OK, because the problem statement is ambiguous, I'd rather try to prove a more general and--presumably--more useful result.
I shall prove that, if #a/b# is rational and #x# is irrational, then their sum #a/b + x# is irrational. (#a# and #b# are integers and #b!=0#)
I'll use a proof by contradiction: let's assume that the sum of the two numbers is rational. This means that the sum can be written as a ratio of two integers #c# and #d#, #d!=0#:
#a/b + x = c/d <=> x = c/d - a/b <=> x = (bc - ad)/(bd) #
Note that, because #a, b, c# and #d# are integers, then #bd# is an integer and #bc - ad# is also an integer. Therefore, #x# is the ratio of two integers, i.e. #x# is rational, which contradicts the hypothesis that #x# is irrational.
Thus, the assumption that #a/b + x# is a rational number led us to a contradiction. Hence, its opposite must be true, that is #a/b + x# is an irrational number.
This kind of reasoning can be applied to any particular cases.
