MathsCBSE Class 9 · The World of Numbers

Irrational Numbers, Proofs and the Square Root Spiral: Class 9

Q: Why is the decimal of an irrational number non-repeating?

A repeating or terminating decimal can always be written as a fraction, so a number whose decimal never repeats cannot be a fraction.

Q: Is every square root irrational?

No. Square roots of perfect squares (and of squares of fractions) are rational; the square root of any prime is irrational.

Q: How many irrational numbers lie between 2 and 3?

Infinitely many, as are the rational numbers between them.

Updated 17 October 2026 · Fact-checked

A rational number can be written as $\frac{p}{q}$ with integers $p$, $q$ and $q \neq 0$; an irrational number cannot. $\sqrt{2}$ and $\sqrt{3}$ are irrational, proved by contradiction: assume $\sqrt{n} = \frac{p}{q}$ in lowest terms and show that $p$ and $q$ share a factor. Sums or products of a non-zero rational and an irrational are irrational, but two irrationals can give a rational result. The square root spiral shows $\sqrt{n}$ as a length.

Understand Irrational Numbers, Proofs and the Square Root Spiral

Rational and irrational numbers. A rational number is one that can be written as $\frac{p}{q}$ with $p, q$ integers and $q \neq = 0$ : integers, fractions, terminating decimals and repeating decimals. An irrational number cannot be written this way; its decimal expansion neither ends nor repeats. $16 = 4$ , $\frac{9}{25} = \frac{3}{5}$ and $\frac{22}{7}$ are rational ( $\frac{22}{7}$ is only an approximation of $π$ ). $18 = 32$ , $12 = 23$ and $0.101001000100001 \dots$ (no repeating block) are irrational.

**Proof that $3$ is irrational.** Suppose $3 = \frac{p}{q}$ with $p$ , $q$ integers having no common factor (lowest terms) and $q \neq = 0$ . Squaring: $p^{2} = 3 q^{2}$ , so $3$ divides $p^{2}$ . Since $3$ is prime, $3$ divides $p$ ; write $p = 3 k$ . Then $9 k^{2} = 3 q^{2}$ , so $q^{2} = 3 k^{2}$ and $3$ divides $q$ . Now $3$ divides both $p$ and $q$ , contradicting that they have no common factor. So $3$ is irrational. The same argument works for $2$ (with the prime $2$ ) and for the square root of any prime.

Sums and products. If $r$ is a non-zero rational and $x$ is irrational, then $r + x$ , $r - x$ , $r x$ and $\frac{x}{r}$ are irrational. For example, if $53$ were rational, say $= r$ , then $3 = \frac{r}{5}$ would be rational, a contradiction; and if $3 + 25 = r$ , then $5 = \frac{r - 3}{2}$ would be rational. Two irrationals can, however, combine into a rational: $(2 + 3) + (2 - 3) = 4$ ; $2 \times 8 = 4$ ; $(3 + 5) (3 - 5) = 9 - 5 = 4$ . But $(2 + 3)^{2} = 5 + 26$ is irrational.

Numbers between two numbers. Between any two distinct real numbers there are infinitely many rational and irrational numbers. Between $2$ and $3$ : the irrational numbers $5 \approx 2.236$ and $7 \approx 2.646$ (since $4 < 5 < 9$ and $4 < 7 < 9$ ). Between $2$ and $3$ ( $1.414$ and $1.732$ ): the rational number $1.5$ . Between $\frac{3}{5}$ and $\frac{4}{5}$ (that is $0.6$ and $0.8$ ): the rational $0.7$ and the irrational $0.70100100010 \dots$

Square root spiral. Start with a right triangle with both legs $1$ : its hypotenuse is $2$ . Build the next right triangle on that hypotenuse with a perpendicular leg of length $1$ : its hypotenuse is $(2)^{2} + 1 = 3$ , and so on. The $n$ th triangle has legs $n$ and $1$ and hypotenuse $n + 1$ . Among the first $15$ triangles the hypotenuse is a whole number when $n + 1$ is a perfect square: $n = 3$ (hypotenuse $2$ ), $n = 8$ (hypotenuse $3$ ) and $n = 15$ (hypotenuse $4$ ). To place a single square root on the number line, use a right triangle: $5 = 2^{2} + 1^{2}$ (legs $2$ and $1$ ), $13 = 3^{2} + 2^{2}$ .

**Marking $5$ on the number line.** Mark $O$ at $0$ and $A$ at $2$ ; at $A$ draw a perpendicular $A B = 1$ ; join $O B$ , so $O B = 5$ ; with centre $O$ and radius $O B$ draw an arc cutting the number line at $P$ , so $O P = 5 \approx 2.236$ .

Exam approach. Classify with a reason (perfect square, non-repeating decimal, $\frac{p}{q}$ form); in proofs use "assume rational in lowest terms, find a common factor, contradiction"; and for "between" questions give an explicit number and a reason it lies in the interval.

Irrational Numbers, Proofs and the Square Root Spiral: formulas to remember

Rational number: $\frac{p}{q}, p, q \in Z, q \neq = 0$; Terminating or repeating decimals.
Irrational number: $non-terminating, non-repeating decimal$; Such as $2$ , $3$ , $π$ .
Square of a prime root: $p prime, p ∣ a^{2} \Rightarrow p ∣ a$; Key step in the proofs.
Rational plus irrational: $r \in Q, x \in / Q \Rightarrow r + x \in / Q$; Also for $r x$ with $r \neq = 0$ .
Spiral lengths: $hypotenuse of triangle n = n + 1$; Legs $n$ and $1$ .
Diagonal of a unit square: $1^{2} + 1^{2} = 2$; Irrational length.

How to solve Irrational Numbers, Proofs and the Square Root Spiral questions

Classify first; prove by contradiction; place numbers using right triangles.

1Simplify the number (take out perfect squares) and check for a repeating decimal or a perfect square.
2To prove a root irrational, assume it equals $\frac{p}{q}$ in lowest terms, square, and use the prime divisibility argument.
3For a sum, product or difference with a rational, assume the result is rational and rearrange to isolate the root.
4For numbers between two numbers, use decimals or squares: choose perfect squares around a surd.
5For the spiral, add one unit leg at a time and square the lengths: $(hyp_{n})^{2} = n + 1$ .
6To place a root on the number line, write the number under the root as a sum of two squares.

Irrational Numbers, Proofs and the Square Root Spiral 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.

An irrational length

Suppose a tile 1 metre by 1 metre needs a decorative strip along its diagonal.

How long is the strip, and why can a tape never show its length exactly?

Try it first, then show the solutionSolution

Diagonal $= 1^{2} + 1^{2} = 2$ m.
$2$ is irrational: its decimal expansion $1.41421 \dots$ neither ends nor repeats, so any tape reading $1.414$ is only an approximation.

Answer: The strip is $2$ m, about 1.414 m, and its exact length cannot be written as a fraction.

Memory hook: Squares of fractions never equal 2.

Tricks to remember Irrational Numbers, Proofs and the Square Root Spiral

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

Prime divides the square
If a prime divides $p^{2}$ it divides $p$ ; this is the key step in every root proof.
Use it when: Contradiction proofs.
Isolate the root
To show $a + b n$ is irrational, assume it is rational and solve for $n$ .
Use it when: Sums and products with surds.
Squares on either side
To squeeze a surd between whole numbers, look for the perfect squares around it.
Use it when: Locating $5$ between 2 and 3.

Irrational Numbers, Proofs and the Square Root Spiral: solved examples

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

Example 1: Classify · 3 marks · Short answerBoard level

State, with reasons, whether each is rational or irrational: (a) $16$ (b) $18$ (c) $0.101001000100001 \dots$ (d) $\frac{22}{7}$ (e) $\frac{9}{25}$ .

Try it first, then show the solutionSolution

(a) $16 = 4 = \frac{4}{1}$ : rational. (½)
(b) $18 = 32$ , a non-zero rational times the irrational $2$ : irrational. (1)
(c) The decimal neither ends nor repeats (the number of zeros keeps increasing): irrational. (½)
(d) $\frac{22}{7}$ is a ratio of integers: rational (it only approximates $π$ ). (½)
(e) $\frac{9}{25} = \frac{3}{5}$ : rational. (½)

Answer: (a), (d), (e) are rational; (b) and (c) are irrational.

Example 2: Numbers between · 3 marks · Short answerBoard level

Insert two irrational numbers between 2 and 3, and a rational number between $2$ and $3$ . Give reasons.

Try it first, then show the solutionSolution

$4 < 5 < 9$ and $4 < 7 < 9$ , so $2 < 5 < 3$ and $2 < 7 < 3$ ; both are irrational. (1½)
$2 \approx 1.414$ and $3 \approx 1.732$ , so $1.5$ lies between them. (1)
Check: $1. 5^{2} = 2.25$ , which is between $2$ and $3$ , so $2 < 1.5 < 3$ . (½)

Answer: For example $5$ and $7$ ; and $1.5$ .

Example 3: Prove $\sqrt{3}$ irrational · 5 marks · Long answerHarder board level

Prove that $3$ is irrational. Hence prove that $53$ and $2 + 3$ are irrational.

Try it first, then show the solutionSolution

Assume $3 = \frac{p}{q}$ with $p, q$ integers, $q \neq = 0$ , no common factor. Then $p^{2} = 3 q^{2}$ . (1)
So $3 ∣ p^{2}$ ; since $3$ is prime, $3 ∣ p$ . Write $p = 3 k$ . (1)
$9 k^{2} = 3 q^{2}$ gives $q^{2} = 3 k^{2}$ , so $3 ∣ q$ . Then $3$ is a common factor of $p$ and $q$ : a contradiction. So $3$ is irrational. (1)
If $53 = r$ were rational, $3 = \frac{r}{5}$ would be rational: contradiction. (1)
If $2 + 3 = r$ were rational, $3 = r - 2$ would be rational: contradiction. (1)

Answer: $3$ , $53$ and $2 + 3$ are irrational.

Example 4: Sum and product of irrationals · 3 marks · Short answerHarder board level

Show by examples that the sum, the product and the difference of two irrational numbers can be rational, but that the square of $2 + 3$ is irrational.

Try it first, then show the solutionSolution

Sum: $(2 + 3) + (2 - 3) = 4$ , rational. (½)
Product: $2 \times 8 = 16 = 4$ ; also $(3 + 5) (3 - 5) = 9 - 5 = 4$ . (1)
Difference: $(5 + 2) - (3 + 2) = 2$ . (½)
$(2 + 3)^{2} = 2 + 3 + 26 = 5 + 26$ , a rational plus a non-zero rational multiple of the irrational $6$ , so irrational. (1)

Answer: Rational results are possible; $5 + 26$ is irrational.

Example 5: The square root spiral · 4 marks · Case-basedHOTS / competency

Case: a square root spiral is built with right triangles. The first has legs 1 and 1; each next triangle has one leg equal to the previous hypotenuse and the other leg equal to 1. (i) Find the hypotenuse of the 8th triangle. (ii) Which triangle has hypotenuse $10$ ? (iii) Find the area of the 4th triangle. (iv) How many of the first 15 triangles have a whole-number hypotenuse?

Try it first, then show the solutionSolution

(i) The hypotenuse of triangle $n$ is $n + 1$ , so for $n = 8$ it is $9 = 3$ . (1)
(ii) $n + 1 = 10$ gives $n = 9$ : the 9th triangle. (1)
(iii) The 4th triangle has legs $4 = 2$ and $1$ : area $= \frac{1}{2} \times 2 \times 1 = 1$ . (1)
(iv) $n + 1 \in {4, 9, 16}$ gives $n = 3, 8, 15$ : three triangles (hypotenuses $2$ , $3$ , $4$ ). (1)

Answer: (i) 3; (ii) the 9th; (iii) 1; (iv) 3 triangles.

Where students lose marks in Irrational Numbers, Proofs and the Square Root Spiral

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

Mistake: Not assuming lowest terms
Missing the contradiction.
Wrong working
$p^{2} = 3 q^{2}$ so $3 ∣ p$ and $3 ∣ q$ , done.
Right working
Assume $p$ and $q$ have no common factor; then $3$ dividing both contradicts it.
Do this: State "in lowest terms" in the first line.
Usually costs: 2 marks.
Mistake: Calling a terminating decimal irrational
Mistaking a long decimal for a non-terminating one.
Wrong working
$3.14159$ is irrational.
Right working
$3.14159$ terminates, so it is rational; $π$ is irrational.
Do this: Irrational means the decimal never ends or repeats.
Usually costs: 1 mark.
Mistake: Assuming sums of irrationals are irrational
Overgeneralising.
Wrong working
$(2 + 3) + (2 - 3)$ is irrational.
Right working
It equals $4$ , which is rational.
Do this: Test by simplifying before classifying.
Usually costs: 1 mark.

Exam tips

Write "assume ... in lowest terms" at the start of every irrationality proof.
Simplify surds before classifying them.
Use perfect squares to place a surd between whole numbers.
In spiral questions, track the squares of the hypotenuses: they go 2, 3, 4, 5, ...

Practice questions on Irrational Numbers, Proofs and the Square Root Spiral

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) $0.3333 \dots$
(b) $0.25$
(c) $12$
(d) $3.14$
Hint
Simplify each number or identify the repeating block.
Answer
(c) $12$ . $12 = 23$ is irrational; $0.333 \dots = \frac{1}{3}$ , $0.25 = 0.5$ and $3.14 = \frac{314}{100}$ are rational.
Q2 · 3 marks · Short answerBoard level
Find a rational and an irrational number between $\frac{1}{7}$ and $\frac{2}{7}$ .
Hint
Write the end values as decimals: $\frac{1}{7} = 0.142857 \dots$ and $\frac{2}{7} = 0.285714 \dots$ .
Answer
$\frac{1}{7} \approx 0.1428$ and $\frac{2}{7} \approx 0.2857$ . Rational: $0.2 = \frac{1}{5}$ , which lies between them. Irrational: $0.2010010001 \dots$ (the number of zeros keeps increasing, so it does not repeat), which also lies between $0.1428$ and $0.2857$ .
Q3 · 3 marks · Short answerHarder board level
Given that $5$ is irrational, prove that $3 + 25$ is irrational.
Hint
Assume it is rational and isolate $5$ .
Answer
Suppose $3 + 25 = r$ with $r$ rational. Then $5 = \frac{r - 3}{2}$ , and a rational number divided by 2 after subtracting 3 is rational. So $5$ would be rational, contradicting the given fact. Hence $3 + 25$ is irrational.
Q4 · 5 marks · Long answerHarder board level
Describe how to represent $5$ on the number line using a ruler and compass, and prove that your length is $5$ . Then state which two whole numbers it lies between and why.
Hint
Write $5 = 2^{2} + 1^{2}$ and use the Baudhāyana–Pythagoras theorem.
Answer
Mark $O = 0$ and $A = 2$ on the line. At $A$ draw a perpendicular $A B$ with $A B = 1$ . Join $O B$ . By the Baudhāyana–Pythagoras theorem $O B^{2} = O A^{2} + A B^{2} = 4 + 1 = 5$ , so $O B = 5$ . With centre $O$ and radius $O B$ draw an arc meeting the number line at $P$ ; then $O P = 5$ . Since $2^{2} = 4 < 5 < 9 = 3^{2}$ , $5$ lies between 2 and 3 (about 2.236).
Q5 · 4 marks · Case-basedHOTS / competency
Case: in a square root spiral the 3rd right triangle has legs $3$ and $1$ . (i) Find its hypotenuse. (ii) Find its perimeter (use $3 = 1.732$ ). (iii) Find its area. (iv) Is the area rational? Give a reason.
Hint
The third triangle has legs $3$ and $1$ .
Answer
(i) $3 + 1 = 2$ . (ii) Perimeter $= 3 + 1 + 2 = 3 + 3 \approx 4.732$ units. (iii) Area $= \frac{1}{2} \times 3 \times 1 = \frac{3}{2} \approx 0.866$ square units. (iv) No: if $\frac{3}{2}$ were rational, $3$ would be rational, but $3$ is irrational.

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.Is $16$ rational?Show answer
Yes, it equals 4.
2.Is $π$ rational?Show answer
No, it is irrational; $\frac{22}{7}$ is only an approximation.
3.Key step in proving $3$ irrational?Show answer
A prime that divides $p^{2}$ divides $p$ .
4.Hypotenuse of the nth spiral triangle?Show answer
$n + 1$ .
5.Can two irrationals multiply to a rational?Show answer
Yes, for example $2 \times 8 = 4$ .

Irrational Numbers, Proofs and the Square Root Spiral: frequently asked questions

Why is the decimal of an irrational number non-repeating?

A repeating or terminating decimal can always be written as a fraction, so a number whose decimal never repeats cannot be a fraction.

Is every square root irrational?

No. Square roots of perfect squares (and of squares of fractions) are rational; the square root of any prime is irrational.

How many irrational numbers lie between 2 and 3?

Infinitely many, as are the rational numbers between them.