which means that 3 divides b
2.
herefore, 3 divides b. If 3 divides a and 3 divides b, then this contradicts the assumption that
T
HCF(a, b) = 1. This completes the proof.
__
__
Constructing
2 and √ 3
√
__
__
√ 2 and √
3 cannot be written as fractions, but can be constructed.
__
Construct √
2
.Let the line segment AB be of length 1 unit.
1
4.Mark the intersection, C, of the circle and m.
C
m
B
A
2.Construct a line m perpendicular to [AB] at B.
m
A
B
B
A
5.Draw the line
__ segment CA.
|AC| = √
2
C
m
3. Construct a circle with centre B and radius [AB].
m
A
A
B
B
11
1
Proof: |AB| = |BC| = 1 (radii of circle)
Real Numbers
|AB|2 + |BC|2 = |AC|2 (Theorem of Pythagoras)
2 + 12 = |AC|2
1
|AC|2 = 2
∴ |AC| = √
2
__
__
Construct √
3
1.Let the line segment AB be of length 1 unit.
B
A
5.Draw the line
__ segment [CD].
|CD| = √
3
C
2.Construct a circle with centre A and radius
length |AB|.
E
A
B
B
A
D
3.Construct a circle with centre B and radius
length |AB|.
Proof: CD is the perpendicular bisector of [AB]
(Construction).
1
∴ |AE| = |EB| = __
2
|AC| = |BC| = 1 (Construction)
B
A
4.Mark the intersection of the two circles as C
and D.
C
B
A
D
12
|AE|2 + |EC|2 = |AC|2(Theorem of
Pythagoras)
( )
12
__
+ |EC|2 = 12
2
1 3
|EC|2= 1 – __
= __
4 4
__
__
√4
3
√
3
∴ |EC| = __
= ___
|CD| = 2 |EC|
3
√
= 2 ___
2
( )
__
__
2
⇒ |CD| = √
3
1
Exercise 1.3
__
1
2
1
90º
1.5 90º
1
1
90º
1
0.5
0
–1.5 –1 –0.5
0 0.5
–0.5
90º
1
1.5
2
2.5
Real Numbers
1. Copy the diagram and use it to show √
5 on the numberline.
__
2. Using a diagram similar to that used in Question 1, show √
7 on the numberline.
__
3. Prove by contradiction that √
2 is irrational.
4. p, the ratio of the circumference of a circle to its diameter, is also an irrational number. The British
mathematician John Wallis (1616–1703) discovered the following formula for approximating p:
(2n)2
22
62
42
× _____
× ... × ______________
= _____
× _____
2 (1)(3) (3)(5) (5)(7)
(2n – 1)(2n + 1)
p
__
(i) Use the first seven terms of the product to approximate p to four decimal places.
(ii) Use your calculator to approximate p to four decimal places.
(iii) Assuming that your calculator has given the true approximation to four decimal places, calculate
the percentage error in using seven terms of Wallis’ formula (answer to two decimal places).
________
__
_______
__
2
5. Show that x = √
3 + 2√ 2
–√
3 – 2√ 2
is a rational number. ( Hint: find x . )
6. Two quantities are in the golden ratio, if the ratio of the sum of the quantities to the larger quantity
is equal to the ratio of the larger quantity to the smaller one. In the rectangle below a > b and
a + b __
a
_____
= ; therefore, the ratio a : b is the golden ratio.
a
b
This ratio is a constant and also irrational. It can be shown that
a
b
the ratio is the positive solution to the equation:
1
r = 1 + __
r
a
(i) Solve the equation to find the golden ratio.
a+b
(ii) Use your calculator to approximate the ratio to four
decimal places.
1.4 Rounding and Significant Figures
Rounding to Decimal Places
In geometry, the number of times the diameter of a circle divides into the circumference is called p.
We normally substitute 3.14 for p in these calculations. However, 3.14 is just an approximation:
There are infinitely many decimal places in p.
p is 3.141592654 to nine decimal places.
For simplicity, we often write p as 3.14, i.e. to two decimal places.
Engineers use 3.1416 as an approximation for p.
13
Wo rke d E x am p l e 1 . 1 0
R
Write the following correct to one decimal place:
Real Numbers
1
(i) 2.57
(ii) 39.32
Solution
(i) 2.57
When rounding to one decimal place, we look at the second number after the decimal
point. If this number is 5 or greater, we round up to 2.6. Otherwise, the corrected answer is
2.5. As 7 is the second number after the decimal point, we round up to 2.6.
Answer = 2.6
(ii) 39.32
Here, the second number after the decimal point is 2, which is less than 5. Therefore, the
number rounded to one decimal place is 39.3.
Answer = 39.3
Significant Figures
Wo rke d Ex am p l e 1 . 1 1
R
Correct the following numbers to two significant figures:
(i) 3.67765
(ii) 61,343
(iii) 0.00356
Solution
(i) 3.67765
The first significant figure in a number is the
first non-zero digit in the number. In this number,
3 is the first significant figure in the number.
We need to correct to two significant figures, so we
look at the third significant digit. If this number
is 5 or greater, we round up the second digit.
The third digit is 7, so the corrected number is 3.7.
14
3.6|7765
1st sig. figure
2nd sig. figure
Answer = 3.7
(ii) 61,343
61,|343
1st sig. figure
Answer = 61,000
Note that all other digits after the rounded digit change
to zero.
2nd sig. figure
(iii) 0.00356
0.0035|6
The third significant digit is 6. Therefore, the
rounded number is 0.0036.
1st sig. figure
2nd sig. figure
Answer = 0.0036
Real Numbers
Here, the third digit is 3, which is less than 5.
Therefore, the rounded number is 61,000.
1
1.5 Orders of Magnitude
and Scientific Notation
When doing calculations, scientists often use very large numbers or very small numbers. For example,
the speed of light is about 300,000,000 metres per second.
Very large or very small numbers can be awkward to write down. So, scientists use scientific notation to
write down these numbers.
scientific notation
Wo rke d Ex a m p l e 1 . 1 2
R
Write the following numbers in scientific notation:
(i) 725,000,000,000
(iii) 0.0000056
(ii) 980,000
(iv) 0.000000034
Solution
(i)First, note that dividing a number N by 10n, where n ∈ N, moves the decimal point n places to
the left.
144.25
For example, _______
= 1.4425 (Decimal point moves two places to the left)
102
725,000,000,000
725,000,000,000 = _______________
× 1011
1011
= 7.25 × 1011
15
1
980,000
(ii) 980,000 = ________
× 105
105
= 9.8 × 105
Real Numbers
(iii)Note that dividing a number N by 10n, where n is a negative integer, moves the decimal n places
to the right.
0.00146
1
For example, ________
= 0.00146 × ____
–3
10–3
10
= 0.00146 × 103 (Rules of indices)
= 1.46 (Decimal point moves three places to the right)
0.0000056
0.0000056 = __________
× 10–6
10–6
= 5.6 × 10–6
0.000000034
(iv) 0.000000034 = ____________
× 10–8
10–8
= 3.4 × 10–8
Orders of Magnitude
Orders of magnitude are generally used to make very approximate comparisons. If two numbers differ
by one order of magnitude, one is about ten times larger than the other.
order of magnitude
Wo rke d Ex am p l e 1 . 1 3
R
By how many orders of magnitude does 345,632 differ from 567,123,423?
Solution
Write both numbers in scientific notation:
345,632 = 3.45632 × 105
567,123,423 = 5.67123423 × 108
≈ 1 × 105
≈ 10 × 108
= 100 × 105
= 101 × 108
= 105
= 109
109
____
5 = 109 – 5 = 104
16
10
Therefore, both numbers differ by four orders of magnitude.
1
Exercise 1.4
5.Calculate each of the following, giving your
answers as decimal numbers:
(i) 5.1456
(iv) 62.1235321
(i) 3.4 × 103 + 2.8 × 10 3
(ii) 7.2983
(v) 23.7654
(ii) 5.2 × 109 + 3.5 × 10 9
(iii) 17.8943
(vi) 0.07893
6.The following numbers are written in
scientific notation. Rewrite the numbers in
ordinary form.
(b)Write these numbers correct to two
decimal places:
(i) 1.263
(iv) 21.3
(i) 2 × 106
(iv) 6.47 × 105
(ii) 5.9876
(v) 22
(ii) 1.69 × 104
(v) 6.12 × 101
(iii) 21.456
(vi) 0.00391
(iii) 2.48 × 103
(vi) 9.43 × 10 5
2. (a)Write these numbers correct to two
significant figures:
7.Write the following in the form a × 10n,
where 1 ≤ a < 10, a ∈ R, n ∈ Z:
(i) 0.00985
(vi) 0.000849
(i) 0.000036
(iv) 0.00063
(ii) 0.00234
(vii) 0.238
(ii) 0.0005613
(v) 0.0078
(iii) 0.0125
(viii) 52.00285
(iii) 0.0345
(vi) 0.0011
(iv) 0.000000785
(ix) 52.487
(v) 1.000034
(x) 967,333
(b)Write these numbers correct to one
significant figure:
8.Write the following numbers in the form
a × 10n, where 1 ≤ a < 10:
(i) 0.00068
(iv) 0.0000000097
(ii) 0.0000328
(v) 0.00000056
(iii) 0.0657
(vi) 0.0030307
(i) 32.14
(iv) 1,698
(ii) 3.857
(v) 5,965
(iii) 19,345
(vi) 999
9. The following numbers are written in
scientific notation. Rewrite the numbers in
ordinary form.
3. Write these numbers in scientific notation:
(i) 34,000,000
(vi) 0.000032
(ii) 0.25
(vii) 5,000,000
(iii) 4,570
(viii) 0.6464
(iv) 0.0001258
(ix) 532,600
(v) 7,206
(x) 5,000
4. Write these as decimal numbers:
Real Numbers
1. (a)Write these numbers correct to three
decimal places:
(i) 1.5 × 10 –3
(iii) 3.5 × 10 –5
(ii) 2.54 × 10 –4 (iv) 6.67 × 10 –6
10. By how many orders of magnitude do the
following numbers differ:
(i) 868, 932, 145 and 284
(ii) 453, 987, 312 and 3548
(iii) 767, 894, 567,000 and 23,000,000
(i) 2.65 × 102
(vi) 4 × 10 –2
(iv) 0.1 and 0.00042
(ii) 4.53 × 10 –3
(vii) 2.64 × 107
(v) 1.8 and 234
(iii) 7.2 × 106
(viii) 7.612 × 103
(iv) 1.7 × 10 –5
(ix) 2.76 × 108
(v) 3 × 10 2
(x) 3.02 × 10 –9
17
Real Numbers
1
Revision Exercises
1. The German mathematician Christian
Goldbach conjectured that every odd positive
integer greater than 5 is the sum of three
primes. Verify this conjecture for each of the
following odd integers:
9.Write the following numbers correct to two
significant places:
(i) 11 (ii) 33 (iii) 97 (iv) 199 (v) 17
(i) 852,233
(iv) 0.000054
(ii) 0.134 (v) 652,494
(iii) 2.00062
(vi) 0.000814
2. Two bikers are riding in a circular path.
The first rider completes a circuit in 12 minutes.
The second rider completes a circuit in
18 minutes. They both started at the same
place and at the same time and go in the same
direction. After how many minutes will they
meet again at the starting point?
10.Given a line segment of length one unit, show
clearly how
__ to construct a line segment of
length √
2 units.
3. Use the properties of integers to prove the
following:
12. (a) (i)Find the values of the primes p and
q, if p3 × 13 × q = 1,768.
(ii)Find the values of the primes m and n,
if 24× m × n = 3,192.
(iii)Using your answers to parts (i) and
(ii), find the HCF and LCM of 1,768
and 3,192.
(i) 5 × –3 = –15
(ii) –5 × –3 = 15
4. Show that no integer of the form n3 + 1 is
prime, other than 2 = 1
3+ 1.
5. (i) List the smallest five consecutive integers
that are composite.
(ii)Find 10,000 consecutive integers that are
composite.
6.Express each of the following numbers as the
product of prime factors, and hence, find the
LCM and HCF of each pair:
(i) 68 and 102
(iii) 104 and 351
(ii) 69 and 123
(iv) 123 and 615
11. Given a line segment of length one unit, show
clearly how
__ to construct a line segment of
length √
3 units.
(b)Evaluate the following, giving your answer
in scientific notation:
3
1
2
3 __ × 2__
+ 1 ___
7
5
10
_______________
3 × 104
(
13. (i)P represents a pointer on a gauge.
What is the decimal value shown by
the pointer?
P
7. (a) What fraction when added to _ 14 gives _ 13 ?
(b)A mathematician states that her children’s
ages are all prime numbers that multiply
together to give 7,429. She also says that
two of her children are teenagers.
(i) How many children does she have?
(ii) What are their ages?
8. A palindromic number is a number that reads
the same forwards and backwards. For example,
52,325 is a palindromic number. All four-digit
palindromic numbers have 11 as a prime factor.
18
(i)Find the prime factorisations of the
palindromic numbers 2,332 and 6,776.
(ii)Hence, find the HCF and LCM of 2,332
and 6,776.
)
0
0.01
(ii)A computer shop buys a batch of iPod
nanos for €99 each and marks the price
up by _ 13 . The goods fail to sell so they are
included in the next sale where all prices
are reduced by _ 14 .
What price would you pay for an iPod
nano in the sale?
(iii)John calculates correctly 85 × 142 = 12,070.
Using this information, what should his
answer be for 12,070 ÷ 850?
__
14. Prove that √
5 is irrational.
16. A man died, leaving some money to be
divided among his children in the following
manner:
1
of what
€x to the first child plus __
16
remains.
1
€2x to the second child plus __
16
of what
then remains.
1
€3x to the third child plus __
16
of what then
remains and so on.
When all the money was distributed, each
child received the same amount of money
and no money was left over.
1
Real Numbers
15. (i) A single-celled protozoa is about onetenth of a millimetre in diameter. Write
down its diameter in metres, giving your
answer in scientific notation.
(ii) A bag contains 350 disks. Seán takes
four-fifths of the disks out of the bag and
divides them into seven equal groups.
How many disks are in each group?
How many children did the man have?
17. The digits 1, 2, 3, 4 and 5 are each used once to create a five-digit number vwxyz, which satisfies
the following conditions:
The three-digit number vwx is odd.
The three-digit number wxy is divisible by 5.
The three-digit number xyz is divisible by 3.
Determine the six five-digit numbers that satisfy all three conditions.
18. A well known series for the irrational p is:
1 × 2 × 3 _____________
1×2×3×4
p
1 1 × 2 _________
__ = 1 + __
+ ______
+
+
+ ...
2
3 3×5 3×5×7 3×5×7×9
By summing the first 10 terms of the series, find an approximation for p to two decimal places.
19
R
1
chapter
Real Numbers
Activity 1.1
1. If p × q = r, then we say that p is a factor of r and q is a factor of r.
For example, 23 × 27 = 210 implies that 23 is a factor of 210 and 27 is a factor of 210.
20 × 25
25
21 × 24
25
22 × 23
25
Using the table above, write down the factors of 25. Write the factors in the form 2n.
2. Find one prime factor of each of the following numbers:
Number
Prime factor
12
35
62
51
221
203
Example: Write 12 as a product of prime factors.
Solution
The work can be summarised more neatly as follows:
Step 1 From Note 1 above, 12 has a
prime factor. 2 is a prime factor of 12.
Therefore, 12 = 2 × 6.
Step 2 The number 6 has a prime factor 2.
Therefore, 6 = 2 × 3.
Step 3 12 = 2 × 2 × 3 = 22 × 3
Factor trees
Repeated division
by primes
2
12
2
6
3
3
12
OR
2
6
1
12 = 22 × 3
2
3
12 = 22 × 3
ACTIVE MATHS – ACTIVITIES
1
1
3. Write the following as a product of their prime factors:
60
36
27
ReAl NumbeRs
book 1
24
R
Activity 1.2
1. The highest common factor (HCF) of two natural numbers is the largest natural number that divides evenly into
both numbers. In this activity you will find the HCF of 36 and 48.
(i) Write 36 and 48 as a product of primes. Write your answers in the form 2m × 3n.
36
48
(ii) What is the highest power of 2 that divides into both 36 and 48?
(iii) What is the highest power of 3 that divides into both 36 and 48?
(iv) Using your answers from parts (ii) and (iii), write down the HCF of 36 and 48.
2. The lowest common multiple (LCM) of two natural numbers is the smallest natural number into which
both numbers divide evenly.
(i) What is the LCM of 29 and 23? Give your answer in the form 2n.
(ii) What is the LCM of 58 and 52? Give your answer in the form 5n.
(iii) Using your results from (i) and (ii), write down the LCM of 29 × 52 and 23 × 58.
2
ACTIVE MATHS – ACTIVITIES
1
3. Write 50 and 160 as a product of prime factors, and hence, find the LCM of 50 and 160.
50
160
book 1
ReAl NumbeRs
LCM of 50 and 160 =
R
Activity 1.3
In this Activity you will prove that there are infinitely many prime numbers.
1. How many 6’s are there in 20? There are three 6’s in 20, with 2 left over. We call ‘3’ the quotient and
‘2’ the remainder. Complete the table.
Quotient
Remainder
12 ÷ 9
31 ÷ 5
500 ÷ 50
41 ÷ 10
2. Write down the quotient and remainder in each of the following divisions:
Quotient
Remainder
(2 × 3 + 1) ÷ 3
(2 × 3 × 5 + 1) ÷ 2
(2 × 3 × 5 × 7 + 1) ÷ 7
(2 × 3 × 5 × 7 × 11 + 1) ÷ 11
3. In the following table each p n is a prime number. Complete the table.
Quotient
Remainder
(p1 × p3 + 1) ÷ p1
(p2 × p3 × p4 + 1) ÷ p2
(p1 × p3 × p6 × p7 + 1) ÷ p6
(p6 × p7 × p8 + 1) ÷ p7
4. Does p1 divide evenly into the number N = p1 × p2 × p3 + 1, where each of p1, p2 and p3 are primes?
Explain your answer.
ACTIVE MATHS – ACTIVITIES
3
1
5. Complete the following sentence:
ReAl NumbeRs
book 1
Every natural number greater than 1 has a prime
.
6. Consider the number N = p1 × p2 × p3 ... × pn + 1. Explain why none of the primes p1 to pn divide N.
7. In this section we will prove by contradiction that there are infinitely many primes.
Theorem: There are infinitely many prime numbers.
Proof:
Assume that there are finitely many primes p1, p2, p3, ..., pn, where pn is the largest prime.
Consider the number N = p1 × p2 × p 3 ... × pn + 1.
Either N is prime or it is not prime.
If N is prime, we have just written a prime greater than p, which is a contradiction to our
supposition.
If N is not prime, it must be divisible by some prime of p1, p2, p3, ... pn.
None of the numbers p1, p2, p3, ..., pn divide evenly into N. Why?
However, every number has a prime divisor and we have just produced a number that does not
have a prime divisor. Therefore, our original assumption that there are finitely many primes must
be false.
4
ACTIVE MATHS – ACTIVITIES
