Prove that $\sqrt{3}+\sqrt{5}$ is an irrational number.
Given that $\sqrt{3}+\sqrt{5}$ is an irrational number
Now we have to prove $\sqrt{3}+\sqrt{5}$ is an irrational number
Let $x=\sqrt{3}+\sqrt{5}$ is a rational
Squaring on both sides
$\Rightarrow x^{2}=(\sqrt{3}+\sqrt{5})^{2}$
$\Rightarrow x^{2}=(\sqrt{3})^{2}+(\sqrt{5})^{2}+2 \sqrt{3} \times \sqrt{5}$
$\Rightarrow x^{2}=3+5+2 \sqrt{15}$
$\Rightarrow x^{2}=8+2 \sqrt{15}$
$\Rightarrow \frac{x^{2}-8}{2}=\sqrt{15}$
Now is rational
$\Rightarrow x^{2}$ is rational
$\Rightarrow \frac{x^{2}-8}{2}$ is rational
$\Rightarrow \sqrt{15}$ is rational
But, $\sqrt{15}$ is an irrational
Thus we arrive at contradiction that $\sqrt{3}+\sqrt{5}$ is a rational which is wrong.
Hence $\sqrt{3}+\sqrt{5}$ is an irrational
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