Classic proof by contradiction. Assume √2 = p/q in lowest terms, derive a contradiction.

Prove √2 is Irrational — Contradiction Method

Question

Prove that $\sqrt{2}$ is irrational.

This is a Class 10 NCERT proof that appears in almost every board exam. The full proof is worth 3-4 marks — and students lose marks not because they don’t know the idea, but because they write it sloppily.

Solution — Step by Step

We use proof by contradiction: assume $\sqrt{2}$ IS rational.

That means we can write $\sqrt{2} = \dfrac{p}{q}$ , where $p$ and $q$ are integers, $q \neq 0$ , and $\frac{p}{q}$ is in its lowest terms — meaning $p$ and $q$ share no common factor (their HCF is 1).

\sqrt{2} = \frac{p}{q} \implies 2 = \frac{p^2}{q^2} \implies p^2 = 2q^2

So $p^2$ is even (it equals $2 \times$ something). Here’s the key logical step: if $p^2$ is even, then $p$ itself must be even. We’ll use this fact right now.

Since $p$ is even, write $p = 2m$ for some integer $m$ .

Substitute into $p^2 = 2q^2$ :

(2m)^2 = 2q^2 \implies 4m^2 = 2q^2 \implies q^2 = 2m^2

So $q^2$ is also even, which means $q$ is also even.

We now have: $p$ is even AND $q$ is even.

But we assumed $\frac{p}{q}$ was in lowest terms — meaning $p$ and $q$ have no common factors. If both are even, they’re both divisible by 2. That’s a contradiction.

Our assumption that $\sqrt{2}$ is rational has led to a contradiction. Therefore, $\sqrt{2}$ is irrational. $\blacksquare$

Why This Works

The proof hinges on one number theory fact: if $n^2$ is even, then $n$ is even. This follows because odd $\times$ odd = odd — so if $n$ were odd, $n^2$ would also be odd. That’s the contrapositive, which is equally valid.

We apply this fact twice — once for $p$ , once for $q$ . Both turn out to be even, which breaks the “lowest terms” condition we set up at the start. That forced contradiction is what makes the proof work.

Proof by contradiction is powerful here because there’s no direct way to “show” a number is irrational. Instead, we show that assuming rationality creates an impossible situation.

Alternative Method

You can frame the same argument using the Fundamental Theorem of Arithmetic (prime factorisation uniqueness).

If $p^2 = 2q^2$ , look at the prime factorisation of both sides. The left side $p^2$ has every prime appearing an even number of times (since squaring doubles all exponents). The right side has $2^1 \times q^2$ — which means 2 appears an odd number of times (one extra from the $2^1$ factor).

This violates the uniqueness of prime factorisation. Same contradiction, slightly different language — useful if the examiner asks for a proof without the “lowest terms” setup.

Common Mistake

Not stating “in lowest terms” at the start.

Many students write ” $\sqrt{2} = \frac{p}{q}$ where $p, q$ are integers” and stop there. The entire contradiction depends on $\frac{p}{q}$ being already fully reduced. Without that condition, finding that $p$ and $q$ are both even is not a contradiction — you’d just say “simplify further.” Always write: “where $p$ and $q$ are coprime integers (HCF = 1).”

In board exams, write the conclusion clearly: “This contradicts our assumption that $p$ and $q$ are coprime. Hence $\sqrt{2}$ cannot be rational, so it is irrational.” Examiners specifically look for the explicit contradiction statement — don’t just trail off.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next