Question 30

Prove that $\sqrt{3}+\sqrt{5}$ is irrational

Solution:

Let us suppose that $\sqrt{3}+\sqrt{5}$ is rational
Let $\sqrt{3}+\sqrt{5}=a$, where a is rational
Therefore, $\sqrt{3}=a-\sqrt{5}$
On squaring both sides, we get
${\left(\sqrt{3}\right)}^2={(a-\sqrt{5})2}^{}$
$\Rightarrow 3={\mathrm{a2}}^{}+5-2a\sqrt{5}$ $\text{[}\therefore {(a-b)2}^{}={\mathrm{a2}}^{}+{\mathrm{b2}}^{}-2ab]$
$\Rightarrow 2a\sqrt{5}={\mathrm{a2}}^{}+2$
$Theref\text{or}e,\sqrt{5}=\frac{{\mathrm{a2}}^{}}{+}$ which is contradiction.
As the right hand side is rational number while $\sqrt{5}$ is irrational. Since 3 and 5 are prime number. Hence, $\sqrt{3}+\sqrt{5}$ is irrational.