Question 34

Prove that $\sqrt{p}+\sqrt{q}$ is an irrational, where $p\text{and}q$ are primes.

Solution:

Let us suppose that $\sqrt{p}+\sqrt{q}$ is rational.
Again,let $\sqrt{p}+\sqrt{q}=a$ where a is rationa.
Therefore, $\sqrt{q}=a-\sqrt{p}$
On squareing both sides, we get
$q=a^2+p-2a\sqrt{p}$ $[\therefore {(a-b)2}^{}={\mathrm{a2}}^{}+{\mathrm{b2}}^{}-2ab]$
Therefore, $\sqrt{p}=(\frac{{\mathrm{a2}}^{}}{+}$$\frac{2}{a}$, which is a contraction as the right hand side is rational number while $\sqrt{p}$ is irrational since p and q are prime numbers/ Hence, $\sqrt{p}+\sqrt{q}$ are prime numbers. Hence, $\sqrt{p}\text{and}\sqrt{q}$ is irrational.