Yes.
It can be proved by contradiction.
Let:
a - a rational number
b - an irrational number
c - the sum of a and b
[tex]a+b=c[/tex]
Let assume that c is a rational number. Then a and c can be expressed as fractions with integer numerator and denominator:
[tex]a=\dfrac{d}{e}\\
c=\dfrac{f}{g}\\[/tex] where [tex]d,e,f,g \in \mathbb{Z}[/tex]
[tex]\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}[/tex]
Since [tex]d,e,f,g[/tex] are all integers, then the products [tex]ef,dg,eg[/tex] and the difference [tex]ef-dg[/tex] are integers as well. It means that the number [tex]b[/tex] is a rational number, but this on the other hand contradicts the earlier assumption that [tex]b[/tex] is an irrational number. Therefore [tex]c[/tex] must be an irrational number.
