Question about some properties of the binary operation $a*b=a-b$.

Question

Originally, the question asks whether $\mathbb{Q}^2\to \mathbb{Q}, (a,b) \mapsto a-b$ is associative, commutative, has a neutral element and has an inverse.

1. Associativity and commutativity: In testing for commutativity, we verify whether it is the case that $a*b=b*a$, for associativity whether that $a*(b*c)=(a*b)*c$. Clearly, this isn't true.

2. Inverse element: I am unsure whether this has an inverse, i.e. can we take that $a$ it self so that $a*a=a*a=0$? If so, then it does.

3. Neutral element: It doesn't have one since $a*0\neq 0*a$.

Does it make sense to say that this has an inverse element when it has no identity element?

Answer 1

It does not make sense to say that inverses exist unless there exists an identity. So you should do (3) before even trying (2) (and if there exists no identity, you can skip (2) and just say that it doesn't make sense to ask whether there are inverses). But I'm not sure that what you're saying for (3) is quite right: to say that there is no identity, you need to say that there exists no element $c$ such that for all $a$, $c*a=a*c=a$. This $c$ doesn't necessarily have to be the rational number $0$. Similarly, if you find that such a $c$ does exist, then in (2) you should be looking for $b$ such that $a*b=b*a=c$, not $a*b=b*a=0$.

Answer 2

It could be claimed that the operation has a right side identity and right side inverses, if right side identities and right side inverses are acceptable.

0 is such that all a*0 = a. BUT it isn't true that 0*a = a.

For each a there exists an a'(a itself) such that a*a' = 0. But it is not true that a'*a = 0.

There is no left side identity, 0' such that 0'*a = a for all a. (If there were 0' - a = a => 0' = 2a which is not constant.) With no left side identity there can be no left side inverse.

So the question is are you allowed/supposed to state whether there are one-sided identities/inverses.