Mathematics — CBSE Class 10 board question

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

Then we can find coprime integers $a$ and $b$ ($b \neq 0$) such that:
$$2 + 3\sqrt{5} = \frac{a}{b}$$
$$3\sqrt{5} = \frac{a}{b} - 2 = \frac{a - 2b}{b}$$
$$\sqrt{5} = \frac{a - 2b}{3b}$$

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

But this contradicts the given fact that $\sqrt{5}$ is irrational.

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

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 $\sqrt{5}$ on one side, showing it equals a ratio of integers — which contradicts the given condition.
• Always state clearly: "this contradicts the fact that $\sqrt{5}$ is irrational" — examiners award a mark specifically for this contradiction statement.
• The phrase "given that $\sqrt{5}$ is irrational" means you do not need to prove $\sqrt{5}$'s irrationality; just use it as a fact.