If $p$ and $q$ are distinct prime numbers, prove that $\sqrt{p} + \sqrt{q}$ is an irrational number.
Step-by-Step Solution
Key Concept: Let $x = \sqrt{p} + \sqrt{q}$ be rational. Square both sides to isolate $\sqrt{pq} = \dfrac{x^2 - p - q}{2}$, showing $\sqrt{pq}$ would be rational, a contradiction.
Stepwise Solution:
Let us assume, on the contrary, that $x = \sqrt{p} + \sqrt{q}$ is a rational number. [0.5 Mark]
Squaring both sides: $x^2 = (\sqrt{p} + \sqrt{q})^2 = p + q + 2\sqrt{pq}$. [1.0 Mark]
Rearranging: $x^2 - p - q = 2\sqrt{pq} \Rightarrow \sqrt{pq} = \dfrac{x^2 - p - q}{2}$. [1.0 Mark]
Since $x, p, q$ are rational/integers, $\dfrac{x^2 - p - q}{2}$ is rational, implying $\sqrt{pq}$ is rational. But $p, q$ are distinct primes, so $pq$ is not a perfect square, making $\sqrt{pq}$ irrational. Contradiction! Thus $\sqrt{p} + \sqrt{q}$ is irrational. Proved! [0.5 Mark]
Marking Scheme:
• Assumption $x = \sqrt{p} + \sqrt{q}$ rational: 0.5 Mark
• Squaring both sides correctly: 1.0 Mark
• Isolating $\sqrt{pq} = (x^2 - p - q)/2$: 1.0 Mark
• Deducing contradiction (since $pq$ square-free $\Rightarrow \sqrt{pq}$ irrational): 0.5 Mark
Correct Answer: