Chapter 1 Real Numbers

Real Numbers
Are the set all of the numbers that lie along an infinitely long number line.
1.1 Factors, Multiples and Prime Factors
Natural Numbers



The Natural Numbers are the ordinary counting numbers. This is an infinite set.
The set is denoted by the letter ‘N’.
They commence at the number 1, the first Natural Number, and gone on into infinity.
Factors

Factor
Is any Natural Number that divides into any other number evenly.

The number 1 is a factor of every Natural Number and also every Natural Number is a
factor of itself.
Multiples

Multiple
Is a Natural Number that divides into another number, leaving no remainder.
Prime Numbers

Prime Numbers
Are the Natural Numbers that have only 2 factors.

Examples of Prime Numbers would include 2, 3, 5 etc…

Composite Numbers
Are Natural Numbers that are greater than 1 that are not prime numbers.

Examples of Composite Numbers would include 4, 6, 8 etc…
The Fundamental Theorem of Arithmetic

Every Natural Number greater than 1 is either prime or can be written as a unique
product of primes.

Highest Common Factor (HCF)
Is the largest Natural Number between 2 Natural Numbers that divides evenly into each
without remainders.

Lowest Common Multiple (LCM)
Is the smallest multiple that both numbers share.
Leaving Certificate Project Mathematics – Higher Level
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Chapter 1 Real Numbers
Worked Example
Question:
Express 240 as a product of Prime Factors
Solution:
To express a certain number as a product of Prime Factors you divide into the number in
question Prime Numbers only and keep dividing in until you are unable to divide another
Prime Number into it. Always start with the lowest and work up.
2
2
2
2
2
3
5
240 – Since 2 is the lowest Prime Number we will start with that and work up.
120 But just say that 2 didn’t divide in evenly into 240 we would then go to the
60
next Prime Number.
30
15
5
5
240 = 2 x 2 x 2 x 2 x 3 x 5.
So, 240 written as a product of Prime Numbers is 24 x 3 x 5
Worked Example
Question:
Find (i) the HCF and (ii) the LCM of 512 and 280.
Solution:
From the previous Worked Example, we are now able to express a number as a product of
Prime Factors. In this example we are going to use what knowledge and find the HCF and
LCM of 2 numbers.
2
2
2
2
2
2
2
2
2
512
256
128
64
32
16
8
4
2
2
2
2
5
7
280
140
70
35
7
512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2.
280 = 2 x 2 x 2 x 5 x 7.
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Chapter 1 Real Numbers
So, 512 written as a product of Prime Numbers is 29 and 280 written as a product of Prime
Numbers is 23 x 5 x 7.
(i)
To gain the HCF of 2 or more numbers you look at the common Prime Factors between
the numbers and choose the lowest possible value of the common Prime Numbers, and
in this case that is 𝟐𝟑 , which can be written as 8.
HCF of 512 & 280 = 8.
(ii)
To gain the LCM of 2 or more number you look at the Prime Factors between the
numbers and choose the highest possible value of the Prime Factors, and in this case
that is 29 , 5, 7.
LCM of 512 & 280 = 17,920
Worked Example
Question:
Periodical Cicadas are insects with very long larval period 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:
In this question we are looking for the LCM of 13 and 17 as we are trying to find the next
lowest possible time that both species will meet again.
Since we can see that 13 and 17 are Prime Factors we just multiply each number to get the
LCM.
17 x 13 = 221.
Therefore, both species will meet again in the year 2232 in that location.
Worked Example
Question:
Evaluate (i) 5! and (ii) 5 x 4!
Solution:
The symbol ! is called Factorial. This symbol means that all the Numbers that are previous
to the number that it is in front of are all multiplied together, including the number in
question. An example would include 6! This reads out 1 x 2 x 3 x 4 x 5 x 6.
(i)
(ii)
5! = 1 x 2 x 3 x 4 x 5. 5! = 120.
5 x 4! = 5 x (1 x 2 x 3 x 4) = 120
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Chapter 1 Real Numbers
Infinitude of Primes

Proof by Contradiction
Is a form of proof that states the truth of a suggestion by showing that the suggestion being
false would imply a contradiction.
1.2 Integers and Rational Numbers


Integers are made up of zero and all the positive and negative whole numbers.
In Mathematics, we use the letter ‘Z’ to denote Integers.
Properties of Integers


The following are the Properties or Characteristics of Integers:
 A + B and A x B are integers whenever A and B are integers. This is a Closure
Property. It states that an integer multiplied by an integer or an integer plus an
integer equals an integer as a result.
 A + B = B + A and A x B = B x A. This is a Commutative Property. It states that if there
is a change in the order of the numbers it does not change the result.
 (A + B) + C = A + (B + C) and (A x B) x C = A x (B x C). This is an Associative Property. It
states that if there is a change in the grouping of the number there will be no change
in the results.
 A x (B + C) = (A x B) + (A x C). This is a Distributive Property. It states that
multiplication distributes over addition and subtraction.
 A + 0 = A and A x 1 = A. This is an Identity Element. It states that if A is added to 0 or
multiplied by 1 then A will always be the result.
 For every integer A, there is an integer –A. So, A + (–A) = 0. This is an Additive
Inverse. It states that for every integer A there is a –A of that integer
In previous Examination Papers for Higher Level Mathematics under the new Project
Maths syllabus and also under the Original syllabus, Questions have arisen that
require you to use the Properties of Integers. Refer to Active Maths 4 – Book 1, Page 8,
for examples of these possible Questions.
1.3 Irrational Numbers

Irrational Number
Is a number that cannot be written in fraction form.

Rational Numbers and Irrational Numbers together make up the Real Number System.
The following Proofs and Constructions must be learned for the Leaving Certificate Project
Mathematics Examination at Higher Level only. The Proof that the 𝟑 is Irrational is not
specified on the Project Maths syllabus at Higher Level. It is still suggested that you be
familiar with this Proof. All others are specified and are possibly examinable.
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Chapter 1 Real Numbers
Proof that the 𝟐 and 𝟑 is Irrational
Proof that the 𝟐 is Irrational
To Prove:
2 is irrational
Proof:

a
Let us assume that the 2 is rational and can be written in the formb, where a, b ∈ Z and
b ≠ 0.

a
Let us also assume that the fraction b is written in its simplest form. HCF(a, b) = 1
a
2 = b,
2=
a2
b2
2









2b = a2
Since 2b2 is even a is even and a2 = a.a. A is even.
Thus, a = 2r, where r is a natural number.
Hence, a2 = 4r2 .
Thus, 2b2 = 4r2 . i.e. b2 = 2r 2
Since 2r 2 is even then b2 is even.
Hence, b is even since b2 = b.b.
2b2 = (2r)2
2b2 = 4r 2
b2 = 2r 2
Since a and b are both even they have 2 as a common factor.
This contradicts our earlier assumption that they were co-prime.
Thus, our original assumption that the 2 is rational is incorrect and hence the 2 is
irrational.
Proof that the 𝟑 is Irrational
To Prove:
3 is irrational
Proof:

a
Let us assume that the 3 is rational and can be written in the formb, where a, b ∈ Z and
b ≠ 0.

a
Let us also assume that the fraction b is written in its simplest form. HCF(a, b) = 1
a
3 = b,
3=
a2
b2
3b2 = a2

Since 3b2 is an integers a is an integer and a2 = a.a. A is an integer.
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Chapter 1 Real Numbers


Thus, a = 2r, where r is a natural number.
Hence, a2 = 3r2 .

Thus, 2b2 = 9r2 . i.e. b2 = 3r2

Since 3r2 is even then b2 is even.

Hence, b is even since b2 = b.b.


3b2 = (3r)2
3b2 = 9r2
b2 = 3r2
Since a and b are both even they have 3 as a common factor.
This contradicts our earlier assumption that they were co-prime.

Thus, our original assumption that the 3 is rational is incorrect and hence the 3 is
irrational.
Constructing 𝟐 and 𝟑
Construct 𝟐

Procedure for Construction:
1. Let the line segment AB be of length 1 unit.
2. Construct a line m perpendicular to [AB] at B.
3. Construct a circle with centre B and radius [AB]
4. Mark the intersection, C, of the circle and m.
5. Draw the line segment CA. |AC| = 2
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Chapter 1 Real Numbers
Proof:
|AB| = |BC| = 1 (Radii of Circle)
|AB|2 + |BC|2 = |AC|2
12 + 12 = |AC|2
|AC|2 = 2
|AC| = 2
Construct 𝟑

Procedure for Construction:
1. Let the line segment AB be of length 1 unit.
2. Construct a circle with centre A and radius length |AB|.
3. Construct a circle with centre B and radius length |AB|.
4. Mark the intersection of the 2 circles as C and D.
5. Draw the line segment [CD]. |CD| = 3
Proof:
|AE| = |EB| =
1
2
|AC| = |BC| = 1
|AB|2 + |AB|2 = |AB|2
1 2
(2) + |AB|2 = |AB|2
1
3
|AB|2 = 1 - 4 = 2
3
|AB|2 = √4 =
3
4
|AB|2 = 2 |EC|
3
= 2( 2 )
|CD| = 3
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Chapter 1 Real Numbers
1.4 Rounding and Significant Numbers
Rounding to Decimal Places


We round to Decimal Places so that we can simplify a number as numbers are infinite.
An example would include . The value of  on the calculator is 3.141592654. We use
the approximate value in school mathematics, 3.14, 3 decimal places.
Worked Example
Question:
Write the following correct to one decimal place:
(i) 2.57 (ii) 19.32
Solution:
In Mathematics, when rounding numbers up to a certain amount of places we look at the
numbers after the Decimal Place and look at the number after the number of places we
are asked to round up to. An example would include, round to 2 Decimal Places the
number 4.634. In this example we would look at the 3rd number after the decimal place as
this will dictate whether or not the end result will increase in value or stay the same. If the
number is greater than 5 we would increase the last number. If it is less than 5 we would
leave the number last number. In that example 4.634 correct to 2 Decimal Places that
would be 4.63.
(i)
2.57 correct to 1 Decimal Place is 2.6.
The reason for this is because in the number 2.57 we are asked to correct this number to 1
Decimal Place. Since we are asked to correct to 1 Decimal Place we 1st look at the numbers
after the Decimal Place and look at the 1st number which is 5 and then look at the 2nd
number which will dictate whether or not the number we are asked to correct to 1 Decimal
Place will increase or stay the same in value. Since the 2nd number after the Decimal Place
is 7 this means that the 1st number after the Decimal Place will stay the same as 7 greater
than 5.
(ii)
19.32 correct to 1 Decimal Place is 19.3
The reason for this is because in the number 19.32 we are asked to correct this number to
1 Decimal Place. Since we are asked to correct to 1 Decimal Place we 1st look at the
numbers after the Decimal Place and look at the 1st number which is 3 and then look at the
2nd number which will dictate whether or not the number we are asked to correct to 1
Decimal Place will increase or stay the same in value. Since the 2nd number after the
Decimal Place is 2 this means that the 1st number after the Decimal Place will stay the
same as it less than 5.
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Chapter 1 Real Numbers
Significant Figures

Significant Figures
Are all the digits except zeros at the start and end of a number.

Never count the Zeros at the start of a number when trying to place a number into
Significant Figures.
Worked Example
Question:
Correct the following numbers to 2 Significant Figures:
(i) 3.67765
(ii) 61,343
(iii) 0.00356
Solution:
To correct a Figure to a certain amount of Significant Figures you look at the 1 st Significant
Figure which is the 1st non – zero figure in the number. After that you count the number of
Figures, including the 1st Figure, to which you are asked to correct to Significant Figures.
After that you look at the number that is behind the last number that you are asked to
correct to a certain amount of Significant Figures you then increase the last Significant
Figure by 1 if that number is great than 5.
(i)
3.67765 = 3.67765
1st Significant Figure
2nd Significant Figure
After identifying the Significant Figures you then look at the number that is after the last
Significant Figure and increase it by 1 if that number is greater than 5. In this question the
last Significant Figure is 6 and the number after that is 7 so 6 would increase up to 7
As a result the Figure 3.67765 to 2 Significant Figures is 3.7
(ii)
61,343 = 61,343
1st Significant Figure
2nd Significant Figure
After identifying the Significant Figures you then look at the number that is after the last
Significant Figure and increase it by 1 if that number is greater than 5. In this question the
last Significant Figure is 1 and the number after that is 3 so 1 would remain the same as 3
is less than 5.
As a result the Figure 61,343 to 2 Significant Figures is 61,000
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Chapter 1 Real Numbers
(iii)
0.00365 = 0.00356
1st Significant Figure
2nd Significant Figure
First off you look at the 1st Significant Figure which is the 1st non – zero figure in the
number and then identify your Significant Figures.
After identifying the Significant Figures you then look at the number that is after the last
Significant Figure and round it up. In this question the last Significant Figure is 5 and the
number after that is 6 so 5 would be rounded up to 6.
As a result the Figure 0.00365 to 2 Significant Figures is 0.0036
1.5 Orders of Magnitude and Scientific Notation

Scientific Notation
Is when a number is written in the form a x 10n , where 1  a  10 and n  N.




In Mathematics, Scientific Notation can also be referred to as Standard Form, Index
Notation or Exponential Notation.
When writing a number in Scientific Notation there are 2 rules that must be followed
rigorously:
1. The Number that stands for the value of a must be between 1 and 10, but NOT the
number 10.
2. And multiplied by 10n, where n ∈ Z.
This is often written in the form of a x 10n, where 1 ≤ a < 10 and n ∈ Z.
Refer to the Junior Certificate Mathematics Course & New Concise Maths 2 for Notes
on Addition, Subtraction, Multiplication & Division of Scientific Notation.
Worked Example
Question:
Write the following numbers in Scientific Notation:
(i) 725,000,000,000 (ii) 0.0000056 (iii) 980,000 (iv) 0.000000034
Solution:
In Mathematics, we are aware at this stage in our mathematical life that numbers go on
into infinity. i.e. They are never ending. The way in which they stop a number expanding
into infinity is because there is an Imaginary Decimal Point at the end of it. An example
would include 3. The number 3 actually has an Imaginary Decimal Point at the end of it. It
looks like this, 3. But naturally in Mathematics it would look like this 3.000000000….
1st of we place the number in question into a number between 1 and under 10. The way
we do this is by moving the imaginary Decimal Place which is at the end of the number in
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Chapter 1 Real Numbers
question either to the Right Hand Side or the Left Hand Side depending whether the
number we are about to create is between 1 but below 10.
(i)
725,000,000,000 in Scientific Notation is 7.25 x 1011.
The reason for this is because the number 725,000,000,000 has an imaginary Decimal
Place at the bottom of it and we must move it to get a new number that will result in the
formation of a new number that is between 1 and 10 but not the number 10. The new
number that we are about to create does not necessarily have to be a whole number.
The number 725,000,000,000 has the imaginary Decimal Place at the back which looks like
this 725,000,000,000. And we must move this Decimal Place to a new location so that we
will have a number that is between 1 and 10 but not on the number 10.
When we move that imaginary Decimal Place to its new location so that we can form a
new number that is between 1 and 10 but not on the number 10 we will be left with the
new number 7.25. This is the number in Decimal Form but we are not finished yet we must
now show how many places we moved the imaginary Decimal Place to form that new
number. To do this we just count the number of places that we moved the imaginary
Decimal Place and that is how many places we moved. This amount is 11 places.
(ii)
0.0000056 in Scientific Notation 5.6 x 10-6
The reason for this is because the number 0.0000056 has no imaginary Decimal Place in
the number as we are able to see it in the number itself. In this type of a number we still
move the Decimal Place to form a new number between 1 and 10 but not 10. The only
thing that changes as a result is the power that n stands for in a x 10n. When there is zeros
and a Decimal Place in the front of the number we just treat the value of n as a minus
figure.
To get -6 as the value of n we just count the number of spaces that we need to move that
Decimal Place to form a new number. This number is 6. But we treat it as a -6 because we
are counting backwards.
(iii) 980,000 in Scientific Notation is 9.8 x 105
In this type of Question we just follow the same steps and procedures as we saw in part (i)
of this Worked Example
(iv)
0.000000034 in Scientific Notation is 3.4 x 10-8
In this type of Question we just follow the same steps and procedures as we saw in part (ii)
of this Worked Example
Orders of Magnitude

Order of Magnitude
Is a number rounded to the nearest power of 10.
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Chapter 1 Real Numbers
Worked Example
Question:
By How many Orders of Magnitude does 345,632 differ from 567,132,423?
Solution:
In Mathematics, to find out how many Orders of Magnitude a certain amount of numbers
differs you must 1st place both numbers into Scientific Notation from.
When the numbers above are placed into Scientific Notation form they look like this as a
result.
345,632 in Scientific Notation is 3.45632 x 105.
567,123,423 in Scientific Notation is 5.67123423 x 108.
After both numbers have been placed into Scientific Notation form you then must add 1 to
the value of n if the value of a of a number in the form a x 10 n is greater than 5 or 5. But if
the value of a is less than 5 the value of n of a number on the form a x 10 ndoes not
increase by 1.
108 x 101 = 109
105 x 100 = 105
Then we divide each of the above numbers.
109
105
= 104
The difference is 4 Orders of Magnitude.
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