Prove that if m|a and m|b then m|a+b I need help proving that if m|a and m|b, then m|a + b, and m|ac for any int c.
I was thinking that m|a and m|b => a = km and b = lm. Thus a + b = km + lm = (k + l)m. Since this is a multiple of m then m|(k + l)m and m|a + b. But I'm not sure if that proves it since I haven't proved m|a => m|ac and am not sure how to. 
Sorry I don't know how to make everything look all math-y.
 A: As far as I know $$a|m$$ is defined to mean $$m=ka$$ for some integer $k$. So your proof is excellent.
A: You answer to the first part was correct. Here is to make to more 'mathy'. You gotta remember that 'mathy' means that each step is a logical consequence of the previous ones, in other words, you don't pull any assumptions out of thing air. Also, you write in 'mathy' talk of course; need to employ one of the methods of proofs. Here we will use direct proof: we proof if $A$ then $B$ by supposing $A$ is true and arriving at the conclusion that $B$ is true.

Part 1. Suppose $m|a$ and $m|b$ (suppose $A$ is true). By definition of divisibility we have $mk=a$, and $ml=b$, where $k,$ and $l$ are some integers (note how we need to specify everything clearly). Adding the two together we have $a + b = mk + ml = m(k+l)$, where the last step is by the distributive property. Since both $k$ and $l$ are integers then so is $k+l$. Then, again by definition, we conclude $m|a+b$ (so we have arrived at $B$ and this completes our proof).

Part 2. Suppose $c$ is some integer, and $m|a$. So by definition $mk = a$ where $k$ is some integer. Now multiple by $c$ and we have $mkc = ac$. By associativity we have $m(kc) = ac$, and since both $k$ and $c$ are integers so is $kc$. Since $c$ is arbitrary, then by definition of divisibility we conclude $m|ac$ where $c$ is any integer.
