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

Proof: Assume, to the contrary, that $\sqrt{3}$ is rational.

Then we can find integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{3} = \dfrac{a}{b}$, where $a$ and $b$ are coprime (no common factor other than 1).

So, $b\sqrt{3} = a$.

Squaring both sides: $3b^2 = a^2$

Therefore, $a^2$ is divisible by 3, and by Theorem 1.2, $a$ is also divisible by 3.

So, let $a = 3c$ for some integer $c$.

Substituting: $3b^2 = 9c^2 \Rightarrow b^2 = 3c^2$

This means $b^2$ is divisible by 3, so $b$ is also divisible by 3.

Therefore, $a$ and $b$ have at least 3 as a common factor. But this contradicts the fact that $a$ and $b$ are coprime.

This contradiction arose because of our incorrect assumption. Hence, $\sqrt{3}$ is irrational. $\blacksquare$

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.
• The key tool used is Theorem 1.2: if a prime $p$ divides $a^2$, then $p$ divides $a$.
• The critical step students miss: after substituting $a = 3c$, show that 3 also divides $b$ — this creates the contradiction with "coprime."
• Write the conclusion clearly: state the contradiction and then state that $\sqrt{3}$ is irrational.
• Examiners award marks for: assumption (½), algebraic steps (1), applying Theorem 1.2 twice (½), contradiction + conclusion (1).