Prove that

Let us assume that $\sqrt{5}+\sqrt{3}$ is rational .Then, there exist positive co primes $a$ and $b$ such that

$\sqrt{5}+\sqrt{3}=\frac{a}{b}$

$\sqrt{5}=\frac{a}{b}-\sqrt{3}$

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

$5=\left(\frac{\mathrm{a}}{\mathrm{b}}\right)^{2}-\frac{2 a \sqrt{3}}{b}+3$

$\Rightarrow \quad 5-3=\left(\frac{\mathrm{a}}{\mathrm{b}}\right)^{2}-\frac{2 a \sqrt{3}}{b}$

$\Rightarrow \quad 2=\left(\frac{\mathrm{a}}{\mathrm{b}}\right)^{2}-\frac{2 a \sqrt{3}}{b}$

$\Rightarrow \quad\left(\frac{\mathrm{a}}{\mathrm{b}}\right)^{2}-2=\frac{2 a \sqrt{3}}{b}$

$\Rightarrow \quad \frac{a^{2}-2 b^{2}}{b^{2}}=\frac{2 a \sqrt{3}}{b}$

$\Rightarrow \quad\left(\frac{a^{2}-2 b^{2}}{b^{2}}\right)\left(\frac{b}{2 a}\right)=\sqrt{3}$

$\Rightarrow \quad \sqrt{3}=\left(\frac{a^{2}-2 b^{2}}{2 a b}\right)$

Here we see that $\sqrt{3}$ is a rational number which is a contradiction as we know that $\sqrt{3}$ is an irrational number.

Hence $\sqrt{5}+\sqrt{3}$ is irrational