Question

Prove that √2 +√5 is irrational

Answer 1

We have to prove that $\sqrt{2}+\sqrt{5}$ is irrational. We can prove this statement by contradiction.

Let us assume that $\sqrt{2}+\sqrt{5}$ is a rational number. Therefore, we can express:

$a=\sqrt{2}+\sqrt{5}$

Let us represent this equation as:

$a-\sqrt{2}=\sqrt{5}$

Upon squaring both the sides:

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

Since a has been assumed to be rational, therefore, $\frac{a^{2}-3}{2a}$ must as well be rational.

But we know that $\sqrt{2}$ is irrational, therefore, from equation $\sqrt{2}=\frac{a^{2}-3}{2a}$ the expression $\frac{a^{2}-3}{2a}$ must be irrational, which contradicts with our claim.

Therefore, by contradiction,  $\sqrt{2}+\sqrt{5}$ is irrational.