Question

Prove that $\sqrt{2}+\sqrt{3}$ is an irrational number .

Answer 1

Let us assume that $\sqrt{2}+\sqrt{3}$ is rational . Then, there exist co-prime positive integers a and b such that
$$\begin{gathered} \sqrt{2}+\sqrt{3}=\frac{a}{b} \\ \Rightarrow \sqrt{3}=\frac{a}{b}-\sqrt{2} \end{gathered}$$
Squaring on both sides, we get
$$\begin{gathered} 3=\frac{a^2}{b^2}-\frac{2a}{b}\sqrt{2}+2 \\ \Rightarrow \frac{a^2}{b^2}-1=\frac{2a}{b}\sqrt{2} \\ \Rightarrow \frac{a^2-b^2}{2ab}=\sqrt{2} \end{gathered}$$
$\Rightarrow \sqrt{2}$ is a rational number $\left(a,b\mathrm{are}\mathrm{integers},\mathrm{so}\frac{a^2-b^2}{2ab}\mathrm{is}\mathrm{rational}\right)$
This contradicts the fact that $\sqrt{2}$ is a rational number. So our assumption was incorrect.
Hence, $\sqrt{2}+\sqrt{3}$ is an irrational number.