Question

Prove that $\sqrt{5}-\sqrt{3}$ is not a rational number.

Answer 1

Let $\sqrt{5}-\sqrt{3}$ be a rational number of form $\frac{a}{b}$ ,where $b\ne 0$

Squaring on both sides

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

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

$5+3+2\sqrt{1}5=\frac{a^2}{b^2}$

$8+2\sqrt{1}5=\frac{a^2}{b^2}$

$2\sqrt{1}5=\frac{a^2}{b^2}-8$

$\sqrt{1}5=\frac{a^2-8b^2}{2b^2}$

since $\sqrt{1}5$ is irrational ,$\frac{a^2-8b^2}{2b^2}$ is rational

Since LHS $\ne$ RHS, contradiction arises,

Therefore $\sqrt{5}-\sqrt{3}$ is irrational