Zero is divisible by every integer, but other integers are not divisible by zero

Question

I have two propositions to prove:

1. $0$ is divisible by every integer. Here is my strategy:

Proof: Let $j,m\in\mathbb Z$. Now, we multiply to get $0$: $j \cdot m = 0$. Since $0$ can also be written as $0 \cdot m$, we now simplify $m$ from both sides and get $j = 0$. Thus, $0$ is divisible by every integer $m$ and the division always gives $0$ ($j$).

However, what about $0/0$ since the proposition states "every integer"? Isn't $0/0$ the indiscriminate form? Is it valid?

2. If $m$ is an integer not equal to $0$, then $m$ is not divisible by $0$. Here is my strategy:

Proof: Let $m \in\mathbb Z$\{$0$} and $j \in\mathbb Z$. Now, we multiply $j$ by $0$ to get $m$. $m = j \cdot 0$. Since $0$ multiplied by any integer gives $0$, we simplify. $m = 0$. However, $m \ne 0$. Hence, if $m$ is an integer not equal $0$, $m$ is not divisible by $0$.

I would greatly appreciate the community's feedback. I am learning how to perform proofs and how to write them more elegantly. Thank you!

Answer 1

Re: $0/0$. The usual definition of divisibility does not rely on division but is as follows.

Let $a,b$ be integers. Then $b$ is divisible by $a$ if and only if there exists an integer $k$ such that $b=ka$.

Taking $a=b=0$, is there an integer $k$ such that $0=k0$? Yes there is, in fact, you can take any integer you like for $k$. Therefore $0$ is a multiple of $0$.