MathsICSE Class 9 · Rational and Irrational Numbers

Rational and Irrational Numbers: Recurring Decimals to p/q and the Proof that √2 Is Irrational (ICSE Class 9)

Q: Is the √2 proof in Class IX?

Yes, the chapter includes the proof by contradiction.

Q: Can the sum of two irrationals be rational?

Yes, for example √2 and −√2.

Updated 17 October 2026 · Fact-checked

A rational number can be written as p/q (q ≠ 0); its decimal terminates or recurs. An irrational number has a non-terminating, non-recurring decimal. To convert a recurring decimal, multiply by 10ⁿ (n digits in the repeating block), subtract, and divide. √2 is irrational, proved by assuming it equals p/q in lowest terms and reaching a contradiction.

Rational decimals stop or repeat; irrational ones like √2 and π never stop and never repeat.

Illustration drawn for QuestionBuddy.

Understand Rational and Irrational Numbers, Recurring Decimals and the Irrationality of √2

Rational and irrational. A rational number is $\frac{p}{q}$ with integers $p, q$ , $q \neq = 0$ . Its decimal expansion either terminates ( $\frac{3}{8} = 0.375$ ) or recurs ( $\frac{1}{3} = 0. \overline{3}$ ). An irrational number such as $2$ , $7$ , $π$ or $0.101001000 \dots$ has a decimal that neither terminates nor recurs. Together they make the real numbers. Note $49 = 7$ and $0.09 = 0.3$ are rational; $n$ is irrational unless $n$ is a perfect square (of an integer or a rational).

Pure recurring decimal. $x = 0. \overline{36}$ : two digits repeat, so $100 x = 36. \overline{36}$ . Subtract: $99 x = 36$ , $x = \frac{36}{99} = \frac{4}{11}$ . In general, a block of $n$ repeating digits gives denominator $99 \dots 9$ ( $n$ nines).

Mixed recurring decimal. $x = 0.7 \overline{3}$ : multiply by 10 to bring the repeating part right after the point: $10 x = 7. \overline{3}$ and $100 x = 73. \overline{3}$ . Subtract: $90 x = 66$ , $x = \frac{66}{90} = \frac{11}{15}$ . Shortcut: $\frac{( all digits ) - ( non-repeating digits )}{9 \dots 90 \dots 0}$ , with as many 9s as repeating digits and as many 0s as non-repeating digits. Check by dividing: $\frac{11}{15} = 0.7333 \dots$

Irrational between two numbers. Between $2$ and $3$ : $5$ and $6$ (since $4 < 5 < 6 < 9$ ).

Combining. Rational $+$ irrational is irrational; non-zero rational $\times$ irrational is irrational. Irrational $+$ irrational can be rational ( $2 + (- 2) = 0$ ).

**Proof that $2$ is irrational (contradiction).** Suppose $2 = \frac{p}{q}$ with integers $p, q$ having no common factor. Then $p^{2} = 2 q^{2}$ , so $p^{2}$ is even, so $p$ is even: $p = 2 k$ . Then $4 k^{2} = 2 q^{2}$ , so $q^{2} = 2 k^{2}$ is even and $q$ is even. Now $p$ and $q$ are both even, contradicting 'no common factor'. So $2$ is irrational.

Level. Class IX is a school-exam class. Mixed recurring decimals and the contradiction proof (with a follow-up like $3 + 2$ ) are at the hardest school-exam level.

Rational and Irrational Numbers, Recurring Decimals and the Irrationality of √2: formulas to remember

Rational: $\frac{p}{q}, q \neq = 0$; Terminating or recurring decimal.
Pure recurring: $0. \overline{ab} = \frac{ab}{99}$; Two repeating digits.
Mixed recurring: $0. a \overline{b} = \frac{ab - a}{90}$; One non-repeating, one repeating.
Rational + irrational: $rational + irrational = irrational$; Always.

How to solve Rational and Irrational Numbers, Recurring Decimals and the Irrationality of √2 questions

Name the repeating block, multiply to align it, subtract, then simplify the fraction.

1Write the decimal as $x$ and identify the repeating block.
2Multiply by $1 0^{n}$ so that the block lines up; for a mixed decimal multiply twice.
3Subtract to cancel the repeating tail.
4Solve for $x$ and cancel common factors.
5Check by dividing the fraction.
6For proofs, state the assumption, derive the contradiction, conclude.

Rational and Irrational Numbers, Recurring Decimals and the Irrationality of √2 in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Repeating digits in a unit price

Suppose a shopkeeper's calculator shows a price per item as $0. \overline{36}$ rupees after dividing a sum equally.

Write the exact fraction.

Try it first, then show the solutionSolution

$100 x - x = 36. \overline{36} - 0. \overline{36} = 36$ , so $99 x = 36$ .
$x = \frac{36}{99} = \frac{4}{11}$ .

Answer: $\frac{4}{11}$ rupee.

Memory hook: Multiply by 100, subtract.

A cyclic remainder pattern

Imagine dividing 1 by 7 on a calculator and seeing 0.142857142857...

Write it as a fraction using the block.

Try it first, then show the solutionSolution

The block has 6 digits: $x = \frac{142857}{999999}$ .
Dividing both by 142857 gives $\frac{1}{7}$ .

Answer: $\frac{1}{7}$

Memory hook: Six digits repeating means six 9s.

Tricks to remember Rational and Irrational Numbers, Recurring Decimals and the Irrationality of √2

Short memory aids. Say each one aloud once, then use it on a question the same day.

n digits, n nines
A repeating block of n digits goes over n nines.
Use it when: Pure recurring decimals.
Take away the tail
Subtract so the infinite tail vanishes.
Use it when: Every conversion.
Assume, square, evens
Contradiction proof: p² = 2q² makes both even.
Use it when: Irrationality of √2.

Rational and Irrational Numbers, Recurring Decimals and the Irrationality of √2: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Recurring decimal · 1 mark · MCQBoard level

$0. \overline{3}$ is equal to:
(a) $\frac{3}{10}$
(b) $\frac{1}{3}$
(c) $\frac{3}{100}$
(d) $\frac{33}{100}$

Try it first, then show the solutionSolution

$10 x - x = 3$ , so $x = \frac{1}{3}$ .

Answer: (b)

Example 2: Pure recurring · 3 marks · Short answerBoard level

Express $0. \overline{36}$ in the form $\frac{p}{q}$ .

Try it first, then show the solutionSolution

$x = 0. \overline{36}$ ; $100 x = 36. \overline{36}$ .
$99 x = 36$ , so $x = \frac{36}{99} = \frac{4}{11}$ .

Answer: $\frac{4}{11}$

Example 3: Mixed recurring · 3 marks · Short answerHarder board level

Express $0.7 \overline{3}$ as a fraction in lowest terms and check by division.

Try it first, then show the solutionSolution

$x = 0.7333 \dots$ ; $10 x = 7. \overline{3}$ ; $100 x = 73. \overline{3}$ .
$100 x - 10 x = 66$ , so $90 x = 66$ and $x = \frac{11}{15}$ .
Check: $11 \div 15 = 0.7333 \dots$ ✓

Answer: $\frac{11}{15}$

Example 4: Classification and insertion · 4 marks · Short answerHarder board level

State, with reasons, which of $49$ , $0.09$ , $7$ , $\frac{22}{7}$ are rational. Insert two irrational numbers between $2$ and $3$ .

Try it first, then show the solutionSolution

$49 = 7$ and $0.09 = 0.3 = \frac{3}{10}$ are rational; $\frac{22}{7}$ is already $\frac{p}{q}$ , so rational; $7$ is not a perfect square, so irrational.
$2 = 4$ and $3 = 9$ , so $5$ and $6$ lie between them and are irrational.

Answer: Rational: $49$ , $0.09$ , $\frac{22}{7}$ . Irrational numbers between 2 and 3: $5$ , $6$ .

Example 5: Proof of irrationality · 5 marks · Long answerHOTS / competency

Prove that $2$ is irrational. Hence prove that $3 + 2$ is irrational.

Try it first, then show the solutionSolution

Suppose $2 = \frac{p}{q}$ with integers $p, q$ having no common factor and $q \neq = 0$ .
Then $p^{2} = 2 q^{2}$ , so $p^{2}$ is even, hence $p$ is even; let $p = 2 k$ .
$4 k^{2} = 2 q^{2}$ gives $q^{2} = 2 k^{2}$ , so $q$ is even too.
Both $p$ and $q$ are even, contradicting 'no common factor'. So $2$ is irrational.
Suppose $3 + 2 = \frac{a}{b}$ is rational. Then $2 = \frac{a}{b} - 3 = \frac{a - 3 b}{b}$ is rational, a contradiction. So $3 + 2$ is irrational.

Answer: Both proved by contradiction.

Where students lose marks in Rational and Irrational Numbers, Recurring Decimals and the Irrationality of √2

The slips examiners see most often, and what to write instead.

Mistake: Wrong denominator for a mixed decimal
The non-repeating digit adds a zero.
Wrong working
$0.7 \overline{3} = \frac{73}{99}$ .
Right working
$\frac{73 - 7}{90} = \frac{11}{15}$ .
Do this: Multiply twice, subtract.
Usually costs: Whole answer.
Mistake: Not simplifying
Leave the fraction in lowest terms.
Wrong working
$\frac{36}{99}$ .
Right working
$\frac{4}{11}$ .
Do this: Divide by the HCF.
Usually costs: One mark.
Mistake: Skipping the contradiction
The final line must say which assumption fails.
Wrong working
$p$ and $q$ are both even.
Right working
This contradicts that p and q have no common factor, so √2 is irrational.
Do this: Always conclude.
Usually costs: Final mark.

Exam tips

Class IX is examined in school papers; the proof of √2 is a standard 4 to 5 mark question.
State "p and q have no common factor" as the assumption.
Write the subtraction step in full when converting decimals.
Check conversions by dividing the fraction.

Practice questions on Rational and Irrational Numbers, Recurring Decimals and the Irrationality of √2

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which of the following is irrational?
(a) $25$
(b) $0.25$
(c) $\frac{22}{7}$
(d) $7$
Hint
7 is not a perfect square.
Answer
(d) $7$
Q2 · 3 marks · Short answerHarder board level
Write $2. \overline{18}$ as a fraction in lowest terms.
Hint
$2 + 0. \overline{18}$ .
Answer
$0. \overline{18} = \frac{18}{99} = \frac{2}{11}$ , so $2. \overline{18} = \frac{24}{11}$ .
Q3 · 4 marks · Short answerHarder board level
Express $0.2 \overline{18}$ in the form $\frac{p}{q}$ .
Hint
Multiply by 10 and by 1000.
Answer
$1000 x - 10 x = 218 - 2 = 216$ , so $990 x = 216$ and $x = \frac{216}{990} = \frac{12}{55}$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The sum of a rational number and an irrational number is irrational.
Reason (R): If $r + 2$ were rational, then $2$ would be a difference of two rationals and so rational.
(a) Both (A) and (R) are true and (R) explains (A).
(b) Both are true but (R) does not explain (A).
(c) (A) is true, (R) is false.
(d) (A) is false, (R) is true.
Hint
Contradiction argument.
Answer
(a)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: A student lists three decimals from a worksheet: $0.285714285714 \dots$ , $0.8333 \dots$ and $0.2020020002 \dots$ (the number of zeros after each 2 grows).
(a) Express $0. \overline{285714}$ as a fraction in lowest terms.
(b) Express $0.8 \overline{3}$ as a fraction in lowest terms.
(c) Say, with a reason, whether the third number is rational.
Hint
Look at the pattern of the repeating block.
Answer
(a) $\frac{285714}{999999} = \frac{2}{7}$ (b) $\frac{83 - 8}{90} = \frac{75}{90} = \frac{5}{6}$ (c) Irrational: the decimal never terminates and no block repeats because the gaps keep growing.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define a rational number.Show answer
A number of the form p/q with q not zero.
2.How does the decimal of an irrational number behave?Show answer
It neither terminates nor recurs.
3.0.(3) as a fraction?Show answer
1/3.
4.Is √49 rational?Show answer
Yes, it equals 7.

Rational and Irrational Numbers, Recurring Decimals and the Irrationality of √2: frequently asked questions

Is the √2 proof in Class IX?

Yes, the chapter includes the proof by contradiction.

Can the sum of two irrationals be rational?

Yes, for example √2 and −√2.

Why does an even square mean an even number?

An odd number squared is odd.