Prove a mod b = b mod a iff a=b I'm just starting to learn the three types of proofs and I came across this question. 
                 a mod b = b mod a iff a=b

I tried looking for solution to prove this but couldn't find it. Most examples I have are of the same divisor like: a mod n = b mod n. But i couldn't find anything on this. 
I assumed it would be by using a contradiction proof but what got me confused is the if and only if condition.
Thanks for any help you send this way!!
 A: Suppose $a\neq b$.  Then, without loss of generality we can assume that $a>b$.  In that case $b\pmod a$ is $b$.  But $a\pmod b$ is the remainder of $a$ on division by $b$, hence it is less than $b$.
A: When you have a statement with an equivalence (an "if and only if") you have to prove two things. If you have

A iff B

You must prove $A \implies B$ and $B \implies A$. When proving those two things you can then use a proof by contradiction. For each of the things, separately. Or you can mix in two types of proofs, one for each implication.
A: Since this is an if and only if proof, we have to do it in two directions. As is often the case in if and only if proofs, one direction is much easier to do than the other. We will do the easy direction first.
Proof.
⇐ We assume a=b and prove a mod b = b mod a. But this is clearly true, since x mod x= 0 for all x, and a and b are equal.
⇒ Here we have to prove that if a mod b = b mod a, then a=b.  Try to find a direct proof, and then give up and do the contrapositive:  assume a ≠ b, and prove that a mod b ≠ b mod a. There are two possible cases: a < b and b < a (we’ve already assumed a≠b).
Case 1: a < b Then a mod b=a.  But b mod a, as we noted, is strictly less than a, so it cannot be equal to a and hence can’t be equal to a mod b.
Case 2: a > b Then b < a, and the proof is the same as above with a replaced by b and b replaced by a.
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