Originally Posted by
Jhevon I don't remember what problem i had with the first one, but now with the second one i realize there are two ways that (m - s)/a can be rational if a is irrational. we can have (m - s) irratonal, or (m - s) = 0 which is rational. however, in the case where (m - s) is 0, we would have r = 0, so we would have
ar + s = m = s
=> 0 + s = s
which is trivial, but nevertheless, i think the proof should account for that in some way