This is the big one.

'Prove that $\sqrt{2}$ is irrational' is a CBSE board exam staple.

The proof has a clear, logical structure:

Assume rational in simplest form
Square both sides
Apply Theorem 1 twice (once for $a$ , once for $b$ )
Contradict co-primality

Every step must be explicitly justified.

Focus Area	What the Examiner Looks For
Terminology	Specific mathematical phrases
Citations	Reference to Theorem 1.3

Let us make sure you can produce this proof fluently and correctly.

1. Filter -- Complete sqrt(2) Proof from Memory

Let us see if you can produce the entire proof without prompting. This is exactly what the board exam requires.

You may use Theorem 1 without proof: If $p$ is prime and $p$ divides $a^{2}$ , then $p$ divides $a$ .

✍️ Question

The Big Challenge

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

Tip: Remember to structure your argument clearly and justify each step as if you were answering a board exam question.

Here is the proof with annotations showing what the examiner looks for.

[Earns marks: clear assumption]

"Assume, for contradiction, that $\sqrt{2}$ is rational."

✍️ MCQ

Choose one

In a proof by contradiction, why do we start by assuming

\sqrt{2}

is rational?

"Let $\sqrt{2} = \frac{a}{b}$ , where $a, b$ are positive integers, $b \neq 0$ , and $\frac{a}{b}$ is in simplest form ( $a$ and $b$ are co-prime, $HCF (a, b) = 1$ )."

✍️ FIB

Fill in the blank

If

a

and

b

are co-prime, what is their Highest Common Factor (HCF)?

[Earns marks: correct algebra]

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

✍️ MCQ

Choose one

What is the result of squaring

\sqrt{2}

?

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

✍️ Yes/No

Yes or No?

If

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

, is

a^2 = 2b^2

the correct rearrangement?

[Earns marks: Theorem 1 with primality cited]

From (i), $a^{2} = 2 b^{2}$ , so 2 divides $a^{2}$ .

✍️ Yes/No

Yes or No?

If

a^2 = 2b^2

, does it mean that

a^2

is a multiple of

2

?

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

✍️ MCQ

Choose one

According to Theorem 1, if a prime

p

divides

a^2

, then

p

must also divide:

[Earns marks: correct substitution]

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

✍️ MCQ

Choose one

If 2 divides

a

, which of the following is correct?

"Substituting in (i): $2 b^{2} = (2 c)^{2} = 4 c^{2}$ ."

⚠️ Note: $(2 c)^{2} = 4 c^{2}$ , NOT $2 c^{2}$ . This is the most common error.

✍️ MCQ

Choose one

Calculate

(4x)^2

.

"So $b^{2} = 2 c^{2}$ ."

✍️ Yes/No

Yes or No?

If

b^2 = 2c^2

, is

b^2

an even number?

[Earns marks: second application]

Recall from our last step that $b^{2} = 2 c^{2}$ . This mathematically proves that 2 divides $b^{2}$ .

✍️ MCQ

Choose one

If the prime number

2

divides

b^2

, then according to Theorem 1:

Since 2 is prime, by Theorem 1, 2 divides $b$ .

The Contradiction Emerges

2 divides both $a$ and $b$ , so 2 is a common factor...

✍️ Yes/No

Yes or No?

If

a

and

b

have a common factor of

2

, can their HCF be

1

?

...contradicting $HCF (a, b) = 1$ .

Hence $\sqrt{2}$ is irrational.

2. Guided Step-by-Step -- Filling in the Proof

Let us build the proof one step at a time.

I will give you the structure
You fill in the critical details

This is your chance to practice the flow and logic we just reviewed.

📋 Given Info

Proof Skeleton: $\sqrt{2}$ is Irrational

Assume $\sqrt{2}$ is rational. Let $\sqrt{2} = \frac{a}{b}$ in simplest form, where $a$ and $b$ are co-prime integers.

Squaring: [BLANK 1]
From this equation: [BLANK 2 — what divides what?]
By Theorem 1: [BLANK 3]
Substitution: Let $a = 2 c$ . Substituting gives: [BLANK 4]
Simplifying: [BLANK 5]
By Theorem 1: [BLANK 6]
Conclusion: [BLANK 7 — State the contradiction]

Reminder: Each step must be explicitly justified to earn full marks from the examiner.

✍️ Question

Fill in all seven blanks.

Let's fill in BLANK 1.

We start with the assumption: $\sqrt{2} = \frac{a}{b}$ . To eliminate the square root, we square both sides: $(\sqrt{2})^{2} = {(\frac{a}{b})}^{2} ⟹ 2 = \frac{a^{2}}{b^{2}}$

Now, to clear the fraction, we multiply both sides by $b^{2}$ : $2 b^{2} = a^{2}$

✍️ MCQ

Choose one

If

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

, what is the rearranged equation after squaring and multiplying by

b^2

?

BLANK 2: Interpreting the equation.

From $2 b^{2} = a^{2}$ , we can see that $a^{2}$ is equal to $2$ times an integer ( $b^{2}$ ).

This means that 2 divides $a^{2}$ .

In other words, $a^{2}$ must be an even number.

✍️ Yes/No

Yes or No?

Is "2 divides

a^2

" the same as saying "

a^2

is a multiple of 2"?

Applying Theorem 1

BLANK 3: To use Theorem 1, we must satisfy two specific conditions:

The divisor must be a prime number (Is 2 prime? YES).
The divisor must divide the square of a number (Does 2 divide $a^{2}$ ? YES).

Since both are true, by Theorem 1, 2 divides $a$ .

✍️ Yes/No

Yes or No?

Does Theorem 1 apply if the divisor is a composite number, like

6

?

Substitution and Squaring

BLANK 4: Since we just proved 2 divides $a$ , we can write $a$ as a multiple of 2: $a = 2 c$

Now, let's substitute this back into our earlier equation, $2 b^{2} = a^{2}$ : $2 b^{2} = (2 c)^{2}$

When we square $(2 c)$ , we must square both the coefficient and the variable: $(2 c)^{2} = 2^{2} \times c^{2} = 4 c^{2}$

So, our equation becomes: $2 b^{2} = 4 c^{2}$ .

✍️ MCQ

Choose one

If

a = 3c

, then

a^2

is equal to:

BLANK 5: Divide $2 b^{2} = 4 c^{2}$ by 2: $b^{2} = 2 c^{2}$ .

✍️ Yes/No

Yes or No?

If

b^2 = 2c^2

, is

b^2

an even number?

BLANK 6: From $b^{2} = 2 c^{2}$ : 2 divides $b^{2}$ . Since 2 is prime, by Theorem 1, 2 divides $b$ .

✍️ MCQ

Choose one

Based on our proof so far, which number is a common factor of both

a

and

b

?

BLANK 7: 2 divides $a$ (from BLANK 3) and 2 divides $b$ (from BLANK 6). So 2 is a common factor of $a$ and $b$ .

✍️ Yes/No

Yes or No?

If

a

and

b

have a common factor of

2

, is it possible for their HCF to be

1

?

But we assumed $a$ and $b$ are co-prime (HCF = 1). Having a common factor of 2 contradicts this.

Hence $\sqrt{2}$ is irrational.

3. Adapting the Proof to sqrt(3)

The Real Test of Understanding 🎯

You've just proved that $\sqrt{2}$ is irrational. That's a significant achievement!

But here's the thing — the real test of whether you truly understand the proof isn't whether you can repeat it.

It's whether you can adapt it to a different prime without just mechanically replacing numbers.

Given Information

You have already proved that $\sqrt{2}$ is irrational.
The proof for $\sqrt{3}$ has exactly the same structure.

Your Task: Let's see if you can write the complete proof for $\sqrt{3}$ and articulate what changes versus what stays the same.

✍️ Question

Your Challenge ✍️

Write the complete proof that $\sqrt{3}$ is irrational.

After completing the proof, identify and state:

(a) What changed compared to the $\sqrt{2}$ proof?
(b) What stayed exactly the same?

The $\sqrt{3}$ proof is exactly the $\sqrt{2}$ proof with '2' replaced by '3'.

✍️ MCQ

Choose one

What is the first step in the proof for

\sqrt{3}

?

Let us write it out and compare.

By comparing the two, you will see that the logic of Theorem 1 and the substitution steps remain perfectly consistent.

✍️ MCQ

Choose one

What is the result of squaring

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

and rearranging?

Side-by-Side Comparison: $\sqrt{2}$ vs $\sqrt{3}$

$\sqrt{2}$ proof: Assume $\sqrt{2} = \frac{a}{b}$ (simplest form).
Square: $2 b^{2} = a^{2}$ .
So $2$ divides $a^{2}$ , by Theorem 1, $2$ divides $a$ .

$\sqrt{3}$ proof: Assume $\sqrt{3} = \frac{a}{b}$ (simplest form).
Square: $3 b^{2} = a^{2}$ .
So $3$ divides $a^{2}$ , by Theorem 1, $3$ divides $a$ .

✍️ FIB

Fill in the blank

If

3

divides

a^2

, then

3

also divides:

The Substitution Step

$\sqrt{2}$ Case:
Let $a = 2 c$ .
Then $2 b^{2} = 4 c^{2}$ , so $b^{2} = 2 c^{2}$ .
Thus, $2$ divides $b$ .

$\sqrt{3}$ Case:
Let $a = 3 c$ .
Then $3 b^{2} = 9 c^{2}$ , so $b^{2} = 3 c^{2}$ .
Thus, $3$ divides $b$ .

✍️ Yes/No

Yes or No?

Does having a common factor of

3

contradict the assumption that

\text{HCF}(a, b) = 1

?

Conclusion

In both cases, we reach a contradiction.

In the $\sqrt{2}$ proof, the common factor is 2.
In the $\sqrt{3}$ proof, the common factor is 3.

Since our initial assumption (that the number is rational) leads to a logical impossibility, the assumption must be wrong. Therefore, $\sqrt{2}$ and $\sqrt{3}$ are irrational.

Seeing the Pattern

See the pattern? The numbers in bold are the only things that change:

The prime itself: $2 \to 3$
Its square in the substitution step: $4 \to 9$

✍️ MCQ

Choose one

If we were proving

\sqrt{5}

is irrational, what number would replace the 4 or 9 in the substitution

(5c)^2

?

Everything else — the assumption, the squaring, the two applications of Theorem 1, the substitution, the contradiction — is identical.

✍️ Yes/No

Yes or No?

Does the logical structure of the proof change when we switch from

\sqrt{2}

to

\sqrt{3}

?

This is why mathematicians write the general proof with ' $p$ ' instead of a specific number: it handles ALL primes at once.

✍️ MCQ

Choose one

In the general proof for

\sqrt{p}

being irrational, what does

p

represent?

Memorise the template with ' $p$ ', and you can handle any prime the board exam throws at you.

✍️ Yes/No

Yes or No?

If you know the proof for

\sqrt{p}

is irrational, can you use the exact same logic to prove

\sqrt{7}

is irrational?

Executing the sqrt(2) Proof Step by Step

1. Filter -- Complete sqrt(2) Proof from Memory

The Big Challenge

The Contradiction Emerges

2. Guided Step-by-Step -- Filling in the Proof

Proof Skeleton: 2\sqrt{2}2​ is Irrational

Applying Theorem 1

Substitution and Squaring

3. Adapting the Proof to sqrt(3)

The Real Test of Understanding 🎯

Given Information

Your Challenge ✍️

Side-by-Side Comparison: 2\sqrt{2}2​ vs 3\sqrt{3}3​

The Substitution Step

Conclusion

Seeing the Pattern

Proof Skeleton: $\sqrt{2}$ is Irrational

Side-by-Side Comparison: $\sqrt{2}$ vs $\sqrt{3}$