Proving that a set is closed with respect to subtraction I have read that an equivalent definition for a subring is the following:


  B is a subring of A if and only if B is closed with respect to subtraction and multiplication


If I was trying to prove that a set is closed with respect to subtraction, would I just need to prove that for $a, b \in B$ that $a-b\in B$ or would I also need to prove that $b-a\in B$?
They are equivalent.  

As $(a,b)$ ranges over all of $B^2$, $(b,a)$ also ranges over all of $B^2$.  So, by the time you have checked every $a-b$ you have also checked every $b-a$.
Proving that for any $a,b\in B$ we have that $a-b\in B$ would be sufficient. The key word here is "ANY".

Once you prove this property in general, and assuming this subset $B$ is nonempty, you can take any $a\in B$ and apply this property to $a$ and $a$, thus showing that $0=a-a\in B$. Then you can can apply it to $0$ and $a$ to show that $-a=0-a\in B$. This is why proving only one direction, i.e. only $a-b$, is sufficient, because containing additive inverses follows.