Prove that $\dfrac{1}{\sqrt{2}}$ is an irrational number.
Step-by-Step Solution
Key Concept: Assume $\dfrac{1}{\sqrt{2}} = \dfrac{a}{b}$ (rational), so $\sqrt{2} = \dfrac{b}{a}$ (rational), contradicting irrationality of $\sqrt{2}$.
Let us assume, on the contrary, that $\dfrac{1}{\sqrt{2}}$ is rational. Then $\dfrac{1}{\sqrt{2}} = \dfrac{a}{b}$ where $a, b \in \mathbb{Z}, b
eq 0, a
eq 0$ and $\text{gcd}(a,b) = 1$. [0.5 Mark]
Reciprocating both sides: $\sqrt{2} = \dfrac{b}{a}$. [0.5 Mark]
Since $a$ and $b$ are integers ($a
eq 0$), $\dfrac{b}{a}$ is a rational number. This implies $\sqrt{2}$ is rational. [0.5 Mark]
But this contradicts the fact that $\sqrt{2}$ is irrational. Hence, $\dfrac{1}{\sqrt{2}}$ is irrational. Proved! [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Contradiction setup: 0.5 Mark
Reciprocating to get $\sqrt{2} = b/a$: 0.5 Mark
Arguing $b/a$ is rational: 0.5 Mark
Deducing contradiction and conclusion: 0.5 Mark
Correct Answer: