Prove that $2 + 3\sqrt{5}$ is an irrational number, given that $\sqrt{5}$ is an irrational number.
Step-by-Step Solution
Key Concept: Assume $2 + 3\sqrt{5} = r$ (rational), isolate $\sqrt{5} = \dfrac{r - 2}{3}$, showing LHS (irrational) = RHS (rational), a contradiction.
Let us assume, on the contrary, that $2 + 3\sqrt{5}$ is rational. Then $2 + 3\sqrt{5} = \dfrac{a}{b}$ where $a, b \in \mathbb{Z}, b
eq 0$ and $\text{gcd}(a,b) = 1$. [0.5 Mark]
Rearranging: $3\sqrt{5} = \dfrac{a}{b} - 2 = \dfrac{a - 2b}{b} \Rightarrow \sqrt{5} = \dfrac{a - 2b}{3b}$. [1.0 Mark]
Since $a$ and $b$ are integers, $\dfrac{a - 2b}{3b}$ is a rational number. This implies that $\sqrt{5}$ is rational. [1.0 Mark]
But this contradicts the given fact that $\sqrt{5}$ is irrational. Hence, $2 + 3\sqrt{5}$ is irrational. Proved! [0.5 Mark]
---
🎯 Official CBSE Marking Scheme:
Rationality assumption: 0.5 Mark
Isolating $\sqrt{5}$: 1.0 Mark
Arguing RHS is rational: 1.0 Mark
Contradiction statement and conclusion: 0.5 Mark
Correct Answer: