Question

Show that $\sqrt{p}+\sqrt{q}$ is an irrational number, where $p,q$ are primers.

Answer 1

First, we'll assume that $\sqrt{p}$ and $\sqrt{q}$ is rational , where p and q are distinct primes

$\sqrt{p}+\sqrt{q}=x$, where x is rational

Rational numbers are closed under multiplication, so if we square both sides, we still get rational numbers on both sides.

$(\sqrt{p}+\sqrt{q}{)}^2=x^2$

$p+2\sqrt{pq}+q=x^2$

$2\sqrt{pq}=x^2-p-q$

$\sqrt{pq}=\frac{(x^2-p-q)}{2}$

Now, $x,x^2,p,q,\&2$ are all rational, and rational numbers are closed under subtraction and division.

So, $\frac{(x^2-p-q)}{2}$ is rational.

But since p and q are both primes, then pq is not a perfect square and therefore $\sqrt{pq}$is not rational. But this is contradiction. Original assumption must be wrong.

So, $\sqrt{p}$ and $\sqrt{q}$ is irrational, where p and q are distinct primes.