Question

Prove the following are irrational.
$\sqrt{3}+\sqrt{5}$

Answer 1

Let us assume that $\sqrt{3}+\sqrt{5}$ is rational.

That is, we can find two integers $a$ and $b(b\ne 0)$ such that,

$\sqrt{3}+\sqrt{5}=\frac{a}{b}$ where $a$ and $b$ are co-prime integers.

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

Squaring on both sides we get,

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

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

$⟹\sqrt{3}=\frac{a^2-2b^2}{2ab}$

Since, $a$ and $b$ are integers, thus $\frac{a^2-2b^2}{2ab}$ is rational. So, $\sqrt{3}$ is also rational.

But, this contradicts the fact that $\sqrt{3}$ is irrational.

Thus, our assumption is wrong that $\sqrt{3}+\sqrt{5}$ is rational.

Hence, $\sqrt{3}+\sqrt{5}$ is an irrational.