...but the answers to (a) and (b) are different...

These types of questions (you are given a strange multiplication in this way) are relatively easy to do once you know how, and are a gift in an exam!

Firstly, look to see if the multiplication is commutative. This is usually quite easy to spot if it is.

Secondly, you need to find an identity element. So, let be an arbitrary element. We want to find an and a such that . So, just use the multiplication given to work out what is.

This is your identity element. You know it is an identity on the right, but unless the multiplication is commutative you do not know if it is an identity on the left. So, knowing what is check to see if . Usually (always if you have been asked to find an identity or an inverse not a right identity or a right inverse (alternatively, left identity etc.)) this will be so. But do it all the same as it is part of the proof. Once you have shown this you have found your identity element.

Now you need to find your inverse. You want to find an element such that . Again, just use the fact that you know what the multiplication is to solve for .

You will usually find a generic inverse, but this is not always an inverse for all elements, as with question (a) here. Usually if this is the case you will be dividing by zero.