Question

Prove that √3 +√5 is irrational.

Answer 1

(proof by contradiction)

Suppose $\sqrt{5}+\sqrt{3} = \dfrac{a}{b}$ is rational.
Multiplying both sides by $\sqrt{5}-\sqrt{3}$ gives
$2=\dfrac{a}{b}(\sqrt{5}-\sqrt{3})\implies\sqrt{5}-\sqrt{3}=\dfrac{2b}{a}$

Now $(\sqrt{5}+\sqrt{3})+(\sqrt{5}-\sqrt{3})=\dfrac{a}{b}+\dfrac{2b}{a}$
$2\sqrt{5}=\dfrac{a^2+2b^2}{ab}$
$\sqrt{5}=\dfrac{a^2+2b^2}{2ab}$

Which is a contradiction because $\sqrt{5}$ is not rational. 

The only resolution to this contradiction is that the starting assumption is wrong. That is, $\sqrt{5}+\sqrt{3}$ is irrational.

Answer 2

√3 +√5 let a is an rational number 
√3 +√5 = asquaring on both sides
(√3 +√5 )² = a²3+5+2√3 √5  = a²
8+2√15 = a²
a²-8 = 2√15
$\frac{ a^{2} -8}{2}$  = √15
if a is rational then $\frac{ a^{2} -8}{2}$ is also a rational
but √15 is a irrational
it is contradict to our assumption.
so √3 +√5 is not a rational