Mathematics — CBSE Class 10 board question

Assume, to the contrary, that $2 - 5\sqrt{3}$ is rational.

Then there exist coprime integers $a$ and $b$ ($b \neq 0$) such that:

$$2 - 5\sqrt{3} = \frac{a}{b}$$

$$\Rightarrow 5\sqrt{3} = 2 - \frac{a}{b} = \frac{2b - a}{b}$$

$$\Rightarrow \sqrt{3} = \frac{2b - a}{5b}$$

Since $a$ and $b$ are integers, $\dfrac{2b-a}{5b}$ is rational, which means $\sqrt{3}$ is rational.

But this contradicts the given fact that $\sqrt{3}$ is irrational.

Therefore, our assumption is wrong and $2 - 5\sqrt{3}$ is irrational.

Source: Chapter 1, Section 1.3 — Revisiting Irrational Numbers

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• This is a proof by contradiction question — always start by assuming the opposite of what you want to prove.
• The key step is isolating $\sqrt{3}$ on one side to show it would have to be rational — contradicting the given condition.
• Examiners award marks for: (1) correct assumption, (2) algebraic rearrangement to isolate $\sqrt{3}$, (3) stating the contradiction clearly. Don't skip the contradiction statement.