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
You're correct. The only way this proof can work is if r != 0. In the first proof, that is taken care of, because if r = 0 then we have division by 0, which we cannot do. But it must be specified that r != 0 in every one of these proofs for any of these proofs to work.