Prove that 3–√+5–√ is an irrational number.

Solution:

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