Question

Prove that $\sqrt{2}+\sqrt{5}$ is irrational number

Answer 1

Solve for the required proof:

Let us assume $\sqrt{2}+\sqrt{5}$ is not a irrational number, that is $\sqrt{2}+\sqrt{5}$ is a rational number

So $\sqrt{2}+\sqrt{5}$ can be expressed as $\frac{p}{q}$ where $p,q\in \mathrm{\mathbb{Z}}$ and $q\ne 0$

$$\begin{gathered} \therefore \sqrt{2}+\sqrt{5}=\frac{p}{q} \\ \Rightarrow {\left(\sqrt{2}+\sqrt{5}\right)}^2={\left(\frac{p}{q}\right)}^2 \\ \Rightarrow 2+2\sqrt{10}+5={\left(\frac{p}{q}\right)}^2\left[\because {\left(a+b\right)}^2=a^2+b^2+2ab\right] \\ \Rightarrow 2\sqrt{10}={\left(\frac{p}{q}\right)}^2-7 \\ \Rightarrow \sqrt{10}=\frac{p^2-7q^2}{2q^2} \end{gathered}$$

Since, $p,q\in \mathrm{\mathbb{Z}}$ and $q\ne 0$, then $\frac{p^2-7q^2}{2q^2}$ must be a rational number.

But $\sqrt{10}$ is irrational .

Hence contradiction

So our assumption $\sqrt{2}+\sqrt{5}$ is a rational number is incorrect

Therefore $\sqrt{2}+\sqrt{5}$ is a irrational number.

Hence proved