MathsCBSE Class 10 · Real Numbers

Proving Numbers Irrational: Class 10 Real Numbers

Updated 14 October 2026 · Fact-checked

Proving a number irrational means showing it cannot be written as p over q, with p and q integers and q not zero. You prove root 2, root 3 and root 5 irrational by contradiction, then expressions like 3 plus 2 root 5. Questions can appear as 3-mark proofs, MCQs, assertion-reason items or part of a case study.

Follow the arrows: assuming √p = a/b makes p divide both a and b, which clashes with co-prime, so the assumption fails (p must be prime).

Illustration drawn for QuestionBuddy.

Understand Proving Numbers Irrational

A rational number can be written as $\frac{p}{q}$ where p and q are integers and $q \neq = 0$ . An irrational number is one that can never be written this way, however clever the fraction. Numbers like $2$ , $3$ and $5$ are irrational, and in this topic you prove it.

You cannot check every fraction, so you use proof by contradiction. You assume the opposite (the number is rational), follow the logic carefully, and reach something impossible. Since correct steps from a true assumption cannot give an impossible result, the assumption must be false.

The key tool comes from the Fundamental Theorem of Arithmetic: **if a prime p divides $a^{2}$ , then p divides a** (Theorem 1.2 in NCERT). The reason is that the primes in $a^{2}$ are exactly the primes in a, each with double the power. Squaring cannot bring in a new prime. For $p$ : assume $p = \frac{a}{b}$ with a, b co-prime. Squaring gives $a^{2} = p b^{2}$ , so p divides a. Put $a = p c$ , get $b^{2} = p c^{2}$ , so p divides b. Now p divides both a and b, contradicting that they were co-prime.

This method uses the fact that p divides $a^{2}$ implies p divides a, which Class 10 proves for a prime p. It fails for non-primes such as 4 or 9. For $4$ : 4 divides $a^{2}$ does not force 4 to divide a (take $a = 2$ ), and indeed $4 = 2$ is rational.

Once you know $5$ is irrational, you can prove $3 + 25$ irrational. Assume it equals $\frac{a}{b}$ , then isolate the surd: $5 = \frac{a - 3 b}{2 b}$ . The right side is rational, but the left is irrational. Contradiction. Behind this are two facts: rational + irrational is irrational, and non-zero rational × irrational is irrational.

Beware: two irrationals can combine to give a rational, such as $3 \times 12 = 6$ or $(2 + 3) + (2 - 3) = 4$ . Such pairs make good assertion-reason questions.

Proving Numbers Irrational: formulas to remember

Definition of rational number: $x = \frac{a}{b}, a, b \in Z, b \neq = 0, HCF (a, b) = 1$; In proofs always take a and b co-prime; the contradiction comes from breaking this.
Prime divides a square (Theorem 1.2): $p prime, p ∣ a^{2} \Rightarrow p ∣ a$; Used twice in every proof for the square root of a prime. State it, do not prove it.
Irrationality of root of a prime: $2, 3, 5 are irrational$; Proved by contradiction; quote as given only when the question says 'given that'.
Rational plus irrational: $r \in Q, x \in / Q \Rightarrow r + x \in / Q$; Also r − x and x − r are irrational.
Non-zero rational times irrational: $r \in Q, r \neq = 0, x \in / Q \Rightarrow r x \in / Q$; If r = 0 the product is 0, which is rational.
Isolating the surd: $\frac{a}{b} = m + n p \Rightarrow p = \frac{a - mb}{nb}$; m, n integers, n not 0. The right side is rational, giving the contradiction.

How to solve Proving Numbers Irrational questions

Every irrationality proof follows the same skeleton. Learn it and adapt it.

1Write the assumption: 'Let us assume, to the contrary, that the number is rational.'
2Write it as $\frac{a}{b}$ with a, b integers, $b \neq = 0$ , and a, b co-prime.
3For a single root $p$ : square to get $a^{2} = p b^{2}$ , say p divides $a^{2}$ , so p divides a (theorem).
4Substitute $a = p c$ , simplify to $b^{2} = p c^{2}$ , conclude p divides b.
5For an expression: rearrange so the surd is alone, e.g. $5 = \frac{a - 3 b}{2 b}$ , and say the right side is rational because a, b are integers.
6State the contradiction clearly: p is a common factor of co-prime a and b, or a rational equals an irrational.
7Conclude: 'Our assumption is wrong, hence the number is irrational.'

Proving Numbers Irrational 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.

Proving expressions like 3 + 2√5 are irrational

Imagine your family's kitchen garden is a rectangle 6 m long on one side and 12 m on the other. You plan to run a string of festival lights along one diagonal of the garden and then a further 5 m straight to the gate. Your cousin says the shop can cut the string to an exact fraction of a metre.

Find the exact total length of the light string in simplest surd form. Given that $5$ is irrational, prove that this total length is irrational.

Try it first, then show the solutionSolution

Diagonal $= 6^{2} + 1 2^{2} = 36 + 144 = 180$ m (Pythagoras theorem).
$180 = 36 \times 5 = 65$ m.
Total length $L = 5 + 65$ m.
Assume, to the contrary, $5 + 65 = \frac{a}{b}$ , where a, b are co-prime integers and $b \neq = 0$ .
Then $65 = \frac{a}{b} - 5 = \frac{a - 5 b}{b}$ .
So $5 = \frac{a - 5 b}{6 b}$ .
Since a and b are integers and $6 b \neq = 0$ , the right side is rational, so $5$ would be rational.
This contradicts the given fact that $5$ is irrational, so our assumption is wrong and $5 + 65$ is irrational.

Answer: The total length is $5 + 65$ m, which is irrational, so no exact fraction of a metre equals it.

Memory hook: Garden diagonal plus a straight strip: rational + surd is still a surd.

Proving that √2, √3 and √5 are irrational

Suppose you are making a rangoli frame for Diwali in the shape of an equilateral triangle of side 4 dm. You want to mark the height from one corner to the middle of the opposite side. A friend claims the height can be marked exactly as a fraction of a decimetre on a ruler.

Find the height of the triangle in simplest surd form. Prove that $3$ is irrational and hence prove that the height is irrational.

Try it first, then show the solutionSolution

The height meets the base at its midpoint, so it forms a right triangle with hypotenuse 4 dm and base 2 dm.
Height $= 4^{2} - 2^{2} = 12 = 23$ dm.
Assume $3 = \frac{a}{b}$ , a, b co-prime integers, $b \neq = 0$ .
Squaring: $a^{2} = 3 b^{2}$ , so 3 divides $a^{2}$ ; 3 is prime, hence 3 divides a.
Let $a = 3 c$ : $9 c^{2} = 3 b^{2}$ , so $b^{2} = 3 c^{2}$ , hence 3 divides $b^{2}$ and so 3 divides b.
3 divides both a and b, contradicting that they are co-prime, so $3$ is irrational.
Now assume $23 = \frac{p}{q}$ , p, q integers, $q \neq = 0$ .
Then $3 = \frac{p}{2 q}$ , which is rational since p, q are integers.
This contradicts that $3$ is irrational, so $23$ is irrational.

Answer: Height $= 23$ dm, which is irrational, so it cannot be marked as an exact fraction of a decimetre.

Memory hook: Rangoli triangle height hides a root 3, so no ruler fraction fits it.

Tricks to remember Proving Numbers Irrational

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

ACSDC
Assume, Co-prime, Square, Divides twice, Contradiction.
Use it when: The five beats of every proof for root of a prime.
Isolate and expose
Get the surd alone; the other side is 'all integers', so rational.
Use it when: Proofs of expressions like $4 - 32$ .
Prime only
Use p divides $a^{2}$ implies p divides a only when p is a prime; it fails for 4 and 9.
Use it when: Spotting wrong proofs like one for $4$ .

Proving Numbers Irrational: solved examples

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

Example 1: Proof of root 2 · 3 marks · Short answerBoard level

Prove that $2$ is irrational.

Try it first, then show the solutionSolution

Assume $2$ is rational: $2 = \frac{a}{b}$ , a, b co-prime integers, $b \neq = 0$ . (½)
Squaring: $a^{2} = 2 b^{2}$ , so 2 divides $a^{2}$ , hence 2 divides a. (1)
Let $a = 2 c$ : $4 c^{2} = 2 b^{2}$ , so $b^{2} = 2 c^{2}$ , hence 2 divides b. (1)
2 is a common factor of a and b, contradiction; so $2$ is irrational. (½)

Answer: $2$ is irrational.

Example 2: Expression with root 5 · 3 marks · Short answerBoard level

Given that $5$ is irrational, prove that $7 - 35$ is irrational.

Try it first, then show the solutionSolution

Assume $7 - 35 = \frac{a}{b}$ , a, b co-prime integers, $b \neq = 0$ . (½)
$35 = 7 - \frac{a}{b} = \frac{7 b - a}{b}$ . (½)
$5 = \frac{7 b - a}{3 b}$ . (½)
a, b integers, so $\frac{7 b - a}{3 b}$ is rational, so $5$ is rational. (1)
This contradicts $5$ irrational, so $7 - 35$ is irrational. (½)

Answer: $7 - 35$ is irrational.

Example 3: Two-part long answer · 5 marks · Long answerBoard level

(i) Prove that $3$ is irrational.
(ii) Hence prove that $\frac{2}{3}$ is irrational.

Try it first, then show the solutionSolution

(i) Assume $3 = \frac{a}{b}$ , a, b co-prime, $b \neq = 0$ . (½)
$a^{2} = 3 b^{2}$ , so 3 divides $a^{2}$ , hence 3 divides a. (1)
Put $a = 3 c$ : $9 c^{2} = 3 b^{2}$ , $b^{2} = 3 c^{2}$ , so 3 divides b. (1)
Contradiction with co-primeness; $3$ is irrational. (½)
(ii) Assume $\frac{2}{3} = \frac{p}{q}$ , p, q integers, $p, q \neq = 0$ (the number is not 0). (½)
Then $3 = \frac{2 q}{p}$ , which is rational. (1)
Contradicts (i), so $\frac{2}{3}$ is irrational. (½)

Answer: Both $3$ and $\frac{2}{3}$ are irrational.

Example 4: Assertion-reason on sums · 1 mark · Assertion-reasonBoard level

Assertion (A): $(5 + 2) + (5 - 2)$ is rational.
Reason (R): The sum of two irrational numbers is always irrational.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

Sum $= 10$ , which is rational, so A is true.
Both terms are irrational yet their sum is rational, so R is false.
Hence option (c). (1)

Answer: (c) A is true but R is false.

Example 5: Find the error · 4 marks · Long answerHOTS / competency

A student 'proves' $9$ is irrational: assume $9 = \frac{a}{b}$ , a, b co-prime; $a^{2} = 9 b^{2}$ , so 9 divides $a^{2}$ , hence 9 divides a; write $a = 9 c$ , so $81 c^{2} = 9 b^{2}$ , $b^{2} = 9 c^{2}$ , so 9 divides b; contradiction.
(i) Why must the proof be wrong?
(ii) Which step is wrong? Give a counter-example.
(iii) Does the same argument work with 3 in place of 9 for $3$ ? Why?

Try it first, then show the solutionSolution

(i) $9 = 3 = \frac{3}{1}$ , which is rational, so the conclusion is false. (1)
(ii) The step '9 divides $a^{2}$ , hence 9 divides a' is wrong; the theorem needs a prime. (½)
Counter-example: $a = 3$ , $a^{2} = 9$ is divisible by 9 but 3 is not. (½)
(iii) Yes: 3 is prime, so 3 divides $a^{2}$ does imply 3 divides a. (1)
The rest of the argument is valid, giving the contradiction. (1)

Answer: (i) $9 = 3$ is rational. (ii) '9 divides $a^{2}$ so 9 divides a' is false, e.g. a = 3. (iii) Yes, because 3 is prime.

Example 6: Root 5, hence 3 + 2√5 · 5 marks · Long answerBoard level

(i) Prove that $5$ is irrational.
(ii) Hence prove that $3 + 25$ is irrational.

Try it first, then show the solutionSolution

(i) Assume, to the contrary, $5 = \frac{a}{b}$ , a, b co-prime integers, $b \neq = 0$ . (½)
Squaring: $a^{2} = 5 b^{2}$ , so 5 divides $a^{2}$ ; since 5 is prime, 5 divides a (Theorem 1.2). (1)
Let $a = 5 c$ : $25 c^{2} = 5 b^{2}$ , so $b^{2} = 5 c^{2}$ ; 5 divides $b^{2}$ , hence 5 divides b. (1)
5 is a common factor of a and b, contradicting co-primeness; so $5$ is irrational. (½)
(ii) Assume $3 + 25 = \frac{p}{q}$ , p, q co-prime integers, $q \neq = 0$ . (½)
$25 = \frac{p}{q} - 3 = \frac{p - 3 q}{q}$ . (½)
$5 = \frac{p - 3 q}{2 q}$ . (½)
p, q are integers, so the right side is rational, making $5$ rational; this contradicts part (i), so $3 + 25$ is irrational. (½)

Answer: $5$ and $3 + 25$ are both irrational.

Example 7: Why rational + irrational is irrational · 2 marks · Short answerBoard level

Let r be a rational number and x an irrational number. Prove that $r + x$ is irrational.

Try it first, then show the solutionSolution

Assume, to the contrary, $r + x = s$ where s is rational. Write $r = \frac{a}{b}$ and $s = \frac{c}{d}$ with a, b, c, d integers, $b, d \neq = 0$ . (½)
Then $x = s - r = \frac{c}{d} - \frac{a}{b} = \frac{b c - a d}{b d}$ , where $b c - a d$ and $b d$ are integers and $b d \neq = 0$ , so x is rational. (1)
This contradicts that x is irrational, so $r + x$ is irrational. (½)

Answer: $r + x$ is irrational whenever r is rational and x is irrational.

Example 8: Sum of two roots · 3 marks · Short answerHarder board level

Given that $15$ is irrational, prove that $3 + 5$ is irrational.

Try it first, then show the solutionSolution

Assume, to the contrary, $3 + 5 = r$ , where r is rational. (½)
Squaring both sides: $3 + 5 + 215 = r^{2}$ , so $8 + 215 = r^{2}$ . (1)
$15 = \frac{r ^{2} - 8}{2}$ . (½)
r is rational, so $r^{2} - 8$ is rational and so is $\frac{r ^{2} - 8}{2}$ ; thus $15$ is rational. (½)
This contradicts that $15$ is irrational, so $3 + 5$ is irrational. (½)

Answer: $3 + 5$ is irrational.

Example 9: Rationalise first, then prove · 3 marks · Short answerHarder board level

Given that $5$ is irrational, prove that $\frac{1}{2 - 5}$ is irrational.

Try it first, then show the solutionSolution

Multiply numerator and denominator by $2 + 5$ . (½)
$\frac{2 + 5}{4 - 5} = \frac{2 + 5}{- 1} = - 2 - 5$ . (½)
Assume, to the contrary, $- 2 - 5 = \frac{a}{b}$ , a, b co-prime integers, $b \neq = 0$ . (½)
$5 = - 2 - \frac{a}{b} = \frac{- ( a + 2 b )}{b}$ . (½)
a, b integers, so the right side is rational, making $5$ rational. (½)
This contradicts that $5$ is irrational, so $- 2 - 5$ , which equals $\frac{1}{2 - 5}$ , is irrational. (½)

Answer: $\frac{1}{2 - 5} = - 2 - 5$ , which is irrational.

Example 10: Case study: rangoli on a square floor · 4 marks · Case-basedHOTS / competency

Your school is making a rangoli on a square floor of side 4 m. A ribbon is laid along one diagonal of the square. A second square is then drawn with this diagonal as its side. Finally, the ribbon is extended by 3 m along the floor edge, so the full ribbon is the diagonal plus 3 m.
(i) Find the length of the diagonal in simplest surd form. (1)
(ii) Is the area of the second square rational? Justify. (1)
(iii) Given that $2$ is irrational, prove that the full ribbon length is irrational. (2)

Try it first, then show the solutionSolution

(i) Diagonal $= 4^{2} + 4^{2} = 32 = 42$ m. (1)
(ii) Area $= (42)^{2} = 16 \times 2 = 32$ m², which equals $\frac{32}{1}$ , so it is rational. (1)
(iii) Ribbon length $= 3 + 42$ m. Assume $3 + 42 = \frac{a}{b}$ , a, b co-prime integers, $b \neq = 0$ . (½)
$42 = \frac{a - 3 b}{b}$ , so $2 = \frac{a - 3 b}{4 b}$ . (½)
a, b integers, so the right side is rational, making $2$ rational. (½)
This contradicts that $2$ is irrational, so $3 + 42$ m is irrational. (½)

Answer: (i) $42$ m. (ii) Yes, the area is 32 m², a rational number. (iii) The ribbon length $3 + 42$ m is irrational.

Where students lose marks in Proving Numbers Irrational

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

Mistake: Not stating a and b are co-prime
The contradiction depends on it.
Wrong working
Let $3 = \frac{a}{b}$ , $b \neq = 0$ . ... 3 divides a and b.
Right working
Let $3 = \frac{a}{b}$ , a, b co-prime integers, $b \neq = 0$ . ... 3 divides both, contradicting co-primeness.
Do this: Write 'co-prime' in the first line.
Usually costs: ½ to 1 mark
Mistake: Quoting the rule instead of proving
'Prove' needs the contradiction argument.
Wrong working
3 is rational and $25$ is irrational, so $3 + 25$ is irrational.
Right working
Assume $3 + 25 = \frac{a}{b}$ ; then $5 = \frac{a - 3 b}{2 b}$ is rational, contradiction.
Do this: Always write the assumption and isolate the surd.
Usually costs: 1 to 2 marks
Mistake: Wrong rearrangement
Algebra slips when isolating the surd.
Wrong working
$3 + 25 = \frac{a}{b} \Rightarrow 5 = \frac{a - 3}{2 b}$
Right working
$25 = \frac{a - 3 b}{b} \Rightarrow 5 = \frac{a - 3 b}{2 b}$
Do this: Take a common denominator before dividing.
Usually costs: ½ to 1 mark
Mistake: Missing the final contradiction and conclusion
The examiner awards the last mark for it.
Wrong working
... so 5 divides b. Hence proved.
Right working
... so 5 divides b. Thus 5 is a common factor of a and b, contradicting co-primeness. Hence $5$ is irrational.
Do this: End with what is contradicted and what follows.
Usually costs: ½ mark

Exam tips

A 3-mark question can ask for a direct proof of $2$ , $3$ or $5$ ; learn one clean version and adapt the prime.
If the question says 'given that $5$ is irrational', you may use it directly; otherwise prove it first.
In 'hence' questions, part (ii) must use part (i); say 'this contradicts part (i)'.
For assertion-reason, test sums and products of irrationals with a quick counterexample like $2 \times 2$ .
Name the theorem 'if a prime p divides $a^{2}$ then p divides a' when you use it.

Practice questions on Proving Numbers Irrational

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

Q1 · 3 marks · Short answerHarder board level
Given that $3$ is irrational, prove that $\frac{2 - 3}{2 + 3}$ is irrational.
Hint
Multiply numerator and denominator by $2 - 3$ first, then isolate $3$ .
Answer
Rationalising: $\frac{( 2 - 3 ) ^{2}}{4 - 3} = 4 - 43 + 3 = 7 - 43$ . Assume $7 - 43 = \frac{a}{b}$ , a, b co-prime, $b \neq = 0$ . Then $3 = \frac{7 b - a}{4 b}$ , which is rational. This contradicts $3$ irrational, so the number is irrational.
Q2 · 3 marks · Short answerBoard level
Given that $2$ is irrational, prove that $\frac{3 + 4 2}{5}$ is irrational.
Hint
Multiply by 5 first, then isolate $2$ .
Answer
Assume it equals $\frac{a}{b}$ . Then $3 + 42 = \frac{5 a}{b}$ , so $2 = \frac{5 a - 3 b}{4 b}$ , rational. Contradiction, so it is irrational.
Q3 · 3 marks · Short answerHarder board level
Prove that $2 + 3$ is irrational, given that $6$ is irrational.
Hint
Assume it is r, square both sides.
Answer
Let $2 + 3 = r$ (rational). Squaring: $5 + 26 = r^{2}$ , so $6 = \frac{r ^{2} - 5}{2}$ , rational. Contradiction, so $2 + 3$ is irrational.
Q4 · 1 mark · MCQBoard level
Which of the following is rational?
(a) $(5 + 3) (5 - 3)$
(b) $(2 + 3)^{2} - 6$
(c) $(5 + 3)^{2}$
(d) $23 + 12$
Hint
Expand or simplify each option fully before deciding.
Answer
(a) 2, since $(5)^{2} - (3)^{2} = 5 - 3 = 2$ . Option (b) is $5 + 26 - 6 = 5 + 6$ , (c) is $8 + 215$ and (d) is $43$ , all irrational.
Q5 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a craft teacher cuts a right-angled triangle from card, with the two shorter sides $3$ dm and $6$ dm. You may use the fact that $2$ is irrational.
(i) Find the hypotenuse. Is it rational? (1 mark)
(ii) Find the area in simplest surd form and decide whether it is rational. Justify. (1 mark)
(iii) Prove that $3 + 6$ is irrational, and hence decide whether the perimeter is rational. (2 marks)
Hint
For (iii), square $3 + 6 = r$ and simplify $18$ .
Answer
(i) Hypotenuse $= 3 + 6 = 9 = 3$ dm, which is rational even though both legs are irrational. (1)
(ii) Area $= \frac{1}{2} 36 = \frac{18}{2} = \frac{3 2}{2}$ dm². If it equalled a rational $s$ , then $2 = \frac{2 s}{3}$ would be rational, a contradiction; so the area is irrational. (1)
(iii) Assume $3 + 6 = r$ , rational. Squaring: $9 + 218 = r^{2}$ , so $9 + 62 = r^{2}$ and $2 = \frac{r ^{2} - 9}{6}$ , which is rational, a contradiction. So $3 + 6$ is irrational. The perimeter $3 + (3 + 6)$ is rational + irrational, so it is irrational. (2)
Q6 · 1 mark · Assertion-reasonHOTS / competency
Assertion (A): For every rational number r, the number $r 2$ is irrational.
Reason (R): The product of a non-zero rational number and an irrational number is irrational.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Test the special rational number $r = 0$ .
Answer
(d) A is false but R is true. For $r = 0$ , $r 2 = 0$ , which is rational, so A fails; R is a correct fact because it excludes 0.
Q7 · 5 marks · Long answerHarder board level
(i) Prove that $3$ is irrational.
(ii) Hence prove that $\frac{1}{2 3 - 3}$ is irrational.
Hint
In (ii), rationalise the denominator first; then isolate $3$ .
Answer
(i) Assume $3 = \frac{a}{b}$ with a, b co-prime, $b \neq = 0$ . Then $a^{2} = 3 b^{2}$ , so 3 divides $a^{2}$ and, 3 being prime, 3 divides a. Put $a = 3 c$ : $b^{2} = 3 c^{2}$ , so 3 divides b. This contradicts co-primeness, so $3$ is irrational. (3 marks)
(ii) $\frac{1}{2 3 - 3} \times \frac{2 3 + 3}{2 3 + 3} = \frac{2 3 + 3}{12 - 9} = 1 + \frac{2}{3} 3$ . Assume this equals a rational $r$ . Then $3 = \frac{3 ( r - 1 )}{2}$ , which is rational, contradicting (i). So the number is irrational. (2 marks)
Q8 · 3 marks · Short answerHarder board level
Given that $2$ is irrational, show that $(2 + \frac{1}{2})^{2} - 2$ is rational but $2 + \frac{1}{2}$ is irrational.
Hint
Write $\frac{1}{2}$ as $\frac{2}{2}$ .
Answer
$(2 + \frac{1}{2})^{2} = 2 + 2 + \frac{1}{2} = \frac{9}{2}$ , so the expression is $\frac{9}{2} - 2 = \frac{5}{2}$ , which is rational. But $2 + \frac{1}{2} = 2 + \frac{2}{2} = \frac{3}{2} 2$ . If this equalled a rational $r$ , then $2 = \frac{2 r}{3}$ would be rational, a contradiction. So it is irrational: squaring an irrational number can give a rational one.

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 an irrational number.Show answer
A number that cannot be written as $\frac{p}{q}$ with p, q integers and $q \neq = 0$ .
2.Which theorem is used in proving $p$ irrational?Show answer
If a prime p divides $a^{2}$ , then p divides a.
3.What method of proof is used?Show answer
Proof by contradiction.
4.Is $2 \times 8$ rational?Show answer
Yes, it equals 4.
5.Rational + irrational is?Show answer
Always irrational.
6.Is $0 \times 3$ irrational?Show answer
No, it is 0, which is rational.

Proving Numbers Irrational: frequently asked questions

Why do we assume a and b are co-prime?

Any fraction can be reduced to lowest terms, so this costs nothing. The proof then finds a common factor, which breaks the co-prime condition and gives the contradiction.

Can I just say root 2 is non-terminating, so it is irrational?

No. Showing a decimal pattern is not a proof and gets no marks for a 'prove' question. Use the contradiction method.

Do I need to prove the theorem that p divides a squared implies p divides a?

No. You may state and use it. Just make sure p is prime when you use it.

Is the product of two irrationals always irrational?

No. For example root 3 times root 12 equals 6. Such counterexamples are good for assertion-reason questions.