Question

Is the sum of an irrational number and rational number always irrational?

Answer 1

Yes.

It can be proved by contradiction.

Let:
a - a rational number
b - an irrational number
c - the sum of a and b

$a+b=c$

Let assume that c is a rational number. Then a and c can be expressed as fractions with integer numerator and denominator:
$a=\dfrac{d}{e}\\ c=\dfrac{f}{g}$      where $d,e,f,g \in \mathbb{Z}$

$\dfrac{d}{e}+b=\dfrac{f}{g}\\ b=\dfrac{f}{g}-\dfrac{d}{e}\\ b=\dfrac{ef}{eg}-\dfrac{dg}{eg}\\ b=\dfrac{ef-dg}{eg}$

Since $d,e,f,g$ are all integers, then the products $ef,dg,eg$ and the difference $ef-dg$ are integers as well. It means that the number $b$ is a rational number, but this on the other hand contradicts the earlier assumption that $b$ is an irrational number. Therefore $c$ must be an irrational number.