Prove that $\sqrt{3} + \sqrt{5}$ is an irrational number.
Step-by-Step Solution
Key Concept: Square $x = \sqrt{3} + \sqrt{5} \Rightarrow x^2 = 8 + 2\sqrt{15} \Rightarrow \sqrt{15} = (x^2 - 8)/2$.
Let $x = \sqrt{3} + \sqrt{5} \in \mathbb{Q} \Rightarrow x^2 = 3 + 5 + 2\sqrt{15} = 8 + 2\sqrt{15}$. [1.0 Mark]
$\sqrt{15} = (x^2 - 8)/2$. RHS is rational, so $\sqrt{15}$ would be rational. [1.0 Mark]
Since $15 = 3 \times 5$ is a product of two distinct primes, $\sqrt{15}$ is irrational. Contradiction! Hence $\sqrt{3} + \sqrt{5}$ is irrational. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Squaring $x = \sqrt{3} + \sqrt{5}$: 1.0 Mark
Isolating $\sqrt{15}$: 1.0 Mark
Contradiction argument for $\sqrt{15}$: 1.0 Mark
Correct Answer: