Prove that $2 + 3\sqrt{5}$ is an irrational number given that $\sqrt{5}$ is irrational.
Generated by claude-sonnet-4-6 · 2026-06-15 10:28 · grounding rag
Model Answer
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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Explanation
- 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.