Question

Prove that ${\displaystyle \sqrt{3}+\sqrt{5}}$ is irrational.

Answer 1

Let us suppose ${\displaystyle \sqrt{3}+\sqrt{5}}$ is rational.

Let ${\displaystyle \sqrt{3}+\sqrt{5}}$ = a, where a is rational.

Therefore, ${\displaystyle \sqrt{3}=a-\sqrt{5}}$ 

On squaring both sides, we get

${\displaystyle {\left(\sqrt{3}\right)}^2={(a-\sqrt{5})}^2}$

${\displaystyle \Rightarrow }$ 3 = ${\displaystyle a^2+5-2a\sqrt{5}}$   ......[∵ (a – b)² = a² + b² – 2ab]

${\displaystyle \Rightarrow }$ ${\displaystyle 2a\sqrt{5}=a^2+2}$

Therefore, ${\displaystyle \sqrt{5}=\frac{a^2+2}{2a}}$, which is a contradiction as the right-hand side is rational number while ${\displaystyle \sqrt{5}}$ is irrational.

Hence, ${\displaystyle \sqrt{3}+\sqrt{5}}$ is irrational.