Question

Prove that $3+\sqrt{5}$ is an rational number.

Answer 1

Let $r=3+\sqrt{5}$ be a rational number

Squaring both sides, we have

${(3+\sqrt{5})}^2=r^2$

$9+5+6\sqrt{5}=r^2$

$\Rightarrow 14+6\sqrt{5}=r^2$

$\Rightarrow \sqrt{5}=\frac{r^2-14}{6}$

Now,

$\frac{r^2-14}{16}$ is a rational number and $\sqrt{5}$ is an irrational number.

Since a rational number cannot be equal to an irrational number.

Thus our assumption that $3+\sqrt{5}$ is rational wrong.

Hence proved that $3+\sqrt{5}$ is an irrational number.