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

Proof: Assume, to the contrary, that $\sqrt{5}$ is rational.

Then we can find integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{5} = \dfrac{a}{b}$, where $a$ and $b$ are coprime.

Squaring both sides: $5b^2 = a^2$

So 5 divides $a^2$. By Theorem 1.2, 5 divides $a$.

Write $a = 5c$ for some integer $c$. Substituting:

$5b^2 = 25c^2 \Rightarrow b^2 = 5c^2$

So 5 divides $b^2$, and therefore 5 divides $b$.

Thus 5 is a common factor of $a$ and $b$, contradicting the fact that $a$ and $b$ are coprime.

This contradiction arose because we assumed $\sqrt{5}$ is rational. Hence, $\sqrt{5}$ is irrational. $\blacksquare$

Source: Chapter 1, Section 1.3 – Revisiting Irrational Numbers

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• This is a standard proof by contradiction. Always start by assuming the opposite of what you want to prove.
• The key tool is Theorem 1.2: if a prime $p$ divides $a^2$, then $p$ divides $a$. Examiners expect you to cite or use this explicitly.
• The contradiction must target the coprime condition — that is where the marks are.
• Write the substitution step ($a = 5c$) clearly; missing it loses a mark.