Is $(x,y) \mapsto 0$ on $\mathbb{Q}\backslash\{0\}$ associative and commutative?

Question

I have the following definition of operations on the following sets:

1. $(x,y) \mapsto 9xy$ on $\mathbb{Z}$
2. $(x,y) \mapsto 0$ 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.?

Answer 1

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.

Answer 2

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.

Answer 3

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.