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.
