(i) Prove that if $p$ is a prime number, then $\sqrt{p}$ is irrational. [3 Marks]
(ii) Using the result of (i), deduce whether $\sqrt{p} + \sqrt{q}$ is irrational when both $p$ and $q$ are distinct prime numbers. [2 Marks]
Step-by-Step Solution
Key Concept: (i) Proof by contradiction for $\sqrt{p}$; (ii) Let $x = \sqrt{p} + \sqrt{q}$, square both sides to isolate $\sqrt{pq}$, which is irrational since $pq$ is square-free.
(i) Suppose $\sqrt{p} = \dfrac{a}{b}$ where $a, b \in \mathbb{Z}, b
eq 0$, and $\text{gcd}(a,b) = 1$. Then $a^2 = p b^2 \Rightarrow p \mid a^2 \Rightarrow p \mid a$. Let $a = p k$. Then $(pk)^2 = p b^2 \Rightarrow p^2 k^2 = p b^2 \Rightarrow b^2 = p k^2 \Rightarrow p \mid b^2 \Rightarrow p \mid b$. Thus $p$ is a common factor of $a$ and $b$, contradicting $\text{gcd}(a,b) = 1$. Hence $\sqrt{p}$ is irrational. [3.0 Marks]
(ii) Let $x = \sqrt{p} + \sqrt{q}$ be rational. Then $x^2 = p + q + 2\sqrt{pq} \Rightarrow \sqrt{pq} = \dfrac{x^2 - p - q}{2}$. Since $x, p, q$ are integers/rationals, $\dfrac{x^2 - p - q}{2}$ is rational, implying $\sqrt{pq}$ is rational. But $p, q$ are distinct primes, so $pq$ has no square factors $>1$, making $\sqrt{pq}$ irrational by part (i). Contradiction! Hence $\sqrt{p} + \sqrt{q}$ is irrational. [2.0 Marks]
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🎯 Official CBSE Marking Scheme:
Part (i) Complete proof by contradiction for $\sqrt{p}$: 3.0 Marks (1.0 for setup, 1.0 for $p \mid a$, 1.0 for $p \mid b$ and contradiction)
Part (ii) Setting up $x = \sqrt{p} + \sqrt{q}$ and squaring: 1.0 Mark
Part (ii) Isolating $\sqrt{pq}$ and applying part (i) to conclude irrationality: 1.0 Mark
Correct Answer: