Prove that√3 + √5 is irrational.

Let us suppose that √3 + √5 is rational.

Let √3 + √5 = a, where a is rational.

Therefore, $\sqrt{3}=a-\sqrt{5}$

On squaring both sides, we get

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

$\Rightarrow$ $3=a^{2}+5-2 a \sqrt{5}$ $\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$

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

Therefore, $\sqrt{5}=\frac{a^{2}+2}{2 a}$ which is contradiction.

As the right hand side is rational number while V5 is irrational. Since, 3 and 5 are prime numbers. Hence, √3 + √5 is irrational.