Before we write the actual proof that $\sqrt{2}$ is irrational, we need to understand the strategy.

Proof by contradiction is like setting a trap.

In this approach, we follow a specific sequence:

Assume the opposite of what we want to prove.
Follow the logical consequences.
Show they lead to something impossible.

The key strategic choice in irrationality proofs is insisting that the fraction $\frac{a}{b}$ is in simplest form.

Feature	Role in the Proof
Assumption	The fraction $a / b$ is in simplest form
Strategy	Prove this assumption cannot be true

This is not a convenience — it is the carefully chosen target that we will later destroy.

Let us make sure you understand why.

1. Filter -- Understanding the Contradiction Setup

Setting Up the Contradiction: Why Simplest Form Is the Key Move 🎯

The proof that $\sqrt{p}$ is irrational uses a strategy called proof by contradiction.

The setup involves two main parts:

Assuming $\sqrt{p} = \frac{a}{b}$
Ensuring the fraction is in simplest form

Before we write the proof, you need to understand exactly what 'simplest form' means in this context and why it is essential to the strategy.

📋 Given Info

Given Information:

To prove $\sqrt{3}$ is irrational, we begin with this assumption:

"Assume $\sqrt{3}$ is rational. Then $\sqrt{3} = \frac{a}{b}$ in simplest form, where $a$ and $b$ are co-prime integers and $b \neq 0$ ."

✍️ Question

Let's check your understanding of this setup

(a) What does 'simplest form' mean here? Define 'co-prime'.
(b) Why is it important that $\frac{a}{b}$ is in simplest form, rather than just any fraction? What would go wrong if we did not insist on simplest form?
(c) After assuming $\sqrt{3} = \frac{a}{b}$ , what equation do you get by squaring both sides and cross-multiplying?

Let us think about WHY the proof is set up this way.

Proof by contradiction means: assume the opposite of what you want to prove, then show that assumption leads to something impossible (a contradiction).

✍️ MCQ

Choose one

What is the initial assumption in a proof by contradiction for

\sqrt{3}

being irrational?

We want to prove: $\sqrt{3}$ is irrational. So we assume the opposite: $\sqrt{3}$ is rational.

✍️ Yes/No

Yes or No?

If a number is rational, can it be written as a fraction

\frac{a}{b}

where

a

and

b

are integers?

If $\sqrt{3}$ is rational, we can write it as a fraction: $\sqrt{3} = \frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$ .

But here is the critical strategic choice: we pick $\frac{a}{b}$ in simplest form (also called lowest terms).

✍️ Yes/No

Yes or No?

Is the fraction

\frac{6}{10}

in its simplest form?

This means $a$ and $b$ are co-prime: their HCF is 1, meaning they share no common factor other than 1.

✍️ FIB

Fill in the blank

What is the HCF of two co-prime numbers?

Why Simplest Form?

Because our proof will eventually show that 3 divides both $a$ and $b$ .

If we had NOT assumed simplest form, this would not be surprising — lots of fractions have a common factor (like $\frac{6}{9}$ where 3 divides both). But with simplest form, any common factor other than 1 is IMPOSSIBLE by assumption.

✍️ Yes/No

Yes or No?

If we find that

3

divides both

a

and

b

, does this agree with our assumption that the fraction

\frac{a}{b}

is in its simplest form?

So when we prove 3 divides both, that IS impossible — a contradiction.

Think of it this way:

Simplest form is like locking a door and claiming it is locked.
The proof then shows someone walked through the locked door.

That is impossible — so the original assumption ( $\sqrt{3}$ is rational) must be wrong.

✍️ MCQ

Choose one

In our analogy, if someone is found inside the 'locked' room, what does that tell us about our original assumption?

Let's Look at the Algebra

Now the algebra: we start with our assumption that $\sqrt{3}$ is rational.

$\sqrt{3} = \frac{a}{b}$

✍️ MCQ

Choose one

To remove the square root from

\sqrt{3} = \frac{a}{b}

, we should:

Square both sides:

$3 = \frac{a^{2}}{b^{2}}$

Multiply both sides by $b^{2}$ :

$3 b^{2} = a^{2}$

This is equation (i) — the starting point for applying Theorem 1.

✍️ MCQ

Choose one

If

3b^2 = a^2

(where

b

is an integer), then

a^2

must be:

2. The Initial Algebra -- Squaring and Cross-Multiplying

Setting Up the Contradiction 🎯

The first algebraic step in the proof is simple but must be done correctly:

Square both sides of the initial assumption:
$\sqrt{p} = \frac{a}{b}$
Rearrange the equation to eliminate the fraction.

This gives you the equation $p b^{2} = a^{2}$ , which is where everything starts.

📋 Given Info

Given Information:

Assume $\sqrt{5} = \frac{a}{b}$ in simplest form, where:

$a$ and $b$ are co-prime positive integers
$b \neq 0$

This means $H C F (a, b) = 1$ , implying they share no common factors other than 1.

✍️ Question

Your Turn ✏️

Starting from $\sqrt{5} = \frac{a}{b}$ :

Square both sides of $\sqrt{5} = \frac{a}{b}$ . What do you get?
Rearrange to eliminate the fraction. What equation do you get?
What does this equation tell you about the relationship between $5$ and $a^{2}$ ?

Let us do the algebra step by step.

We start with: $\sqrt{5} = \frac{a}{b}$ .

✍️ FIB

Fill in the blank

Calculate

(\sqrt{5})^2

:

Step 1: Square both sides.

Left side: $(\sqrt{5})^{2} = 5$ (by definition of square root).
Right side: ${(\frac{a}{b})}^{2} = \frac{a^{2}}{b^{2}}$ (square numerator and denominator separately).

✍️ MCQ

Choose one

When squaring

\frac{a}{b}

, we square:

So: $5 = \frac{a^{2}}{b^{2}}$

Step 2: Eliminate the fraction.

Now that we have $5 = \frac{a^{2}}{b^{2}}$ , we want to make this equation easier to work with. To do that, we need to get rid of the fraction on the right side by multiplying both sides by $b^{2}$ .

✍️ MCQ

Choose one

To clear the fraction in

5 = \frac{a^2}{b^2}

, we multiply both sides by:

Multiplying both sides by $b^{2}$ looks like this:

$5 \times b^{2} = \frac{a^{2}}{b^{2}} \times b^{2}$

$5 b^{2} = a^{2}$

This is the key equation. Label it (i) for reference.

✍️ Yes/No

Yes or No?

Does

5b^2 = a^2

mean

a^2

is a multiple of

5

?

Step 3: Read the equation.

$a^{2} = 5 b^{2} = 5 \times (b^{2})$ .

✍️ Yes/No

Yes or No?

If

a^2 = 5 \times (b^2)

, does that mean

a^2

is a multiple of

5

?

So $a^{2}$ is 5 times something, which means $a^{2}$ is a multiple of 5.

In mathematical language: 5 divides $a^{2}$ .

This is the moment where Theorem 1 kicks in: since 5 is prime and 5 divides $a^{2}$ , we can conclude 5 divides $a$ .

✍️ MCQ

Choose one

If

5

divides

a^2

, then according to the theorem,

5

must also divide:

Common errors to avoid:

Writing ' $5 b = a^{2}$ ' (forgetting to square $b$ ). It must be $5 b^{2} = a^{2}$ .

✍️ MCQ

Choose one

If we square the equation

\sqrt{5} = \frac{a}{b}

, which of these is the correct result?

Writing ' $5 = a^{2} - b^{2}$ ' (subtracting instead of cross-multiplying). You multiply by $b^{2}$ , not subtract.

✍️ MCQ

Choose one

In the equation

5 = \frac{a^2}{b^2}

, what is the correct operation to isolate

a^2

?

3. Recognising the Contradiction Template

You know the setup (assume rational, simplest form) and the algebra (squaring gives $p b^{2} = a^{2}$ ).

The proof then applies Theorem 1 twice to show:

$p$ divides $a$
$p$ divides $b$

This contradicts the co-primality (simplest form) assumption.

Let's check whether you can see this as a template that works for ANY prime.

📋 Given Info

Here are the key equations from three different proofs of irrationality:

Case	First equation	After substitution
$\sqrt{2}$	$2 b^{2} = a^{2}$	$b^{2} = 2 c^{2}$
$\sqrt{3}$	$3 b^{2} = a^{2}$	$b^{2} = 3 c^{2}$
$\sqrt{5}$	$5 b^{2} = a^{2}$	$b^{2} = 5 c^{2}$

Notice the symmetry in the algebra across all three examples.

✍️ Question

Three-part question 🔍

(a) What pattern do you see? If the prime is $p$ , what are the two key equations?

(b) Why does the template always produce a contradiction?

(c) A student tries the template for $\sqrt{4}$ :

$4 b^{2} = a^{2}$ , so $2$ divides $a$ , so $a = 2 c$ , giving $4 b^{2} = 4 c^{2}$ , so $b^{2} = c^{2}$ .

Does a contradiction arise? Why or why not?

Generalizing the Pattern for any Prime $p$

Let us see the pattern clearly. For ANY prime p, the proof follows the same logical path.

Equation 1: $p b^{2} = a^{2}$

From this equation, we see that $p$ divides $a^{2}$ . According to Theorem 1, if a prime $p$ divides the square of a number, it must also divide the number itself.

So, $p$ divides $a$ .

✍️ T/F

True or False?

If a prime

p

divides

a^2

, then

p

must also divide

a

.

Substitution and Equation 2

Since $p$ divides $a$ , we can write $a$ as a multiple of $p$ , say $a = p c$ for some integer $c$ .

Substituting this back into Equation 1: $p b^{2} = (p c)^{2} = p^{2} c^{2}$

Now, divide both sides by $p$ to get: Equation 2: $b^{2} = p c^{2}$

Just like before, this means $p$ divides $b^{2}$ , which implies $p$ divides $b$ .

✍️ MCQ

Choose one

If

b^2 = pc^2

, which of these is true?

The Contradiction and Universality

Contradiction: We have found that $p$ divides both $a$ and $b$ . However, this contradicts our initial assumption that $a$ and $b$ are co-prime (they should have no common factors other than 1).

The template works because Theorem 1 applies to ANY prime. Whether you replace $p$ with 2, 3, 5, 7, or 11, the logic remains identical every time.

✍️ Yes/No

Yes or No?

Does this specific proof logic rely on

p

being a prime number?

Testing the Template: What about $\sqrt{4}$ ?

Now, let's see what happens if we try to apply this same logic to the square root of 4.

If we follow the template, we assume $\sqrt{4} = \frac{a}{b}$ and square it: $4 b^{2} = a^{2}$

✍️ Yes/No

Yes or No?

Is

4

a prime number?

Is 4 prime? NO ( $4 = 2 \times 2$ ).

Because 4 is not prime, we cannot apply Theorem 1 to 4 directly.

However, we can note that since 4 divides $a^{2}$ , then $2$ (a factor of 4) must also divide $a^{2}$ . By Theorem 1, if prime $2$ divides $a^{2}$ , then $2$ divides $a$ .

So, let $a = 2 c$ .

✍️ MCQ

Choose one

If

a = 2c

, then

a^2

is:

Substitute the value back in: $4 b^{2} = (2 c)^{2} = 4 c^{2}$

Now, divide both sides by 4: $b^{2} = c^{2}$

Taking the square root: $b = c$ .

Notice: the equation is $b^{2} = c^{2}$ , NOT $b^{2} = 4 c^{2}$ . There is no '4' left to force $4 ∣ b$ or even $2 ∣ b$ . The template stalls.

✍️ MCQ

Choose one

In the proof for

\sqrt{5}

, we reached

b^2 = 5c^2

. Why was that '5' important?

And indeed: $a = 2 c$ , $b = c$ , so $a / b = 2 c / c = 2$ . We have $\sqrt{4} = 2$ , a perfectly rational number. No contradiction — because there is nothing wrong with $\sqrt{4}$ being rational!

The template fails because 4 is not prime. When we substitute $a = 2 c$ , the 4 from $(2 c)^{2}$ exactly cancels the 4 from $4 b^{2}$ , leaving nothing behind.

✍️ Yes/No

Yes or No?

In the equation

4b^2 = 4c^2

, does any factor of

4

remain after we divide both sides by

4

?

For a prime $p$ , $(p c)^{2} = p^{2} c^{2}$ , and dividing by $p$ still leaves a factor of $p$ — that leftover $p$ is what catches $b$ .

✍️ MCQ

Choose one

What is the simplified equation after dividing

pb^2 = p^2c^2

by

p

?

4. Writing the Complete Proof for sqrt(2)

You understand the strategy and the template. Now let us see if you can write the actual proof for the most commonly examined case:

$\sqrt{2}$ is irrational.

This is a CBSE board favourite — it appears almost every other year.

📋 Given Info

Given Information

Theorem 1: If $p$ is a prime and $p$ divides $a^{2}$ , then $p$ divides $a$ .

Note: You may use this theorem without reproof in your solutions.

✍️ Question

Your Turn ✍️

Write a complete proof that $\sqrt{2}$ is irrational.

Your proof must include:

The initial assumption
The simplest-form setup
The squaring step
Both applications of Theorem 1
The substitution step
The explicit contradiction

Opening (assumption):

'Assume, for the sake of contradiction, that $\sqrt{2}$ is rational. Then $\sqrt{2} = \frac{a}{b}$ , where $a$ and $b$ are positive integers, $b \neq 0$ ...'

✍️ MCQ

Choose one

In the rational fraction

\frac{a}{b}

, what is the condition for

b

?

...and $\frac{a}{b}$ is in simplest form. This means $a$ and $b$ are co-prime: $HCF (a, b) = 1$ .'*

You MUST state: (1) the assumption, (2) that it is in simplest form, (3) that $a$ and $b$ are co-prime. Examiners look for these.

✍️ FIB

Fill in the blank

The HCF of two co-prime numbers is:

Algebra:

'Squaring both sides: $2 = \frac{a^{2}}{b^{2}}$ .'

✍️ MCQ

Choose one

What is the next step to simplify

2 = \frac{a^2}{b^2}

?

'So $2 b^{2} = a^{2}$ . ...(i)'

Label the equation (i) because you will refer back to it.

First application of Theorem 1:

From equation (i), we have $a^{2} = 2 b^{2}$ . This means that $a^{2}$ is twice some integer, which tells us that 2 divides $a^{2}$ .

✍️ MCQ

Choose one

If

a^2 = 2b^2

, then

a^2

is a multiple of which number?

Since 2 is prime and 2 divides $a^{2}$ , we can apply our rule.

By Theorem 1, we conclude that 2 divides $a$ .

Whenever you write this in an exam, remember the three-part citation:

State that 2 is prime.
State that 2 divides $a^{2}$ .
Mention Theorem 1 to reach the conclusion.

✍️ Yes/No

Yes or No?

Is

2

a prime number?

Since we established that $2$ divides $a$ , we can express $a$ in a different way.

Let $a = 2 c$ for some positive integer $c$ .

✍️ MCQ

Choose one

What is

(2c)^2

?

Now, let's plug this into our first equation, which we labeled as (i):

$2 b^{2} = a^{2}$ $2 b^{2} = (2 c)^{2} = 4 c^{2}$

⚠️ Critical: $(2 c)^{2} = 4 c^{2}$ , NOT $2 c^{2}$ .

✍️ MCQ

Choose one

To simplify

2b^2 = 4c^2

, we should:

Dividing by 2:

$b^{2} = 2 c^{2}$

Second application of Theorem 1:

From $b^{2} = 2 c^{2}$ , 2 divides $b^{2}$ .

✍️ MCQ

Choose one

If

2

divides

b^2

and

2

is prime, what is the conclusion from Theorem 1?

'Since 2 is prime and 2 divides $b^{2}$ , by Theorem 1, 2 divides $b$ .'

✍️ MCQ

Choose one

To apply Theorem 1 to the statement '

2

divides

b^2

', we must state that:

The Common Factor

'Thus 2 divides both $a$ and $b$ , making 2 a common factor.'

✍️ MCQ

Choose one

What is the HCF of two co-prime numbers?

The Contradiction

'But $a$ and $b$ are co-prime — contradiction.'

✍️ MCQ

Choose one

The contradiction occurred because:

The Conclusion

'The contradiction arose from assuming $\sqrt{2}$ is rational. Hence, $\sqrt{2}$ is irrational.'

Exam Tip: Always state WHAT contradicts WHAT. Do not just write 'contradiction, hence proved.'

✍️ MCQ

Choose one

Which two statements contradict each other in this proof?

5. The General Case and Boundary -- sqrt(p) for Any Prime

Excellent progress! 🎯

You've successfully proved that $\sqrt{2}$ is irrational using our proof-by-contradiction template:

Assume $\sqrt{2} = \frac{a}{b}$ in simplest form
Square both sides
Apply Theorem 1 twice
Reach contradiction with $H C F (a, b) = 1$

Now let's see if you can generalize this technique and understand its boundaries.

✍️ Question

Two-Part Challenge 📝

(a) Write the general proof for $\sqrt{p}$ where $p$ is any prime number.

Use variables $m$ and $n$ instead of $a$ and $b$ .

(b) A CBSE question asks:

"Prove that $\sqrt{6}$ is irrational."

Can you plug $p = 6$ into your template?
If not, suggest an alternative approach.

General proof (Theorem 6):

Let $p$ be any prime. Assume $\sqrt{p}$ is rational.
Let $\sqrt{p} = \frac{m}{n}$ in simplest form ( $m$ , $n$ co-prime, $n \neq 0$ ).

✍️ Yes/No

Yes or No?

If

m

and

n

are co-prime, do they share any common factor other than

1

?

Squaring: $p = \frac{m^{2}}{n^{2}}$ , so $p n^{2} = m^{2}$ ...(i)

From (i): $m^{2} = p \times n^{2}$ , so $p$ divides $m^{2}$ .
Since $p$ is prime, by Theorem 1, $p$ divides $m$ .

Let $m = p q$ for some positive integer $q$ .
Substitute in (i): $p n^{2} = (p q)^{2} = p^{2} q^{2}$ .
Divide by $p$ : $n^{2} = p q^{2}$ .

✍️ MCQ

Choose one

If

n^2 = p q^2

and

p

is prime, what can we conclude about

n

using Theorem 1?

So $p$ divides $n^{2}$ .
Since $p$ is prime, by Theorem 1, $p$ divides $n$ .

The Contradiction

$p$ divides both $m$ and $n$ , contradicting $HCF (m, n) = 1$ .
Hence $\sqrt{p}$ is irrational. QED

Notice: this is ONE proof that works for ALL primes.

✍️ Yes/No

Yes or No?

Could you use the Theorem 6 approach to prove

\sqrt{11}

is irrational?

Plug in $p = 2$ and you get the $\sqrt{2}$ proof. Plug in $p = 3, 5, 7, 11, 13, 17, \dots$ and you get each specific proof.

✍️ MCQ

Choose one

Which of these numbers is NOT suitable for the

\sqrt{p}

proof template?

For $\sqrt{6}$ :

$6$ is NOT prime ( $6 = 2 \times 3$ ), so you cannot write 'by Theorem 1, $6$ divides $a$ '.

✍️ Yes/No

Yes or No?

Can we apply Theorem 1 directly to the number

6

to show it divides

a

?

But here is a workaround: from $6 b^{2} = a^{2}$ , note that $2$ divides $a^{2}$ (since $6 = 2 \times 3$ , so $2$ divides $6$ , so $2$ divides $6 b^{2} = a^{2}$ ). Since $2$ is prime, by Theorem 1, $2$ divides $a$ .

✍️ MCQ

Choose one

If

3

is also a prime factor of

6

, what can we conclude about

a

?

Similarly, $3$ divides $a$ . Since $2$ and $3$ both divide $a$ , and $\gcd (2, 3) = 1$ , we get $6$ divides $a$ (by the property that if two co-prime numbers each divide a number, their product also divides it).

Let $a = 6 c$ . Then $6 b^{2} = 36 c^{2}$ , so $b^{2} = 6 c^{2}$ .

✍️ Yes/No

Yes or No?

If

b^2 = 6c^2

, does this mean

b^2

is a multiple of

6

?

By the same argument we used for $a$ , we can conclude that $6$ divides $b$ .

The Contradiction: This means both $a$ and $b$ share a common factor of $6$ . This contradicts our assumption that $a$ and $b$ are co-prime.

✍️ Subjective

Your turn

If

a

and

b

are both divisible by

6

, is their HCF equal to

1

?

Type your answer below, or hold Space to speak

This workaround works but requires the extra step of arguing from prime factors $2$ and $3$ separately.

The Rule: The direct template only works when the number under the square root is itself prime.

Setting Up the Contradiction: Why Simplest Form Is the Key Move

1. Filter -- Understanding the Contradiction Setup

Setting Up the Contradiction: Why Simplest Form Is the Key Move 🎯

Why Simplest Form?

Let's Look at the Algebra

2. The Initial Algebra -- Squaring and Cross-Multiplying

Setting Up the Contradiction 🎯

Your Turn ✏️

Step 2: Eliminate the fraction.

Common errors to avoid:

3. Recognising the Contradiction Template

Three-part question 🔍

Generalizing the Pattern for any Prime ppp

Testing the Template: What about 4\sqrt{4}4​?

4. Writing the Complete Proof for sqrt(2)

Given Information

Your Turn ✍️

Opening (assumption):

5. The General Case and Boundary -- sqrt(p) for Any Prime

Excellent progress! 🎯

Two-Part Challenge 📝

General proof (Theorem 6):

The Contradiction

For 6\sqrt{6}6​:

Generalizing the Pattern for any Prime $p$

Testing the Template: What about $\sqrt{4}$ ?

For $\sqrt{6}$ :