Question

Prove That $\sqrt{3}+\sqrt{5}$ Is Irrational

Answer 1

Given: $\sqrt{3}+\sqrt{5}$

We need to prove$\sqrt{3}+\sqrt{5}$ is an irrational number.

Proof:

Let us assume that $\sqrt{3}+\sqrt{5}$ is a rational number.

A rational number can be written in the form of $\frac{p}{q}$ where$p$, and $q$ are integers and $q\ne 0$

$\sqrt{3}+\sqrt{5}=\frac{p}{q}$

On squaring both sides we get,

$$\begin{gathered} \Rightarrow {(\sqrt{3}+\sqrt{5})}^2={\left(\frac{p}{q}\right)}^2 \\ \Rightarrow {\left(\sqrt{3}\right)}^2+{\left(\sqrt{5}\right)}^2+2\left(\sqrt{5}\right)\left(\sqrt{3}\right)=\frac{p^2}{q^2} \\ \Rightarrow 3+5+2\sqrt{15}=\frac{p^2}{q^2} \\ \Rightarrow 8+2\sqrt{15}=\frac{p^2}{q^2} \\ \Rightarrow 2\sqrt{15}=\frac{p^2}{q^2}–8 \\ \Rightarrow \sqrt{15}=\frac{(p²-8q²)}{2q^2} \end{gathered}$$

$p$, $q$ are integers then $\frac{(p²-8q²)}{2q^2}$ is a rational number.

Then $\sqrt{15}$ is also a rational number.

But this contradicts the fact that $\sqrt{15}$ is an irrational number.

Our assumption is incorrect

Thus, $\sqrt{3}+\sqrt{5}$ is an irrational number.