Question

Show that 5 + $\sqrt{7}$ is an irrational number.

Answer 1

Let us assume that 5 + $\sqrt{7}$ is a rational number.

$$\begin{gathered} \Rightarrow 5+\sqrt{7}=\frac{p}{q},\mathrm{where}p\mathrm{and}q\mathrm{are}\mathrm{two}\mathrm{integers}\mathrm{and}q\ne 0 \\ \Rightarrow \sqrt{7}=\frac{p}{q}-5=\frac{p-5q}{q} \end{gathered}$$

Since, $p,q\mathrm{and}5\mathrm{are}\mathrm{integers},\mathrm{so}\frac{p-5q}{q}\mathrm{is}\mathrm{a}\mathrm{rational}\mathrm{number}$.

$\Rightarrow$ $\sqrt{7}$ is also a rational number.

But this contradicts the fact that $\sqrt{7}$ is an irrational number.

This contradiction has arisen due to our assumption that 5 + $\sqrt{7}$ is a rational number.

Hence, 5 + $\sqrt{7}$ is an irrational number.