Prove that √2 is irrational

Question

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

Solution — Step by Step

We use proof by contradiction: assume that $\sqrt{2}$ IS rational, and show this leads to an impossibility.

If $\sqrt{2}$ is rational, it can be written as:

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

where $p$ and $q$ are integers, $q \neq 0$ , and $\frac{p}{q}$ is in its lowest terms (meaning $p$ and $q$ have no common factors — their HCF is 1).

Squaring both sides:

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

\Rightarrow p^2 = 2q^2

This tells us $p^2$ is even (it equals 2 times something). If $p^2$ is even, then $p$ itself must be even. (Because if $p$ were odd, $p^2$ would be odd — odd × odd = odd.)

So we can write $p = 2m$ for some integer $m$ .

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

(2m)^2 = 2q^2

4m^2 = 2q^2

q^2 = 2m^2

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

We’ve shown that both $p$ and $q$ are even. This means they share a common factor of 2.

But we assumed at the start that $\frac{p}{q}$ is in its lowest terms — meaning $p$ and $q$ have no common factors. Having both be even contradicts this assumption.

We have a contradiction.

Our assumption that $\sqrt{2}$ is rational led to a contradiction. Therefore, our assumption is false.

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

Why This Works

This style of proof is called Reductio Ad Absurdum (reduction to absurdity) — one of the most elegant tools in mathematics. We don’t try to show something directly; instead, we show that the opposite is impossible.

The key insight is the relationship between even squares and even numbers: if $n^2$ is even, then $n$ must be even. This follows because even numbers have 2 as a prime factor, and in $n^2 = n \times n$ , that factor of 2 must come from $n$ itself (prime factors of a square are just doubled powers of the prime factors of the original).

This exact proof — with $\sqrt{2}$ being irrational — appears in CBSE Class 10 (Number Systems) and is a 3-mark question. The proof structure works for $\sqrt{3}$ , $\sqrt{5}$ , and any $\sqrt{p}$ where $p$ is prime. For $\sqrt{6}$ , you’d show both $p$ and $q$ are divisible by 6, or use the fact that $6 = 2 \times 3$ and track divisibility by 2 and 3 separately.

Alternative — Why Can’t We Just Use Decimals?

You might think: ” $\sqrt{2} = 1.41421356...$ , and this non-terminating non-repeating decimal proves it’s irrational.” That’s true, but it’s circular — we’d need to prove the decimal is indeed non-terminating and non-repeating, which requires essentially the same logic. The algebraic proof above is cleaner and more rigorous.

Common Mistake

The most common error in this proof is forgetting to state that $\frac{p}{q}$ is in lowest terms at the beginning. Without this assumption, the contradiction at Step 4 doesn’t work — of course two integers can share factors in general. The assumption of lowest terms is what makes the contradiction powerful. CBSE marks are deducted if this crucial starting condition is omitted.

Question

Solution — Step by Step

Why This Works

Alternative — Why Can’t We Just Use Decimals?

Common Mistake

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