Prove √2 is irrational using proof by contradiction

Question

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

(CBSE 2023 — this proof has appeared in board exams almost every year)

Solution — Step by Step

Suppose $\sqrt{2}$ is rational. Then we can write:

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

where $p$ and $q$ are integers with no common factor (i.e., $p/q$ is in lowest terms, $\gcd(p, q) = 1$ ).

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

p^2 = 2q^2

This tells us $p^2$ is even (since it equals $2 \times$ something). If $p^2$ is even, then $p$ must be even. Why? Because an odd number squared is always odd.

Since $p$ is even, let $p = 2k$ for some integer $k$ .

(2k)^2 = 2q^2

4k^2 = 2q^2

q^2 = 2k^2

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

We’ve shown that both $p$ and $q$ are even — meaning both are divisible by 2. But we started by assuming $p/q$ is in lowest terms (no common factor).

This is a contradiction. Our assumption that $\sqrt{2}$ is rational must be wrong.

Therefore, $\sqrt{2}$ is irrational. $\blacksquare$

Why This Works

The proof hinges on one key fact: if $n^2$ is even, then $n$ is even. This is because odd numbers squared stay odd ( $\text{odd} \times \text{odd} = \text{odd}$ ). So the only way to get an even square is from an even number.

The contradiction method works beautifully here because directly proving “something is irrational” is hard — there’s no formula for it. But proving “if it were rational, something impossible would happen” is clean and logically airtight.

This same structure can be adapted to prove $\sqrt{3}$ , $\sqrt{5}$ , or any $\sqrt{p}$ where $p$ is prime — just replace “even” with “divisible by $p$ ”.

Alternative Method — Using the Fundamental Theorem of Arithmetic

We can also argue using prime factorisation. In $p^2 = 2q^2$ , count the number of times 2 appears as a prime factor on each side. On the left, $p^2$ has an even count of 2’s. On the right, $2q^2$ has an odd count (one extra 2 plus the even count from $q^2$ ). Even $\neq$ odd, contradiction.

CBSE has asked this proof in 2019, 2020, 2022, and 2023. It carries 3-4 marks. The marking scheme specifically awards marks for: (1) correct assumption with $\gcd(p,q) = 1$ , (2) showing $p$ is even, (3) showing $q$ is even, (4) stating the contradiction clearly. Don’t skip any step.

Common Mistake

Many students forget to state that $p/q$ is in lowest terms at the start. Without this assumption, there’s no contradiction at the end — because if $p$ and $q$ could have common factors, both being even wouldn’t be a problem. The " $\gcd(p, q) = 1$ " condition is what makes the proof work.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the Fundamental Theorem of Arithmetic

Common Mistake

Try These Next