Question:

Stepwise Solution:

Let us assume, on the contrary, that $\sqrt{3} + \sqrt{5}$ is rational, say $x = \sqrt{3} + \sqrt{5}$ where $x \in \mathbb{Q}$. [0.5 Mark]

Squaring both sides: $x^2 = (\sqrt{3} + \sqrt{5})^2 = 3 + 5 + 2\sqrt{15} = 8 + 2\sqrt{15}$. [1.0 Mark]

Rearranging: $x^2 - 8 = 2\sqrt{15} \Rightarrow \sqrt{15} = \dfrac{x^2 - 8}{2}$. [1.0 Mark]

Since $x$ is rational, $\dfrac{x^2 - 8}{2}$ is rational, which implies $\sqrt{15}$ is rational. But $15 = 3 \times 5$ is a product of two distinct primes, so $\sqrt{15}$ is irrational. Contradiction! Hence $\sqrt{3} + \sqrt{5}$ is irrational. [0.5 Mark]

Marking Scheme:

• Assumption of rationality $x = \sqrt{3} + \sqrt{5}$: 0.5 Mark
• Squaring both sides correctly: 1.0 Mark
• Isolating $\sqrt{15}$: 1.0 Mark
• Contradiction deduction and final conclusion: 0.5 Mark