Prove that $2 + \sqrt{3}$ is an irrational number, given that $\sqrt{3}$ is an irrational number.
Generated by claude-sonnet-4-6 · 2026-06-15 10:29 · grounding rag
Model Answer
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
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Explanation
- 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.