Im wondering if its possible given x,y irrational, that x-y is rational (other than the case x=y). The reason I am asking this is that I am reading a book on measure theory and they try to construct a non measurable set and they start with an equivalence relation on [0,1} x~y if x-y is rational. Then they construct a set using the axiom of choice which contains exactly 1 element from each equivalence class. I know that the set of all rational numbers in [0,1) is an equivalence class, also each irrational number forms an equivalence class because for each irrational number x, x-x=0 (rational). Is there any other possibilities?