Question

Question 10
Prove that $\sqrt{3}+\sqrt{5}$ irrational

Answer 1

Let us suppose that $\sqrt{3}+\sqrt{5}$ is rational.
Let $\sqrt{3}+\sqrt{5}=a$ is rational.
Therefore, $\sqrt{3}=a-\sqrt{5}$
On squaring both sides, we get,
$\left(\sqrt{3}\right)2=(a-\sqrt{5}{)}^2$
$\Rightarrow 3=a^2+5-2a\sqrt{5}$
$[\because (a-b{)}^2=a^2+b^2–2ab]$
$\Rightarrow 2a\sqrt{5}=a^2+2$
Therefore, $\sqrt{5}=\frac{a^2+2}{2a}$
As the right hand side is rational number while $\sqrt{5}$ is irrational. It contradicts with our assumption.
Hence, $\sqrt{3}+\sqrt{5}$ is irrational.