Mathematics — Standard (041) — CBSE Class 10 board question

Proof: Assume, to the contrary, that $\sqrt{2}$ is rational.

Then we can find integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{2} = \dfrac{a}{b}$, where $a$ and $b$ are coprime (no common factor other than 1).

Squaring both sides: $2b^2 = a^2$

So 2 divides $a^2$. By Theorem 1.2, 2 divides $a$.

Let $a = 2c$ for some integer $c$. Substituting:

$$2b^2 = 4c^2 \implies b^2 = 2c^2$$

So 2 divides $b^2$, which means 2 divides $b$.

Thus 2 is a common factor of both $a$ and $b$, which contradicts the fact that $a$ and $b$ are coprime.

This contradiction arose because of our incorrect assumption. Hence, $\sqrt{2}$ is irrational.

Source: Chapter 1, Section 1.3 – Revisiting Irrational Numbers

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• This is a proof by contradiction — always start by assuming the opposite of what you want to prove.
• Key step: use Theorem 1.2 (if a prime $p$ divides $a^2$, then $p$ divides $a$) — examiners expect you to cite this.
• The contradiction must be clearly stated: $a$ and $b$ were assumed coprime, but both are divisible by 2.
• Write "Hence $\sqrt{2}$ is irrational" as the concluding line — never omit the conclusion.