Inspired by Joe Lee

Suppose on the contrary that there exist some rational $a,b$ and irrational $c$ such that $a+c=b$ .

Then $c=a-b$ . Now recall that the set of rational numbers is closed under subtraction. Which means that the difference of two rational numbers is also a rational number. Therefore, since both $a$ and $b$ are rational, therefore their difference is also rational.

Hence $a-b=c$ is rational. But this is a contradiction ,as we had assumed that $c$ was irrational.

This contradiction show that our original assumption was false and therefore the statement "The sum of a rational and an irrational number is always an irrational number" is $\boxed{true}$