Is $(x,y) \mapsto 0$ on $\mathbb{Q}\backslash\{0\}$ associative and commutative? I have the following definition of operations on the following sets:


*

*$(x,y) \mapsto 9xy$ on $\mathbb{Z}$

*$(x,y) \mapsto 0$ on $\mathbb{Q}\backslash\{0\}$


I have to determine whether the operations on the given sets are associative, commutative, have a neutral element, and have inverse elements.
For $(x,y) \mapsto 9xy$ I have that it is associative, commutative, and has the neutral element $1 \in \mathbb{Z}$, but does not have inverse elements as $(9xy)^{-1} \notin \mathbb{Z}$.
Could you please help me with $(x,y) \mapsto 0$? I don't understand the operation. It always maps $(x,y) \mapsto 0$, so how do I prove if this is associative, commutative etc.?
 A: For the first one, it is indeed associative and commutative because the usual multiplication of integers is som. It does not have a neutral element though, for the following reason: if $u \in \Bbb Z$ is this neutral element, then $9xu = x$ for all $x \in \Bbb Z$. For $x=1$ this would imply $9u = 1$, whence $u = \frac 1 9$ which is not in $\Bbb Z$.
For the second, $\Bbb Q \setminus \{0\}$ is not even closed under the operation $(x,y) \mapsto 0$, so it makes no sense to speak about associativity and the rest.
A: It is also associative because for all $x,y,z$:
$$0=(xy)z=x(yz)=0.$$
And it is also commutative because for all $x,y$:
$$0=xy=yx=0.$$
Edit
This proves that this law is associative and commutative on $\mathbb Q$.
Since the OP is considering this law on $\mathbb Q\setminus \{0\}\to \mathbb Q\setminus \{0\}$, this is not an intern law because of instant $1\cdot 1=0 \notin \mathbb Q\setminus \{0\}$. So the law is not well-defined on those sets.
A: The first one is associative and commutative as these two identities are homogeneous and then 
$$
(x,y)\rightarrow qxy
$$
is such on $\mathbb{Z}\times \mathbb{Z}$ for all $q\in \mathbb{Z}$. Only $q\in \{-1,1\}$ provides neutral.
The second is not even internal.   
