Proof of associativity and commutativity Say I have a set $X$ and an operation $m$ on $X$. Consider the following example.
$(x,y) \mapsto 9xy$ on $\mathbb{Z}$
How do you show that this is associative and commutative? I have already done that with, e.g.,
$(g,f) \mapsto g \circ f$ on $\{f\in Map(\mathbb{Q}, \mathbb{Q}) \text{ | f is bijective}\}$. But the constant $9$ in this example in the term confuses me. Could you please give me a hint?
 A: [This has already, almost, been answered in the comment section above but here goes.]
An operation is just a function as you already know. So to (hopefully) simplify things, write out everything in function-notation that you are used to. In the example that you mention your function is:
$f(x,y)=9xy$.
What does it mean for this to be commutative? Well, you know what it means for the operation $\odot$ to be commutative, i.e that $x\odot y=y\odot x$. This is just written with infix notation, meaning that we place the operands on either side of the function name. Written with a, perhaps, more familiar notation it reads: $\odot(x,y)=\odot(y,x)$. So I think you, also with help from the comments above, have understood that in your example,
$9xy=9yx$,
proves commutativity.
Then what about associativity? Well, again, use the same technique. You know that if $x\odot(y\odot z)=(x\odot y)\odot z$, then it is left associative (doing it from the right of course yields right associativity). So, again, with different notation you have:
$\odot(x,\odot(y,z))=\odot(\odot(x,y),z).$
In your example you would have to prove:
$f(x,f(y,z))=f(f(x,y),z)$
which yields?
(Get back to me if you need more help, but I think the step to convert it is simple enough for you to handle (and better left to you). As was said in the comments above, it's just about getting lost in notation. And again, don't forget to also prove right associativity, which is just as simple as left in this case.)
