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.?

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.
