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

Proof (by contradiction):

Assume $2 + \sqrt{3}$ is rational. Then we can write:

$$2 + \sqrt{3} = \frac{a}{b}, \quad \text{where } a, b \text{ are integers and } b \neq 0$$

Rearranging:

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

Since $a$ and $b$ are integers, $\dfrac{a-2b}{b}$ 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 $2 + \sqrt{3}$ is irrational. $\blacksquare$

Source: Chapter 1, Section 1.3 – Revisiting Irrational Numbers

---

• This is a standard "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 — showing it equals a rational expression, which contradicts the given condition.
• Examiners award marks for: (1) correct assumption, (2) correct rearrangement isolating $\sqrt{3}$, (3) stating the contradiction clearly, (4) conclusion. Don't skip the conclusion line.