O-Level E-Math notes · Chapter 1

Numbers and Their Operations: O-Level E-Math Notes

Every other chapter leans on this one. By the end you can break a number into primes, find HCF and LCM, work with negatives, fractions and calculators, round and estimate sensibly, write very large and very small numbers in standard form, and use the laws of indices with confidence.

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Chapter
1

Figures · Numbers and Their Operations at a glance

Four points on a number line (half-unit ticks).

x < -1 \ \text{or} \ x \geqslant 2

: two separate rays

The set

-3 < x \leqslant 2

(open dot at one end, filled dot at the other).

10 outcomes · 40 worked examples · every step shown ✦

Primes and prime factors

Every whole number above 1 is prime, or a unique product of primes.

HCF, LCM, roots

HCF: common primes, lowest powers. LCM: every prime, highest powers.

Calculator work

Type a long fraction as (top) ÷ (bottom), with brackets. A missing bracket round the bottom is the most common slip.

By the end of this chapter

Write a number as a product of primes and use it to find HCF, LCM, square roots and cube roots.
Work with integers, rational numbers and real numbers, including negative numbers and fractions.
Use your calculator well, and round and estimate to a required accuracy.
Order numbers on a number line and use the symbols < > ⩽ ⩾.
Write numbers in standard form and calculate with them.
Use positive, negative, zero and fractional indices and the laws of indices.

Start here

Maths In Real Life

$Two city buses at a Singapore bus interchange at dawn with a timeline drawn over the picture marking the minutes at which each service leaves and the minutes at which both leave together$

Maths In Real Life 1

Buses that line up

Buses, MRT trains and lifts run on repeating gaps. The lowest common multiple tells a planner when two gaps line up again.

Two bus services leave a Singapore interchange together at the start of the morning. Service A leaves every $14$ minutes and Service B every $21$ minutes (a made-up timetable).

(a)Find how many minutes pass before they next leave together.
(b)How many times do they leave together from the start (time $0$ ) up to and including $3$ hours?
(c)How many trips has each service made by the second joint departure after the start?

Show full solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyfactorise, HCF or LCM

(a)

\displaystyle 14 = 2 \times 7, \quad 21 = 3 \times 7

Factorise both gaps.

\displaystyle \text{LCM} = 42

They meet again after the LCM of the two gaps (every prime once, highest powers).

(b)

3

hours is

180

minutes. The joint departures are at the multiples of

42

that do not go past

180

: 0, 42, 84, 126, 168.

List the multiples of the LCM up to the limit.

Counting the start, that is

5

times.

The start (time zero) counts as the first joint departure.

(c)

After the start, the joint departures are at

42

84

, ... minutes, so the second one after the start is at

84

minutes.

Skip the start at zero; the question counts the departures that come after it.

\displaystyle 84 \div 14 = 6, \quad 84 \div 21 = 4

Divide the time by each gap to count the trips.

Final answer

(a)

42

minutes

(b)

5

times

(c)Service A:

6

trips. Service B:

4

trips.

$A close view of a single hair strand beside tiny dust particles and a bacterium, with a scale bar and the widths drawn over the picture$

Maths In Real Life 2

How many fit across a hair?

Lengths in biology and technology can be tiny. Standard form makes them easy to compare without counting zeros.

A strand of hair is $8 \times 10^{ -5 }$ m wide. A dust particle is $2 \times 10^{ -8 }$ m across and a bacterium is $2.5 \times 10^{ -6 }$ m long (made-up values for practice).

(a)How many dust particles, side by side, would span the width of the hair? Give the answer in standard form.
(b)How many bacteria, end to end, would span the width of the hair? Give the answer in standard form.
(c)Write the width of the dust particle as an ordinary number, in metres.

Show full solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solveindex law or standard formfinal answer

(a)

\displaystyle (8 \times 10^{ -5 }) \div (2 \times 10^{ -8 }) = (8 \div 2) \times 10^{ -5 - ( -8 ) }

Particles across = width of the hair ÷ width of one particle.

\displaystyle = 4 \times 10^{ 3 }

4 is already between 1 and 10, so this is standard form.

(b)

\displaystyle (8 \times 10^{ -5 }) \div (2.5 \times 10^{ -6 }) = (8 \div 2.5) \times 10^{ -5 - ( -6 ) }

Bacteria across = width of the hair ÷ length of one bacterium.

\displaystyle = 3.2 \times 10^{ 1 }

3.2 is already between 1 and 10, so this is standard form.

(c)

-8

is negative, so move the decimal point

8

places to the left.

Negative power: a small number. Positive power: a big number.

\displaystyle 2 \times 10^{ -8 } = 0.00000002

Write the digits in their new places.

Final answer

(a)

4 \times 10^{ 3 }

particles

(b)

3.2 \times 10^{ 1 }

bacteria

(c)

0.00000002

$A busy hawker centre food stall counter with a bill listing seven dishes and three drinks drawn over the picture$

Maths In Real Life 3

The hawker centre bill

Estimating first is a quick way to catch a slip before you pay. The stall prices here are made up for practice.

A family orders $7$ dishes at $\text{\textdollar}4.90$ each and $3$ drinks at $\text{\textdollar}1.85$ each.

(a)Estimate the total by rounding each price to the nearest dollar.
(b)Find the exact total, to the nearest cent.
(c)By how much does the estimate differ from the exact total?
(d)They pay with a fifty-dollar note. Find the change, to the nearest cent.

Show full solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyround, estimate or compare

(a)

\displaystyle 7 \times 5 + 3 \times 2 = 35 + 6 = 41

Round each price to the nearest dollar, then work with the easy numbers.

(b)

\displaystyle 7 \times 4.90 = 34.30, \quad 3 \times 1.85 = 5.55

Work out each part exactly.

\displaystyle 34.30 + 5.55 = \text{\textdollar}39.85

Add. The result is exact in cents.

(c)

\displaystyle 41 - 39.85 = \text{\textdollar}1.15

Estimate minus exact total.

(d)

\displaystyle 50 - 39.85 = \text{\textdollar}10.15

Amount paid minus the bill.

Final answer

(a)

\text{\textdollar}41

(b)

\text{\textdollar}39.85

(c)

\text{\textdollar}1.15

(d)

\text{\textdollar}10.15

Supports N1.1

Primes and prime factorisation

What this chapter covers (SEC G3 Mathematics, K310, topic N1)

N1.1: primes and prime factorisation
N1.2: finding highest common factor (HCF) and lowest common multiple (LCM), squares, cubes, square roots and cube roots by prime factorisation
N1.3: negative numbers, integers, rational numbers, real numbers, and their four operations
N1.4: calculations with calculator
N1.5: representation and ordering of numbers on the number line
N1.6: use of the symbols <, >, ⩽, ⩾
N1.7: approximation and estimation (including rounding off numbers to a required number of decimal places or significant figures and estimating the results of computation)
N1.8: use of standard form A × 10ⁿ, where n is an integer, and 1 ⩽ A < 10
N1.9: positive, negative, zero and fractional indices
N1.10: laws of indices

A prime number has exactly two factors: 1 and itself. So $2$ , $3$ , $5$ , $7$ and $11$ are prime, but $9$ is not (it has the factor $3$ ). The number 1 has only one factor, so it is not prime. A whole number above 1 that is not prime is called composite.

Think About It

Why is 2 the only even prime number?

Show the worked answerHide the worked answer

Take any even number above

2

, for example

10

Pick an example first.

\displaystyle 10 = 2 \times 5

Every even number can be written as 2 times something.

So it has the factors

1

2

and more. That is more than two factors, so it is not prime.

A prime has exactly two factors.

Final answerEvery other even number can be divided by 2, so it has a third factor.

The Core Idea

Every whole number above 1 is prime, or a unique product of primes.

Writing it as that product is called prime factorisation. It is the key to HCF, LCM, square roots and cube roots.

Is it prime? Test with the primes up to the square root

Divide by the primes in order. Stop once the prime's square is bigger than the number: a larger factor would be paired with a smaller one you already tried.

Example 1.1

Is the number prime?

Decide whether each number is prime.

(i) $87$
(ii) $91$
(iii) $97$

Show solutionHide solution

Colour keyfactorise, HCF or LCMfinal answer

(i)

9^2 = 81

and

10^2 = 100

. The number

87

lies between

81

and

100

, so only the primes up to

9

need testing: 2, 3, 5, 7.

A factor bigger than the square root would be paired with a smaller one that we test anyway.

Try them in order: 2 does not go in exactly.

Quick checks: even? digit sum a multiple of 3? ends in 0 or 5?

\displaystyle 87 \div 3 = 29

3 divides 87 exactly, so 87 has a factor besides 1 and itself.

\displaystyle 87 = 3 \times 29

So the number is a product of two smaller whole numbers: it is not prime.

(ii)

9^2 = 81

and

10^2 = 100

. The number

91

lies between

81

and

100

, so only the primes up to

9

need testing: 2, 3, 5, 7.

A factor bigger than the square root would be paired with a smaller one that we test anyway.

Try them in order: 2, 3, 5 do not go in exactly.

Quick checks: even? digit sum a multiple of 3? ends in 0 or 5?

\displaystyle 91 \div 7 = 13

7 divides 91 exactly, so 91 has a factor besides 1 and itself.

\displaystyle 91 = 7 \times 13

So the number is a product of two smaller whole numbers: it is not prime.

(iii)

9^2 = 81

and

10^2 = 100

. The number

97

lies between

81

and

100

, so only the primes up to

9

need testing: 2, 3, 5, 7.

A factor bigger than the square root would be paired with a smaller one that we test anyway.

None of these primes divides 97 exactly.

No small factor means no factor at all, so the number is prime.

Final answer

(i)

87

is not prime:

87 = 3 \times 29

(ii)

91

is not prime:

91 = 7 \times 13

(iii)

97

is prime.

Your Turn 1.1

Decide whether each number is prime.

(i) $119$
(ii) $121$
(iii) $103$

Show worked solutionHide worked solution

Colour keyfactorise, HCF or LCMfinal answer

(i)

10^2 = 100

and

11^2 = 121

. The number

119

lies between

100

and

121

, so only the primes up to

10

need testing: 2, 3, 5, 7.

A factor bigger than the square root would be paired with a smaller one that we test anyway.

Try them in order: 2, 3, 5 do not go in exactly.

Quick checks: even? digit sum a multiple of 3? ends in 0 or 5?

\displaystyle 119 \div 7 = 17

7 divides 119 exactly, so 119 has a factor besides 1 and itself.

\displaystyle 119 = 7 \times 17

So the number is a product of two smaller whole numbers: it is not prime.

(ii)

121 = 11^2

exactly, so only the primes up to

11

need testing: 2, 3, 5, 7, 11.

A factor bigger than the square root would be paired with a smaller one that we test anyway.

Try them in order: 2, 3, 5, 7 do not go in exactly.

Quick checks: even? digit sum a multiple of 3? ends in 0 or 5?

\displaystyle 121 \div 11 = 11

11 divides 121 exactly, so 121 has a factor besides 1 and itself.

\displaystyle 121 = 11^{ 2 }

So the number is a product of two smaller whole numbers: it is not prime.

(iii)

10^2 = 100

and

11^2 = 121

. The number

103

lies between

100

and

121

, so only the primes up to

10

need testing: 2, 3, 5, 7.

A factor bigger than the square root would be paired with a smaller one that we test anyway.

None of these primes divides 103 exactly.

No small factor means no factor at all, so the number is prime.

Final answer

(i)

119

is not prime:

119 = 7 \times 17

(ii)

121

is not prime:

121 = 11^{ 2 }

(iii)

103

is prime.

SGSchoolKaki Tip

Quick checks: a number is divisible by 3 if its digits add to a multiple of 3, by 5 if it ends in 0 or 5, and by 2 if it is even.

Prime factorisation by repeated division

Divide by the smallest prime that goes in exactly, keep the answer, and repeat until you reach 1. The primes you used are the prime factors. Group equal primes with index notation.

Example 1.2

Index form by repeated division

Write $1260$ as a product of prime factors in index form.

Show solutionHide solution

Colour keyfactorise, HCF or LCMindex law or standard form

\displaystyle 1260 \div 2 = 630

Divide by the smallest prime that goes in exactly.

\displaystyle 630 \div 2 = 315

2 goes in again, so divide by 2 again.

\displaystyle 315 \div 3 = 105

2 no longer goes in; try the next prime, 3.

\displaystyle 105 \div 3 = 35

3 goes in again, so divide by 3 again.

\displaystyle 35 \div 5 = 7

3 no longer goes in; try the next prime, 5.

\displaystyle 7 \div 7 = 1

5 no longer goes in; try the next prime, 7.

\displaystyle 1260 = 2 \times 2 \times 3 \times 3 \times 5 \times 7 = 2^{ 2 } \times 3^{ 2 } \times 5 \times 7

Collect equal primes as powers (index form).

Final answer

1260 = 2^{ 2 } \times 3^{ 2 } \times 5 \times 7

Your Turn 1.2

Write $2340$ as a product of prime factors in index form.

Show worked solutionHide worked solution

Colour keyfactorise, HCF or LCMindex law or standard form

\displaystyle 2340 \div 2 = 1170

Divide by the smallest prime that goes in exactly.

\displaystyle 1170 \div 2 = 585

2 goes in again, so divide by 2 again.

\displaystyle 585 \div 3 = 195

2 no longer goes in; try the next prime, 3.

\displaystyle 195 \div 3 = 65

3 goes in again, so divide by 3 again.

\displaystyle 65 \div 5 = 13

3 no longer goes in; try the next prime, 5.

\displaystyle 13 \div 13 = 1

5 no longer goes in, and neither do 7 and 11. The next prime that goes in is 13.

\displaystyle 2340 = 2 \times 2 \times 3 \times 3 \times 5 \times 13 = 2^{ 2 } \times 3^{ 2 } \times 5 \times 13

Collect equal primes as powers (index form).

Final answer

2340 = 2^{ 2 } \times 3^{ 2 } \times 5 \times 13

Example 1.3

Reading prime factors

$N = 2^{ 3 } \times 3^{ 2 } \times 7$ .

(i)Find the value of $N$ .
(ii)Is $N$ divisible by $12$ ? Explain using prime factors.
(iii)Is $N$ divisible by $35$ ? Explain.

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyfactorise, HCF or LCMround, estimate or compare

(i)

\displaystyle N = 2^{ 3 } \times 3^{ 2 } \times 7 = 8 \times 9 \times 7 = 504

Work out each power first, then multiply.

(ii)

\displaystyle 12 = 2^{ 2 } \times 3

Write 12 in prime factors.

N

must contain every prime of

12

at least as many times. It does: yes.

Compare the powers prime by prime.

(iii)

\displaystyle 35 = 5 \times 7

Write 35 in prime factors.

N

must contain every prime of

35

at least as many times. It has too few of

5

Compare the powers prime by prime.

Final answer

(i)

N = 504

(ii)Yes.

12 = 2^{ 2 } \times 3

and

N

contains all those primes.

(iii)No.

35 = 5 \times 7

but

N

does not have enough factors of

5

Your Turn 1.3

$M = 2^{ 2 } \times 3 \times 5^{ 2 } \times 11$ .

(i)Find the value of $M$ .
(ii)Is $M$ divisible by $15$ ? Explain using prime factors.
(iii)Is $M$ divisible by $28$ ? Explain.

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplifyfactorise, HCF or LCMround, estimate or compare

(i)

\displaystyle M = 2^{ 2 } \times 3 \times 5^{ 2 } \times 11 = 4 \times 3 \times 25 \times 11 = 3300

Work out each power first, then multiply.

(ii)

\displaystyle 15 = 3 \times 5

Write 15 in prime factors.

M

must contain every prime of

15

at least as many times. It does: yes.

Compare the powers prime by prime.

(iii)

\displaystyle 28 = 2^{ 2 } \times 7

Write 28 in prime factors.

M

must contain every prime of

28

at least as many times. It has too few of

7

Compare the powers prime by prime.

Final answer

(i)

M = 3300

(ii)Yes.

15 = 3 \times 5

and

M

contains all those primes.

(iii)No.

28 = 2^{ 2 } \times 7

but

M

does not have enough factors of

7

Links To

Want the short version first? Read the matching part of the free E-Math revision page: numbers and arithmetic

Supports N1.2

HCF, LCM, squares, cubes and roots

The highest common factor (HCF) of some numbers is the biggest number that divides all of them. The lowest common multiple (LCM) is the smallest number that all of them divide into. Listing works for small numbers; prime factorisation works for every number.

The Core Idea

HCF: common primes, lowest powers. LCM: every prime, highest powers.

Line the numbers up in index form. A prime missing from a number counts as power 0.

Example 2.1

HCF and LCM of two numbers

Find the HCF and the LCM of $84$ and $90$ .

Show solutionHide solution

Colour keyfactorise, HCF or LCM

\displaystyle 84 = 2^{ 2 } \times 3 \times 7

Factorise each number by repeated division.

\displaystyle 90 = 2 \times 3^{ 2 } \times 5

Same method for the next number.

\displaystyle \text{HCF} = 2 \times 3 = 6

HCF: take only the primes that are in every number, each at its lowest power.

\displaystyle \text{LCM} = 2^{ 2 } \times 3^{ 2 } \times 5 \times 7 = 1260

LCM: take every prime that appears anywhere, each at its highest power.

Final answer

HCF

= 6

, LCM

= 1260

Example 2.2

Three numbers at once

Find the HCF and LCM of $60$ , $126$ and $210$ .

Show solutionHide solution

Colour keyfactorise, HCF or LCM

\displaystyle 60 = 2^{ 2 } \times 3 \times 5

Factorise each number by repeated division.

\displaystyle 126 = 2 \times 3^{ 2 } \times 7

Same method for the next number.

\displaystyle 210 = 2 \times 3 \times 5 \times 7

Same method for the next number.

\displaystyle \text{HCF} = 2 \times 3 = 6

HCF: take only the primes that are in every number, each at its lowest power.

\displaystyle \text{LCM} = 2^{ 2 } \times 3^{ 2 } \times 5 \times 7 = 1260

LCM: take every prime that appears anywhere, each at its highest power.

Final answer

HCF

= 6

, LCM

= 1260

Your Turn 2.2

Find the HCF and LCM of $72$ , $108$ and $180$ .

Show worked solutionHide worked solution

Colour keyfactorise, HCF or LCM

\displaystyle 72 = 2^{ 3 } \times 3^{ 2 }

Factorise each number by repeated division.

\displaystyle 108 = 2^{ 2 } \times 3^{ 3 }

Same method for the next number.

\displaystyle 180 = 2^{ 2 } \times 3^{ 2 } \times 5

Same method for the next number.

\displaystyle \text{HCF} = 2^{ 2 } \times 3^{ 2 } = 36

HCF: take only the primes that are in every number, each at its lowest power.

\displaystyle \text{LCM} = 2^{ 3 } \times 3^{ 3 } \times 5 = 1080

LCM: take every prime that appears anywhere, each at its highest power.

Final answer

HCF

= 36

, LCM

= 1080

Example 2.3

LCM in a timetable

At a bus interchange, Service A leaves every $12$ minutes and Service B every $18$ minutes. They leave together at 0715 (24-hour clock). When do they next leave together?

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyfactorise, HCF or LCM

\displaystyle 12 = 2^{ 2 } \times 3, \quad 18 = 2 \times 3^{ 2 }

Factorise both gaps.

\displaystyle \text{LCM} = 36

They meet again after the LCM of the two gaps (every prime once, highest powers).

Start time 0715 plus 36 minutes gives 0751 on the 24-hour clock.

Add the LCM to the starting time; write the time in four digits.

Final answer

They next leave together at

0751\ \left(\text{after } 36\ \text{minutes}\right)

SGSchoolKaki Tip

HCF is not 'the biggest number in the list', and LCM is not 'the two numbers multiplied'. Check: the HCF divides every number, and every number divides the LCM.

Squares, cubes, square roots and cube roots by prime factorisation

A number is a perfect square when every prime in it has an even power, and a perfect cube when every power is a multiple of 3. To find a square root, halve every power. To find a cube root, divide every power by 3.

Example 2.4

Roots without a calculator

Use prime factorisation to find

(i) $\sqrt{ 7056 }$
(ii) $\sqrt[ 3 ]{ 10648 }$

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyfactorise, HCF or LCMindex law or standard form

(i)

\displaystyle 7056 = 2^{ 4 } \times 3^{ 2 } \times 7^{ 2 }

Prime factorise first.

\displaystyle \sqrt{ 7056 } = 2^{ 2 } \times 3 \times 7

Halve every power.

\displaystyle = 4 \times 3 \times 7 = 84

Multiply the results.

(ii)

\displaystyle 10648 = 2^{ 3 } \times 11^{ 3 }

Prime factorise first.

\displaystyle \sqrt[ 3 ]{ 10648 } = 2 \times 11

Divide every power by 3.

\displaystyle = 22

Multiply the results.

Final answer

(i)

84

(ii)

22

Example 2.5

Making a perfect square or cube

Find the smallest whole number to multiply $392$ by so that the product is

(i)a perfect square.
(ii)a perfect cube.

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyfactorise, HCF or LCMindex law or standard form

(i)

\displaystyle 392 = 2^{ 3 } \times 7^{ 2 }

Prime factorise.

A perfect square needs every power to be a multiple of

2

. The power of

2

3

; the power of

7

2

Look at each power in turn.

Multiply by the smallest primes that fix the powers:

2

(to the power

1

Only fix the powers that are not yet multiples of

2

\displaystyle \text{multiplier} = 2

Multiply the primes you need.

\displaystyle 392 \times 2 = 784 = 28^{ 2 }

Check by taking the root of the product.

(ii)

\displaystyle 392 = 2^{ 3 } \times 7^{ 2 }

Prime factorise.

A perfect cube needs every power to be a multiple of

3

. The power of

2

3

; the power of

7

2

Look at each power in turn.

Multiply by the smallest primes that fix the powers:

7

(to the power

1

Only fix the powers that are not yet multiples of

3

\displaystyle \text{multiplier} = 7

Multiply the primes you need.

\displaystyle 392 \times 7 = 2744 = 14^{ 3 }

Check by taking the root of the product.

Final answer

(i)

2

(ii)

7

Your Turn 2.5

Find the smallest whole number to multiply $600$ by so that the product is

(i)a perfect square.
(ii)a perfect cube.

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplifyfactorise, HCF or LCMindex law or standard form

(i)

\displaystyle 600 = 2^{ 3 } \times 3 \times 5^{ 2 }

Prime factorise.

A perfect square needs every power to be a multiple of

2

. The power of

2

3

; the power of

3

1

; the power of

5

2

Look at each power in turn.

Multiply by the smallest primes that fix the powers:

2

(to the power

1

3

(to the power

1

Only fix the powers that are not yet multiples of

2

\displaystyle \text{multiplier} = 2 \times 3 = 6

Multiply the primes you need.

\displaystyle 600 \times 6 = 3600 = 60^{ 2 }

Check by taking the root of the product.

(ii)

\displaystyle 600 = 2^{ 3 } \times 3 \times 5^{ 2 }

Prime factorise.

A perfect cube needs every power to be a multiple of

3

. The power of

2

3

; the power of

3

1

; the power of

5

2

Look at each power in turn.

Multiply by the smallest primes that fix the powers:

3

(to the power

2

5

(to the power

1

Only fix the powers that are not yet multiples of

3

\displaystyle \text{multiplier} = 3^{ 2 } \times 5 = 45

Multiply the primes you need.

\displaystyle 600 \times 45 = 27000 = 30^{ 3 }

Check by taking the root of the product.

Final answer

(i)

6

(ii)

45

Example 2.6

Finding the missing number

The HCF of two numbers is $12$ and their LCM is $180$ . One of the numbers is $36$ . Find the other number.

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyfactorise, HCF or LCM

\displaystyle \text{HCF} \times \text{LCM} = 12 \times 180 = 2160

For TWO numbers, HCF × LCM = the product of the two numbers.

\displaystyle \text{other number} = 2160 \div 36 = 60

Divide the product by the number you know.

\displaystyle \text{Check: HCF}(36, 60) = 12, \ \text{LCM} = 180

Verify using prime factors.

Final answer

60

SGSchoolKaki Tip

HCF × LCM = product works for exactly TWO numbers, not for three or more.

Links To

Square roots and cube roots also appear when you work with area and volume scale factors. Chapter 2: Ratio and proportion

Supports N1.3

Integers, rational numbers, real numbers and the four operations

Numbers come in families. Integers are whole numbers: positive, negative or zero. Rational numbers can be written as a fraction of two integers. Irrational numbers cannot: their decimals never end and never repeat. Real numbers are all of these together, every point on the number line.

Which family?
Number	Positive integer	Integer	Rational	Irrational	Real
$7$	yes	yes	yes	no	yes
$-3$	no	yes	yes	no	yes
$\frac{3}{4}$	no	no	yes	no	yes
$0.35$	no	no	yes	no	yes
$\sqrt{ 81 }$	yes	yes	yes	no	yes
$\sqrt{ 7 }$	no	no	no	yes	yes
$\pi$	no	no	no	yes	yes

Think About It

Can two irrational numbers add up to a rational number?

Show the worked answerHide the worked answer

\displaystyle \sqrt{ 7 } + \left( -\sqrt{ 7 } \right) = 0

Take an irrational number and its negative.

The sum is

0

, and

0

is rational (it equals

0

over

1

Zero is a fraction of two integers.

Final answerYes, for example a number and its negative.

Example 3.1

Sorting numbers into families

From the list $\displaystyle -5,\ \frac{7}{8},\ \sqrt{ 11 },\ \sqrt{ 64 },\ \pi,\ -\dfrac{ 15 }{ 5 }$ write down

(i)the integers.
(ii)the irrational numbers.

Show solutionHide solution

Colour keyfinal answer

(i)

Simplify each number first, then decide.

Some numbers look unfriendly but simplify, e.g. a root of a perfect square.

\displaystyle -5

This is an integer.

\displaystyle \frac{7}{8}

This is rational but not an integer.

\displaystyle \sqrt{ 11 }

This is irrational: its decimal never ends or repeats.

\displaystyle \sqrt{ 64 } = 8

This is an integer.

\displaystyle \pi

This is irrational: its decimal never ends or repeats.

\displaystyle -\dfrac{ 15 }{ 5 } = -3

This is an integer.

So the integers are

-5

8

and

-3

These are the ones marked integer above.

(ii)

Use the simplified list from part (i):

-5

\frac{7}{8}

\sqrt{11}

8

\pi

-3

No need to simplify again.

So the irrational numbers are

\sqrt{11}

and

\pi

These are the ones marked irrational in part (i).

Final answer

(i)

-5,\ 8,\ -3

(ii)

\sqrt{11},\ \pi

Your Turn 3.1

From the list $\displaystyle 12,\ - \frac{2}{3},\ \sqrt{ 5 },\ \sqrt{ 49 },\ \pi$ write down

(i)the integers.
(ii)the irrational numbers.

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Colour keyfinal answer

(i)

Simplify each number first, then decide.

Some numbers look unfriendly but simplify, e.g. a root of a perfect square.

\displaystyle 12

This is an integer.

\displaystyle - \frac{2}{3}

This is rational but not an integer.

\displaystyle \sqrt{ 5 }

This is irrational: its decimal never ends or repeats.

\displaystyle \sqrt{ 49 } = 7

This is an integer.

\displaystyle \pi

This is irrational: its decimal never ends or repeats.

So the integers are

12

and

7

These are the ones marked integer above.

(ii)

Use the simplified list from part (i):

12

- \frac{2}{3}

\sqrt{5}

7

\pi

No need to simplify again.

So the irrational numbers are

\sqrt{5}

and

\pi

These are the ones marked irrational in part (i).

Final answer

(i)

12,\ 7

(ii)

\sqrt{5},\ \pi

Example 3.2

Negative numbers and the order of operations

Work out $-7 - (-12) \times 3 + 20 \div (-4)$ .

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplify

\displaystyle ( -12 ) \times 3 = -36

Multiply and divide before adding and subtracting.

\displaystyle 20 \div ( -4 ) = -5

Same signs: positive. Different signs: negative.

\displaystyle -7 - (-36) + ( -5 ) = -7 + 36 + ( -5 )

Subtracting a negative is the same as adding the positive.

\displaystyle = 24

Work from left to right.

Final answer

24

Example 3.3

Temperatures that go below zero

A freezer reads $-18^\circ\text{C}$ . A power cut makes it rise by $26^\circ\text{C}$ . After the power returns it falls by $11^\circ\text{C}$ .

(i)Find the temperature after the rise, and after the fall.
(ii)Find the difference between the highest and lowest of the three readings.

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Colour keyadd, subtract, multiply, divide or simplify

(i)

\displaystyle ( -18 ) + 26 = 8

A rise moves up the number line from the starting reading.

\displaystyle 8 - 11 = -3

A fall moves down from the new reading.

(ii)

The three readings are

-18

8

and

-3

degrees.

List all the readings first.

\displaystyle 8 - ( -18 ) = 26

Highest minus lowest; subtracting a negative adds.

Final answer

(i)After the rise:

8^\circ\text{C}

. After the fall:

-3^\circ\text{C}

(ii)

26^\circ\text{C}

Your Turn 3.3

A cold room reads $-9^\circ\text{C}$ , rises by $17^\circ\text{C}$ , then falls by $22^\circ\text{C}$ .

(i)Find the temperature after the rise, and after the fall.
(ii)Find the difference between the highest and lowest of the three readings.

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Colour keyadd, subtract, multiply, divide or simplify

(i)

\displaystyle ( -9 ) + 17 = 8

A rise moves up the number line from the starting reading.

\displaystyle 8 - 22 = -14

A fall moves down from the new reading.

(ii)

The three readings are

-9

8

and

-14

degrees.

List all the readings first.

\displaystyle 8 - ( -14 ) = 22

Highest minus lowest; subtracting a negative adds.

Final answer

(i)After the rise:

8^\circ\text{C}

. After the fall:

-14^\circ\text{C}

(ii)

22^\circ\text{C}

Example 3.4

Fractions and mixed numbers

Work out $2\dfrac{ 1 }{ 3 } \div \frac{ 5 }{ 6 } - 1\dfrac{ 3 }{ 4 } \times \frac{ 2 }{ 7 }$ , giving the answer as a mixed number in its simplest form.

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Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

\displaystyle 2\dfrac{ 1 }{ 3 } = \dfrac{ 7 }{ 3 }, \quad 1\dfrac{ 3 }{ 4 } = \dfrac{ 7 }{ 4 }

Change every mixed number to an improper fraction first.

\displaystyle \dfrac{ 7 }{ 3 } \div \dfrac{ 5 }{ 6 } = \dfrac{ 7 }{ 3 } \times \dfrac{ 6 }{ 5 } = \dfrac{ 14 }{ 5 }

Dividing means multiplying by the reciprocal; cancel before you multiply.

\displaystyle \dfrac{ 7 }{ 4 } \times \dfrac{ 2 }{ 7 } = \dfrac{ 1 }{ 2 }

Cancel common factors, then multiply.

\displaystyle \dfrac{ 14 }{ 5 } - \dfrac{ 1 }{ 2 } = \dfrac{ 28 }{ 10 } - \dfrac{ 5 }{ 10 } = \dfrac{ 23 }{ 10 }

Write both fractions over the common denominator 10, then subtract.

\displaystyle = 2\dfrac{ 3 }{ 10 }

Convert back to a mixed number.

Final answer

2\dfrac{ 3 }{ 10 }

Your Turn 3.4

Work out $3\dfrac{ 1 }{ 2 } \div \frac{ 7 }{ 8 } - 2\dfrac{ 2 }{ 3 } \times \frac{ 9 }{ 10 }$ , giving the answer as a mixed number in simplest form.

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Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

\displaystyle 3\dfrac{ 1 }{ 2 } = \dfrac{ 7 }{ 2 }, \quad 2\dfrac{ 2 }{ 3 } = \dfrac{ 8 }{ 3 }

Change every mixed number to an improper fraction first.

\displaystyle \dfrac{ 7 }{ 2 } \div \dfrac{ 7 }{ 8 } = \dfrac{ 7 }{ 2 } \times \dfrac{ 8 }{ 7 } = 4

Dividing means multiplying by the reciprocal; cancel before you multiply.

\displaystyle \dfrac{ 8 }{ 3 } \times \dfrac{ 9 }{ 10 } = \dfrac{ 12 }{ 5 }

Cancel common factors, then multiply.

\displaystyle 4 - \dfrac{ 12 }{ 5 } = \dfrac{ 20 }{ 5 } - \dfrac{ 12 }{ 5 } = \dfrac{ 8 }{ 5 }

Write both fractions over the common denominator 5, then subtract.

\displaystyle = 1\dfrac{ 3 }{ 5 }

Convert back to a mixed number.

Final answer

1\dfrac{ 3 }{ 5 }

Example 3.5

Rational numbers as decimals

Without long division, decide which of $\frac{ 7 }{ 40 }$ , $\frac{ 5 }{ 12 }$ and $\frac{ 9 }{ 35 }$ are terminating decimals.

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Colour keyfactorise, HCF or LCM

A fraction in lowest terms gives a terminating decimal exactly when its denominator has no prime factor except

2

and

5

Because 10 = 2 × 5, only 2s and 5s can divide a power of 10.

\displaystyle \dfrac{ 7 }{ 40 }: \ 40 = 2^{ 3 } \times 5

Prime factors of the denominator: only 2s and 5s.

\displaystyle \dfrac{ 5 }{ 12 }: \ 12 = 2^{ 2 } \times 3

Prime factors of the denominator: there is a prime other than 2 and 5.

\displaystyle \dfrac{ 9 }{ 35 }: \ 35 = 5 \times 7

Prime factors of the denominator: there is a prime other than 2 and 5.

So:

\frac{ 7 }{ 40 } = 0.175

terminates;

\frac{ 5 }{ 12 } \approx 0.417

does not terminate;

\frac{ 9 }{ 35 } \approx 0.257

does not terminate.

Check on your calculator: a terminating decimal stops, the others keep going.

Final answer

\frac{ 7 }{ 40 } = 0.175

terminates;

\frac{ 5 }{ 12 } \approx 0.417

does not terminate;

\frac{ 9 }{ 35 } \approx 0.257

does not terminate.

Your Turn 3.5

Decide which of $\frac{ 11 }{ 80 }$ , $\frac{ 7 }{ 15 }$ and $\frac{ 13 }{ 28 }$ are terminating decimals.

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Colour keyfactorise, HCF or LCM

A fraction in lowest terms gives a terminating decimal exactly when its denominator has no prime factor except

2

and

5

Because 10 = 2 × 5, only 2s and 5s can divide a power of 10.

\displaystyle \dfrac{ 11 }{ 80 }: \ 80 = 2^{ 4 } \times 5

Prime factors of the denominator: only 2s and 5s.

\displaystyle \dfrac{ 7 }{ 15 }: \ 15 = 3 \times 5

Prime factors of the denominator: there is a prime other than 2 and 5.

\displaystyle \dfrac{ 13 }{ 28 }: \ 28 = 2^{ 2 } \times 7

Prime factors of the denominator: there is a prime other than 2 and 5.

So:

\frac{ 11 }{ 80 } = 0.1375

terminates;

\frac{ 7 }{ 15 } \approx 0.467

does not terminate;

\frac{ 13 }{ 28 } \approx 0.464

does not terminate.

Check on your calculator: a terminating decimal stops, the others keep going.

Final answer

\frac{ 11 }{ 80 } = 0.1375

terminates;

\frac{ 7 }{ 15 } \approx 0.467

does not terminate;

\frac{ 13 }{ 28 } \approx 0.464

does not terminate.

Links To

The next chapters build on these operations: percentages and rates use fractions and decimals all the time. Chapter 3: Percentage

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Notes version 1.0 · Last updated 3 Nov 2026 · Chapter 1 of 19