Prove that $3 + 2\sqrt{5}$ is an irrational number, given that $\sqrt{5}$ is irrational.
Step-by-Step Solution
Key Concept: Assume $3 + 2\sqrt{5}$ is rational $a/b$, rearrange to isolate $\sqrt{5}$, and show LHS (irrational) = RHS (rational), yielding a contradiction.
Let us assume, on the contrary, that $3 + 2\sqrt{5}$ is rational. Then $3 + 2\sqrt{5} = \dfrac{a}{b}$, where $a, b \in \mathbb{Z}, b
eq 0$, and $\text{gcd}(a,b)=1$. [0.5 Mark]
Rearranging: $2\sqrt{5} = \dfrac{a}{b} - 3 = \dfrac{a - 3b}{b} \Rightarrow \sqrt{5} = \dfrac{a - 3b}{2b}$. [1.0 Mark]
Since $a, b$ are integers, $\dfrac{a - 3b}{2b}$ 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, our assumption is false, and $3 + 2\sqrt{5}$ is irrational. [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Assumption of rationality: 0.5 Mark
Correct algebraic isolation of $\sqrt{5}$: 1.0 Mark
Arguing RHS is rational: 1.0 Mark
Contradiction and conclusion: 0.5 Mark
Correct Answer: