Real Numbers

Irrational numbers, how to tell whether a root is rational, where real numbers lie on the number line, how the kinds of number fit together, and how to compare, order and calculate with real numbers.

Before you start

A rational number can be written as $\frac{a}{b}$ , where $a$ and $b$ are integers and $b \neq 0$ . As a decimal it either terminates, like $\frac{3}{8} = 0.375$ , or recurs, like $\frac{1}{3} = 0.333\ldots$ Every terminating decimal and every recurring decimal is a rational number.

$\sqrt{9} = 3$ because $3^{2} = 9$ , and $\sqrt[3]{8} = 2$ because $2^{3} = 8$ . So 9 is a perfect square and 8 is a perfect cube.

On the number line, a number further to the right is greater. Of two negative numbers, the one further from 0 is the smaller: $-3 < -2$ .

Irrational Numbers

No fraction of two integers is equal to $\sqrt{2}$ . As a decimal, $\sqrt{2} = 1.41421356\ldots$ : the digits never end, and they never settle into a block that repeats. A number like this is called irrational.

Definition

An irrational number is a number that cannot be written as $\frac{a}{b}$ , where $a$ and $b$ are integers and $b \neq 0$ . As a decimal, it neither terminates nor recurs.

Other irrational numbers are $\sqrt{3} = 1.73205080\ldots$ , $\sqrt[3]{5} = 1.70997594\ldots$ and $\pi = 3.14159265\ldots$

The square root of a whole number that is not a perfect square is irrational. So is the cube root of a whole number that is not a perfect cube. But $\sqrt{4} = 2$ , $\sqrt[3]{27} = 3$ and $\sqrt{\frac{9}{16}} = \frac{3}{4}$ are rational, so check a root before you decide.

A decimal can follow a pattern and still be irrational. In $0.1010010001\ldots$ the run of 0s before each 1 is one digit longer than the run before it, so the digits never end and no block of digits repeats.

Is $\pi$ equal to $\frac{22}{7}$ ?

No. $\frac{22}{7} = 3.142857\ldots$ and $3.14$ are rational, but $\pi = 3.141592\ldots$ is irrational. So $\frac{22}{7}$ and $3.14$ are only approximations of $\pi$ : values close to it that are easier to calculate with.

Example 1: Rational or irrational?

Which of these numbers are irrational?
$\sqrt{16}$ , $\sqrt{5}$ , $\sqrt[3]{9}$ , $0.\dot{7}$ , $\frac{22}{7}$ , $\pi$

Solution

$\sqrt{16} = 4$ , which is an integer, so it is rational.

5 is not a perfect square ( $2^{2} = 4$ and $3^{2} = 9$ ), so $\sqrt{5}$ is irrational.

9 is not a perfect cube ( $2^{3} = 8$ and $3^{3} = 27$ ), so $\sqrt[3]{9}$ is irrational, although 9 is a perfect square.

$0.\dot{7} = 0.777\ldots$ recurs, so it is rational. (It is equal to $\frac{7}{9}$ .)

$\frac{22}{7}$ is a fraction of two integers, so it is rational. $\pi$ is irrational.

The irrational numbers are $\sqrt{5}$ , $\sqrt[3]{9}$ and $\pi$ .

Is a Root Rational? Look at the Number under It

You can tell whether a square root or a cube root is rational from the number under the root sign, without working out a single decimal place.

Checking the number under the root

A whole number. Its square root is rational if the number is a perfect square, and irrational if it is not. Its cube root is rational if the number is a perfect cube, and irrational if it is not.
Is it a perfect square or a perfect cube? For a small number, look at the squares or cubes on either side of it: 5 lies between $2^{2} = 4$ and $3^{2} = 9$ , so it is not a perfect square. For a larger number, write its prime factorisation. A whole number greater than 1 is a perfect square if every index is even, and a perfect cube if every index is a multiple of 3.
A fraction. Write it in its simplest form first. Its square root is rational if the numerator and the denominator are both perfect squares, and irrational if either of them is not: $\sqrt{\frac{9}{16}} = \frac{3}{4}$ , because $\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$ . Cube roots work in the same way, with perfect cubes.
A decimal. Write it as a fraction, then check the fraction.

Example 2: Checking the number under the root

Is each number rational or irrational? If it is rational, write down its value.
(a) $\sqrt{324}$
(b) $\sqrt{72}$
(c) $\sqrt[3]{216}$
(d) $\sqrt[3]{500}$
(e) $\sqrt{\frac{12}{27}}$
(f) $\sqrt{2.25}$
(g) $\sqrt{0.4}$

Solution

(a) $324 = 2^{2} \times 3^{4}$ . Both indices are even, so 324 is a perfect square. Halve each index: $\sqrt{324} = 2 \times 3^{2} = 18$ .

(b) $72 = 2^{3} \times 3^{2}$ . The index of 2 is 3, which is odd, so 72 is not a perfect square. You can also see this from $8^{2} = 64$ and $9^{2} = 81$ .

(c) $216 = 2^{3} \times 3^{3}$ . Both indices are multiples of 3, so 216 is a perfect cube. Divide each index by 3: $\sqrt[3]{216} = 2 \times 3 = 6$ .

(d) $500 = 2^{2} \times 5^{3}$ . The index of 2 is 2, which is not a multiple of 3, so 500 is not a perfect cube.

(e) Simplify first: $\frac{12}{27} = \frac{4}{9}$ . Both 4 and 9 are perfect squares, so $\sqrt{\frac{4}{9}} = \frac{2}{3}$ . Check: $\frac{2}{3} \times \frac{2}{3} = \frac{4}{9}$ .

(f) $2.25 = \frac{225}{100} = \frac{9}{4}$ . Both 9 and 4 are perfect squares, so $\sqrt{2.25} = \frac{3}{2} = 1.5$ . Check: $1.5 \times 1.5 = 2.25$ .

(g) $0.4 = \frac{4}{10} = \frac{2}{5}$ . Neither 2 nor 5 is a perfect square, so $\sqrt{0.4}$ is irrational. It is not $0.2$ , because $0.2 \times 0.2 = 0.04$ .

(a) Rational: 18
(b) Irrational
(c) Rational: 6
(d) Irrational
(e) Rational: $\frac{2}{3}$
(f) Rational: $1.5$
(g) Irrational

Real Numbers and the Number Line

The rational numbers and the irrational numbers together are called the real numbers.

Definition

A real number is a number that is either rational or irrational. Every real number is represented by a point on the number line, and every point on the number line represents a real number.

Irrational numbers have their places too: $\sqrt{2} = 1.414\ldots$ lies between 1 and 2, and $\pi = 3.141\ldots$ lies just to the right of 3.

You cannot write down every digit of an irrational number, but you can trap it between two numbers that you know. Of two positive numbers, the greater one has the greater square. So, since $9 < 10 < 16$ , we have $\sqrt{9} < \sqrt{10} < \sqrt{16}$ : the number $\sqrt{10}$ lies between 3 and 4.

Estimating a square root without a calculator

Find the perfect squares on either side of the number under the root. Their square roots are the two consecutive integers that the root lies between. Consecutive integers come one after the other, such as 3 and 4.
Square numbers with one decimal place between those two integers, until the number under the root is trapped between two squares.
To give the root correct to 1 decimal place, which means rounded to the nearest tenth, square the number halfway between the two. If that square is less than the number under the root, the root is greater than the halfway number, so round up. If it is more, round down.

A cube root is trapped in the same way, with perfect cubes: $27 < 50 < 64$ , so $\sqrt[3]{50}$ lies between 3 and 4.

Example 3: Estimating a square root without a calculator

(a) Find the two consecutive integers between which $\sqrt{10}$ lies.
(b) Find $\sqrt{10}$ correct to 1 decimal place.

Solution

(a) $3^{2} = 9$ and $4^{2} = 16$ . Since $9 < 10 < 16$ , $\sqrt{10}$ lies between 3 and 4. As 10 is much nearer to 9 than to 16, start just above 3.

(b) $3.1^{2} = 9.61$ , which is less than 10, and $3.2^{2} = 10.24$ , which is more than 10. So $\sqrt{10}$ lies between $3.1$ and $3.2$ .

Halfway between them is $3.15$ , and $3.15^{2} = 9.9225$ , which is less than 10. So $\sqrt{10}$ is greater than $3.15$ : it is nearer to $3.2$ than to $3.1$ .

(a) $\sqrt{10}$ lies between 3 and 4.
(b) $\sqrt{10} = 3.2$ (correct to 1 decimal place). In fact, $\sqrt{10} = 3.1622\ldots$

A negative root is the opposite of the positive one: $-\sqrt{10}$ is as far to the left of 0 as $\sqrt{10}$ is to the right, so $-\sqrt{10}$ lies between $-4$ and $-3$ .

The Kinds of Number

The natural numbers are the positive integers 1, 2, 3, … Together with 0 they are the whole numbers, and the whole numbers together with the negative integers are the integers. In the diagram below, each kind sits inside the next. For example, $-5$ is an integer, so it is also rational ( $-5 = \frac{-5}{1}$ ) and real, but it is not a whole number.

To classify a number, work out its value first if you can. A number belongs to a kind because of its value, not because of the way it is written: $\sqrt{9}$ is the integer 3.

Example 4: Classifying and ordering real numbers

From the list $\frac{0}{4}$ , $-\frac{5}{7}$ , $-123$ , $\sqrt{9}$ , $\sqrt{2}$ , $\sqrt{80}$ , $21$ , write down
(a) the integers,
(b) the rational numbers,
(c) the irrational numbers.
(d) Arrange the numbers in ascending order.

Solution

(a) First work out what you can: $\frac{0}{4} = 0$ and $\sqrt{9} = 3$ , so both are integers.

(b) Every integer is rational, and $-\frac{5}{7}$ is a fraction of two integers.

(d) $1^{2} < 2 < 2^{2}$ , so $\sqrt{2}$ is between 1 and 2. $8^{2} = 64$ and $9^{2} = 81$ , so $\sqrt{80}$ is between 8 and 9. $-\frac{5}{7}$ is between $-1$ and 0.

(a) $-123$ , $\frac{0}{4}$ , $\sqrt{9}$ , $21$
(b) $-123$ , $-\frac{5}{7}$ , $\frac{0}{4}$ , $\sqrt{9}$ , $21$
(c) $\sqrt{2}$ , $\sqrt{80}$
(d) In ascending order: $-123$ , $-\frac{5}{7}$ , $\frac{0}{4}$ , $\sqrt{2}$ , $\sqrt{9}$ , $\sqrt{80}$ , $21$

Comparing and Ordering Real Numbers

Real numbers are compared in the same way as rational numbers: the number further to the right on the number line is the greater. For an irrational number you do not need its decimal. It is enough to trap it between numbers that you know.

Comparing real numbers without a calculator

Work out every root of a perfect square or a perfect cube, and write the fractions as decimals.
Find the two consecutive integers that each irrational number lies between. Often this settles the order.
If two positive numbers lie between the same two integers, compare their squares: the number with the greater square is the greater number.
For two negative numbers, compare them without their signs. The one further from 0 is the smaller.

The square of $\sqrt{6}$ is 6, because that is what the square root means: $\sqrt{6} \times \sqrt{6} = 6$ .

Example 5: Comparing a root with a decimal

Write $<$ or $>$ between the two numbers of each pair.
(a) $\sqrt{6}$ and $2.5$
(b) $-\sqrt{11}$ and $-3.4$
(c) $\pi$ and $\sqrt{10}$

Solution

(a) Both numbers lie between 2 and 3, so compare their squares. The square of $\sqrt{6}$ is 6, and $2.5^{2} = 6.25$ . Since $6 < 6.25$ , $\sqrt{6}$ is the smaller.

(b) Compare $\sqrt{11}$ and $3.4$ first. $3.4^{2} = 11.56$ , which is more than 11, so $3.4$ is greater than $\sqrt{11}$ . Both numbers in the question are negative, and $-3.4$ is the one further from 0, so it is the smaller.

(c) $\pi = 3.14159\ldots$ , which is less than $3.15$ . In Example 3, $3.15^{2} = 9.9225$ , which is less than 10, so $\sqrt{10}$ is greater than $3.15$ .

(a) $\sqrt{6} < 2.5$
(b) $-\sqrt{11} > -3.4$
(c) $\pi < \sqrt{10}$

Example 6: Ordering real numbers

Without a calculator, arrange $\sqrt{17}$ , $4.1$ , $-\pi$ , $4\frac{1}{3}$ , $-3.2$ and $\sqrt[3]{64}$ in descending order.

Solution

Work out what you can: $\sqrt[3]{64} = 4$ and $4\frac{1}{3} = 4.333\ldots$

$16 < 17 < 25$ , so $\sqrt{17}$ lies between 4 and 5. So do $4.1$ and $4.333\ldots$ , so compare squares: $4.1^{2} = 16.81$ , which is less than 17, and $4.3^{2} = 18.49$ , which is more than 17. So $4.1 < \sqrt{17} < 4.3$ , and $4.3$ is less than $4.333\ldots$

The four positive numbers, starting with the greatest, are $4\frac{1}{3}$ , $\sqrt{17}$ , $4.1$ and $\sqrt[3]{64}$ .

The two negative numbers come last. $\pi = 3.14159\ldots$ is less than $3.2$ , so $-3.2$ is further from 0 than $-\pi$ and is the smaller.

In descending order: $4\frac{1}{3}$ , $\sqrt{17}$ , $4.1$ , $\sqrt[3]{64}$ , $-\pi$ , $-3.2$

The Four Operations on Real Numbers

Real numbers are added, subtracted, multiplied and divided by the same rules as rational numbers: the rules of signs and the order of operations stay as they are. What is new is how to write an answer that has an irrational number in it.

Exact answers

An irrational number has no last digit, so any decimal written for it has been cut short or rounded. To give an exact answer, leave $\pi$ or the root sign in it. A question asks for this with the words "give the exact value" or "leave your answer in terms of $\pi$ ", which means with $\pi$ left in the answer.
A number written in front of $\pi$ or of a root multiplies it: $10\pi$ means $10 \times \pi$ , and $4\sqrt{5}$ means $4 \times \sqrt{5}$ . Multiply or divide the other numbers and write the result in front: $2 \times \pi \times 5 = 10\pi$ and $12\pi \div 4 = 3\pi$ .
Multiples of $\pi$ are added and subtracted like any count of the same thing: $10\pi + 6\pi = 16\pi$ , just as 10 tens and 6 tens make 16 tens.
By the meaning of a square root, $\sqrt{5} \times \sqrt{5} = 5$ . In the same way, $(\sqrt{7})^{2} = 7$ .
A root of a perfect square or a perfect cube is worked out at once: write 7 for $\sqrt{49}$ .
Work out what is under a root sign before you take the root: $\sqrt{9 + 16} = \sqrt{25} = 5$ . It is not $3 + 4$ .

When a question gives a value to use, such as "take $\pi = 3.14$ ", put it in at the last step. The answer is then an approximation, close to the exact value but not equal to it.

From primary school: the circumference of a circle is $\pi \times$ diameter, which is $2 \times \pi \times$ radius, and the area of a circle is $\pi \times$ radius $\times$ radius.

Example 7: Exact answers with π

A circular pond has a radius of 5 m.
(a) Find its circumference, leaving your answer in terms of $\pi$ .
(b) Find its area, leaving your answer in terms of $\pi$ .
(c) A path goes round the pond. Its outer edge is a circle of radius 8 m. How much longer is the outer edge of the path than the circumference of the pond? Give the exact value.
(d) Taking $\pi = 3.14$ , find the circumference of the pond.

Solution

(a) Circumference $= 2 \times \pi \times 5 = 10\pi$

(b) Area $= \pi \times 5 \times 5 = 25\pi$

(d) $10\pi$ is $10 \times \pi$ , so with $\pi = 3.14$ the circumference is $10 \times 3.14 = 31.4$ .

(a) The circumference is $10\pi$ m.
(b) The area is $25\pi$ m².
(c) The outer edge is $6\pi$ m longer.
(d) The circumference is about 31.4 m.

Example 8: Roots in a calculation

Evaluate each of the following without a calculator.
(a) $-\sqrt{36} \div \sqrt[3]{8} + \sqrt{\frac{9}{16}}$
(b) $(\sqrt{7})^{2} - 10$
(c) $\sqrt{9 + 16} - \sqrt{9} - \sqrt{16}$
(d) $3\sqrt{5} \times \sqrt{5}$

Solution

(a) Work out the roots first: $\sqrt{36} = 6$ , $\sqrt[3]{8} = 2$ and $\sqrt{\frac{9}{16}} = \frac{3}{4}$ . Then divide before adding: $-6 \div 2 + \frac{3}{4} = -3 + \frac{3}{4} = -2\frac{1}{4}$

(b) $(\sqrt{7})^{2} = 7$ , so the value is $7 - 10 = -3$ .

(c) Under the first root sign, $9 + 16 = 25$ , and $\sqrt{25} = 5$ . So the value is $5 - 3 - 4 = -2$ . This shows that $\sqrt{9 + 16}$ is not equal to $\sqrt{9} + \sqrt{16}$ .

(d) $3\sqrt{5} \times \sqrt{5} = 3 \times \sqrt{5} \times \sqrt{5} = 3 \times 5 = 15$

(a) $-2\frac{1}{4}$
(b) $-3$
(c) $-2$
(d) $15$

Common Mistakes

Treating $\frac{22}{7}$ or $3.14$ as the exact value of $\pi$ . They are approximations of $\pi$ , which is irrational. Use one of them only when a question gives it, as in "take $\pi = 3.14$ ", and the answer is then an approximation too. Do not use one when a question asks for an exact answer or tells you to use the $\pi$ key of a calculator.
Calling every root irrational. $\sqrt{49} = 7$ and $\sqrt[3]{64} = 4$ are rational.
Calling a decimal irrational because it never ends. $0.333\ldots$ never ends, but it recurs, so it is rational. An irrational number neither terminates nor recurs.
Putting negative roots in the wrong order. $\sqrt{10} > 2$ , but $-\sqrt{10} < -2$ , because $-\sqrt{10}$ is further from 0.
Splitting a root over a sum. $\sqrt{9 + 16} = \sqrt{25} = 5$ , not $3 + 4 = 7$ .
Giving a decimal when the exact value is asked for. $10\pi$ is exact, and $31.4$ is only close to it.

Practice

Work each question on paper before you open its solution. No question here needs a calculator.

Question 1

From the list $-5$ , $0$ , $\sqrt{36}$ , $\frac{2}{3}$ , $\sqrt{10}$ , $\sqrt[3]{12}$ , $1.25$ , $\pi$ , write down
(a) the natural numbers,
(b) the integers,
(c) the rational numbers,
(d) the irrational numbers.

ShowHide solution

$\sqrt{36} = 6$ , which is a natural number.

Every integer is rational. $\frac{2}{3}$ is a fraction of two integers and $1.25 = \frac{5}{4}$ , so they are rational too.

10 is not a perfect square, so $\sqrt{10}$ is irrational. 12 is not a perfect cube ( $2^{3} = 8$ and $3^{3} = 27$ ), so $\sqrt[3]{12}$ is irrational. $\pi$ is irrational.

(a) $\sqrt{36}$
(b) $-5$ , $0$ , $\sqrt{36}$
(c) $-5$ , $0$ , $\sqrt{36}$ , $\frac{2}{3}$ , $1.25$
(d) $\sqrt{10}$ , $\sqrt[3]{12}$ , $\pi$

Question 2

Is each statement true or false? Give a reason.
(a) Every integer is a rational number.
(b) Every rational number is an integer.
(c) Every irrational number is a real number.
(d) $\pi$ is equal to $\frac{22}{7}$ .
(e) The decimal $0.272727\ldots$ , in which the digits 27 repeat without end, is an irrational number.
(f) The decimal $0.121221222\ldots$ , in which each group of 2s is one digit longer than the group before it, is an irrational number.

ShowHide solution

(a) An integer can be written as a fraction with denominator 1: for example, $-5 = \frac{-5}{1}$ .

(b) $\frac{1}{2}$ is a rational number, but it is not an integer.

(d) $\frac{22}{7} = 3.142857\ldots$ is rational, and $\pi = 3.141592\ldots$ is irrational. $\frac{22}{7}$ is only an approximation of $\pi$ .

(e) A decimal that recurs is rational. (This one is equal to $\frac{3}{11}$ .)

(f) The digits never end, and no block of digits repeats, because the groups of 2s keep getting longer. A decimal that neither terminates nor recurs is irrational.

(a) True (b) False (c) True (d) False (e) False (f) True

Question 3

Is each number rational or irrational? If it is rational, write down its value.
(a) $\sqrt{441}$
(b) $\sqrt{250}$
(c) $\sqrt[3]{1000}$
(d) $\sqrt[3]{81}$
(e) $\sqrt{\frac{8}{18}}$
(f) $\sqrt{0.09}$

ShowHide solution

(a) $441 = 3^{2} \times 7^{2}$ . Both indices are even, so 441 is a perfect square: $\sqrt{441} = 3 \times 7 = 21$ .

(b) $250 = 2 \times 5^{3}$ . The indices 1 and 3 are odd, so 250 is not a perfect square. (It lies between $15^{2} = 225$ and $16^{2} = 256$ .)

(c) $1000 = 2^{3} \times 5^{3}$ . Both indices are multiples of 3, so 1000 is a perfect cube: $\sqrt[3]{1000} = 2 \times 5 = 10$ .

(d) $81 = 3^{4}$ . The index 4 is not a multiple of 3, so 81 is not a perfect cube, although it is a perfect square.

(e) Simplify first: $\frac{8}{18} = \frac{4}{9}$ . Both 4 and 9 are perfect squares, so $\sqrt{\frac{4}{9}} = \frac{2}{3}$ .

(f) $0.09 = \frac{9}{100}$ . Both 9 and 100 are perfect squares, so $\sqrt{0.09} = \frac{3}{10} = 0.3$ . Check: $0.3 \times 0.3 = 0.09$ .

(a) Rational: 21
(b) Irrational
(c) Rational: 10
(d) Irrational
(e) Rational: $\frac{2}{3}$
(f) Rational: $0.3$

Question 4

(a) Without a calculator, find the two consecutive integers between which $\sqrt{20}$ lies.
(b) Still without a calculator, find $\sqrt{20}$ correct to 1 decimal place.

ShowHide solution

(a) $4^{2} = 16$ and $5^{2} = 25$ . Since $16 < 20 < 25$ , $\sqrt{20}$ is between $\sqrt{16} = 4$ and $\sqrt{25} = 5$ .

(b) 20 is about halfway between 16 and 25, so start near $4.5$ . $4.5^{2} = 20.25$ , which is more than 20, and $4.4^{2} = 19.36$ , which is less than 20. So $\sqrt{20}$ lies between $4.4$ and $4.5$ .

Halfway between them is $4.45$ , and $4.45^{2} = 19.8025$ , which is less than 20. So $\sqrt{20}$ is greater than $4.45$ , and it rounds up.

(a) $\sqrt{20}$ lies between 4 and 5.
(b) $\sqrt{20} = 4.5$ (correct to 1 decimal place)

Question 5

Find the two consecutive integers between which each number lies.
(a) $\sqrt[3]{100}$
(b) $-\sqrt{75}$
(c) $2\pi$

ShowHide solution

(a) $4^{3} = 64$ and $5^{3} = 125$ . Since $64 < 100 < 125$ , $\sqrt[3]{100}$ is between 4 and 5.

(b) $8^{2} = 64$ and $9^{2} = 81$ . Since $64 < 75 < 81$ , $\sqrt{75}$ is between 8 and 9. Its opposite, $-\sqrt{75}$ , is between $-9$ and $-8$ .

(a) Between 4 and 5
(b) Between $-9$ and $-8$
(c) Between 6 and 7

Question 6

The points P, Q, R and S on the number line below represent the numbers $\sqrt{5}$ , $-\sqrt{3}$ , $\pi$ and $-\sqrt{7}$ , but not in that order. Which point represents which number?

ShowHide solution

$1 < 3 < 4$ , so $\sqrt{3}$ lies between 1 and 2, and $-\sqrt{3}$ lies between $-2$ and $-1$ . That is Q.

$4 < 5 < 9$ and $4 < 7 < 9$ , so $\sqrt{5}$ and $\sqrt{7}$ both lie between 2 and 3. So $\sqrt{5}$ is the point between 2 and 3, which is R, and $-\sqrt{7}$ is the point between $-3$ and $-2$ , which is P.

$\pi = 3.14159\ldots$ lies just to the right of 3. That is S.

P is $-\sqrt{7}$ , Q is $-\sqrt{3}$ , R is $\sqrt{5}$ and S is $\pi$ .

Question 7

Without a calculator, write $<$ , $>$ or $=$ between the two numbers of each pair.
(a) $\sqrt{50}$ and $7$
(b) $-\sqrt{17}$ and $-4$
(c) $\sqrt{\frac{9}{16}}$ and $0.75$
(d) $\sqrt{30}$ and $5.5$
(e) $\pi$ and $\frac{22}{7}$

ShowHide solution

(a) $7^{2} = 49$ , which is less than 50, so 7 is less than $\sqrt{50}$ .

(b) $4^{2} = 16$ , which is less than 17, so $\sqrt{17}$ is greater than 4. Both numbers are negative, and $-\sqrt{17}$ is the one further from 0, so it is the smaller.

(d) $5.5^{2} = 30.25$ , which is more than 30, so $5.5$ is greater than $\sqrt{30}$ .

(e) $\pi = 3.14159\ldots$ and $\frac{22}{7} = 3.14285\ldots$ They differ in the third decimal place, where 1 is less than 2.

(a) $\sqrt{50} > 7$
(b) $-\sqrt{17} < -4$
(c) $\sqrt{\frac{9}{16}} = 0.75$
(d) $\sqrt{30} < 5.5$
(e) $\pi < \frac{22}{7}$

Question 8

Without a calculator, arrange the numbers of each list in ascending order.
(a) $\sqrt{30}$ , $-2$ , $\frac{9}{2}$ , $-\sqrt{10}$ , $\sqrt[3]{8}$ , 6
(b) $\sqrt{12}$ , $3.5$ , $-\sqrt{5}$ , $-2.2$ , $\pi$ , $3\frac{1}{3}$

ShowHide solution

(a) First work out what you can: $\sqrt[3]{8} = 2$ and $\frac{9}{2} = 4.5$ .

$5^{2} = 25$ and $6^{2} = 36$ , so $\sqrt{30}$ is between 5 and 6. $3^{2} = 9$ and $4^{2} = 16$ , so $\sqrt{10}$ is between 3 and 4, and $-\sqrt{10}$ is between $-4$ and $-3$ .

So $-\sqrt{10}$ is less than $-2$ , and $\sqrt{30}$ is greater than 4.5 and less than 6.

(b) The two negative numbers: $2.2^{2} = 4.84$ , which is less than 5, so $\sqrt{5}$ is greater than $2.2$ . So $-\sqrt{5}$ is further from 0 than $-2.2$ and is the smaller.

The four positive numbers all lie between 3 and 4. $\pi = 3.14159\ldots$ and $3\frac{1}{3} = 3.333\ldots$ For $\sqrt{12}$ , compare squares: $3.4^{2} = 11.56$ and $3.5^{2} = 12.25$ , so $\sqrt{12}$ is between $3.4$ and $3.5$ .

$3.333\ldots$ is less than $3.4$ , so $\pi < 3\frac{1}{3} < \sqrt{12} < 3.5$ .

(a) $-\sqrt{10}$ , $-2$ , $\sqrt[3]{8}$ , $\frac{9}{2}$ , $\sqrt{30}$ , 6
(b) $-\sqrt{5}$ , $-2.2$ , $\pi$ , $3\frac{1}{3}$ , $\sqrt{12}$ , $3.5$

Question 9

Evaluate each of the following without a calculator. Leave $\pi$ in your answer to (e).
(a) $\sqrt{81} - \sqrt[3]{64} \times \sqrt{\frac{1}{4}}$
(b) $(\sqrt{11})^{2} - 20$
(c) $-\sqrt{\frac{25}{4}} \div \sqrt[3]{8}$
(d) $\sqrt{100 - 36}$
(e) $7\pi - 3\pi + 2\pi$
(f) $2\sqrt{3} \times 5\sqrt{3}$

ShowHide solution

(a) $\sqrt{81} = 9$ , $\sqrt[3]{64} = 4$ and $\sqrt{\frac{1}{4}} = \frac{1}{2}$ . Multiply before subtracting: $9 - 4 \times \frac{1}{2} = 9 - 2 = 7$

(b) $(\sqrt{11})^{2} = 11$ , so the value is $11 - 20 = -9$ .

(c) $\sqrt{\frac{25}{4}} = \frac{5}{2}$ and $\sqrt[3]{8} = 2$ . So the value is $-\frac{5}{2} \div 2 = -\frac{5}{2} \times \frac{1}{2} = -\frac{5}{4} = -1\frac{1}{4}$ .

(d) Work out what is under the root sign first: $100 - 36 = 64$ , and $\sqrt{64} = 8$ . It is not $10 - 6$ .

(e) 7 lots of $\pi$ , take away 3 lots, add 2 lots: $7 - 3 + 2 = 6$ lots of $\pi$ .

(f) Multiply in any order: $2 \times 5 \times \sqrt{3} \times \sqrt{3} = 10 \times 3 = 30$

(a) $7$
(b) $-9$
(c) $-1\frac{1}{4}$
(d) $8$
(e) $6\pi$
(f) $30$

Question 10

A circular flower bed has a radius of 6 m.
(a) Find its circumference, leaving your answer in terms of $\pi$ .
(b) Find its area, leaving your answer in terms of $\pi$ .
(c) A gardener has 37 m of edging strip. Without a calculator, decide whether this is enough to go all the way round the flower bed.
(d) The flower bed is divided into 4 equal parts, one for each of four kinds of flower. Find the exact area of each part.

ShowHide solution

(a) Circumference $= 2 \times \pi \times 6 = 12\pi$

(b) Area $= \pi \times 6 \times 6 = 36\pi$

(c) $\pi = 3.14159\ldots$ is greater than $3.1$ , so $12\pi$ is greater than $12 \times 3.1 = 37.2$ . The circumference is more than 37.2 m, which is more than 37 m.

(d) Area of each part $= 36\pi \div 4 = 9\pi$

(a) The circumference is $12\pi$ m.
(b) The area is $36\pi$ m².
(c) No. The strip is too short.
(d) Each part has an area of $9\pi$ m².

Question 11

A square tile has an area of 45 cm².
(a) Write down the exact length of a side of the tile.
(b) Between which two consecutive whole numbers of centimetres does this length lie?
(c) A square frame measures 6.5 cm along each inside edge. Without a calculator, decide whether the tile fits inside the frame.
(d) Find the exact perimeter of the tile.

ShowHide solution

(a) Side $\times$ side $= 45$ , so the side is $\sqrt{45}$ cm. 45 is not a perfect square, so this number is irrational, and the exact value keeps its root sign.

(b) $6^{2} = 36$ and $7^{2} = 49$ . Since $36 < 45 < 49$ , $\sqrt{45}$ is between 6 and 7.

(c) $6.5^{2} = 42.25$ , which is less than 45, so $6.5$ is less than $\sqrt{45}$ . The side of the tile is longer than 6.5 cm.

(d) Perimeter $= 4 \times \sqrt{45} = 4\sqrt{45}$

(a) A side is $\sqrt{45}$ cm long.
(b) Between 6 cm and 7 cm
(c) No. The tile is too big for the frame.
(d) The perimeter is $4\sqrt{45}$ cm.

Question 12

In a game, four friends each turn over a card with a number on it. Hui Min has $\sqrt{40}$ , Arjun has $6.3$ , Sofia has $2\pi$ and Danish has $6.4$ . The greatest number wins.
(a) Without a calculator, arrange the four numbers in descending order.
(b) Who wins, and who has the smallest number?

ShowHide solution

(a) $6.3^{2} = 39.69$ , which is less than 40, and $6.4^{2} = 40.96$ , which is more than 40. So $6.3 < \sqrt{40} < 6.4$ .

$\pi = 3.14159\ldots$ is less than $3.15$ , so $2\pi$ is less than $2 \times 3.15 = 6.3$ .

So $2\pi < 6.3 < \sqrt{40} < 6.4$ .

(a) In descending order: $6.4$ , $\sqrt{40}$ , $6.3$ , $2\pi$
(b) Danish wins, and Sofia has the smallest number.

Key Points

An irrational number cannot be written as $\frac{a}{b}$ with $a$ and $b$ integers and $b \neq 0$ . Its decimal neither terminates nor recurs.
$\sqrt{2}$ , $\sqrt[3]{5}$ and $\pi$ are irrational; $\sqrt{4}$ , $\sqrt[3]{27}$ and $\sqrt{\frac{9}{16}}$ are not. $\frac{22}{7}$ and $3.14$ are only approximations of $\pi$ .
The square root of a whole number is rational only if the number is a perfect square, and its cube root only if the number is a perfect cube. Simplify a fraction before you check its numerator and denominator.
The real numbers are the rational numbers and the irrational numbers together. Each has its own point on the number line.
Natural numbers are whole numbers, whole numbers are integers, integers are rational numbers, and rational numbers are real numbers.
Trap a root between numbers that you know: $9 < 10 < 16$ , so $\sqrt{10}$ lies between 3 and 4, and $3.1^{2} < 10 < 3.2^{2}$ , so it lies between $3.1$ and $3.2$ .
Of two positive numbers, the one with the greater square is the greater. Of two negative numbers, the one further from 0 is the smaller.
An exact answer keeps $\pi$ or the root sign: $2 \times \pi \times 5 = 10\pi$ and $\sqrt{5} \times \sqrt{5} = 5$ .

Real Numbers

Before you start

Irrational Numbers

Definition

Is π\piπ equal to 227\frac{22}{7}722​?

Example 1: Rational or irrational?

Is a Root Rational? Look at the Number under It

Checking the number under the root

Example 2: Checking the number under the root

Real Numbers and the Number Line

Definition

Estimating a square root without a calculator

Example 3: Estimating a square root without a calculator

The Kinds of Number

Example 4: Classifying and ordering real numbers

Comparing and Ordering Real Numbers

Comparing real numbers without a calculator

Example 5: Comparing a root with a decimal

Example 6: Ordering real numbers

The Four Operations on Real Numbers

Exact answers

Example 7: Exact answers with π

Example 8: Roots in a calculation

Common Mistakes

Practice

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Key Points

Is $\pi$ equal to $\frac{22}{7}$ ?