Prove that √2 + √5 is irrational

Solution:

Let us suppose that √2 + √5 is rational.

Let $\sqrt{2} + \sqrt{5} = \frac{a}{b}$ , where a and b coprime integers and b ≠ 0.

Now $\sqrt{2} + \sqrt{5} = \frac{a}{b}$

$\Rightarrow \frac{a}{b}-\sqrt{2} = \sqrt{5}$

Squaring both side

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

$\Rightarrow \dfrac{(a-\sqrt{2}b)^2}{b^2}=5$

$\Rightarrow \dfrac{a^2+2b^2-2\sqrt{2}ab}{b^2}=5$

$\Rightarrow a^2+2b^2-2\sqrt{2}ab = 5b^2$

$\Rightarrow -2\sqrt{2}ab = 3b^2-a^2$

$\Rightarrow \sqrt{2} = \dfrac{a^2-3b^2}{2ab}$

⇒ √2 is rational, since $\dfrac{a^2-3b^2}{2ab}$ is rational.

This contradict the fact that √2 is an irrational number.

So, our assumption is wrong.

Hence, √2 + √5 is irrational