Real Numbers: Proof of Irrationality of Sum of Radicals

Let us assume to the contrary that $\sqrt{3} + \sqrt{5}$ is rational. Then there exist co-prime integers $a$ and $b$ ($b \neq 0$) such that: $\sqrt{3} + \sqrt{5} = \frac{a}{b}$

$\sqrt{5} = \frac{a}{b} - \sqrt{3}$ Squaring both sides: $(\sqrt{5})^2 = \left( \frac{a}{b} - \sqrt{3} \right)^2$ $5 = \frac{a^2}{b^2} - 2\sqrt{3}\frac{a}{b} + 3$ $5 - 3 - \frac{a^2}{b^2} = -2\sqrt{3}\frac{a}{b} \implies 2 - \frac{a^2}{b^2} = -2\sqrt{3}\frac{a}{b}$ $\frac{a^2 - 2b^2}{b^2} = 2\sqrt{3}\frac{a}{b}$ $\sqrt{3} = \frac{a^2 - 2b^2}{2ab}$

Since $a$ and $b$ are integers, $\frac{a^2 - 2b^2}{2ab}$ is a rational number. This implies that $\sqrt{3}$ is rational, which contradicts the known fact that $\sqrt{3}$ is irrational. Hence, our assumption that $\sqrt{3} + \sqrt{5}$ is rational was false. Therefore, $\sqrt{3} + \sqrt{5}$ is irrational.