I have to prove "If a is congruent a (mod b) then a and b have the same remainders when divided by n.
I'm in a intro to proof class and we havent gone into much detail about this topic. In fact this is the first proof dealing with remainders I have to prove. Not sure if i did it right though?
The Attempt at a Solution
Here is my proof:
Assume n|(a-b) implies ns=a-b where sεZ. Let a=nk+r1 where 0≤n<n. This means that ns=nk+r1-b then nk-ns+r1=b where n(k-s)+r1. Since k-sεZ then a and b both share the same remainder.