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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?

  1. 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!

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    $\begingroup$ This question really seems to be dependent on what kind of axioms you are allowed to use, etc. $\endgroup$ Commented Feb 5, 2015 at 3:18
  • $\begingroup$ It is based on the definition of divisibility and that 0 = m x 0 where m belongs to the set of integers. $\endgroup$
    – Johnathan
    Commented Feb 5, 2015 at 3:19
  • $\begingroup$ $0.m=0, so 0/0=m$, which means it can take any value. $\endgroup$
    – JMP
    Commented Feb 5, 2015 at 3:21
  • $\begingroup$ I can't wrap my mind around 0/0 because of my calculus class... lol $\endgroup$
    – Johnathan
    Commented Feb 5, 2015 at 3:24
  • $\begingroup$ The first solution is not clear. We say that $a$ divides $b$ if there is an integer $q$ such that $b=aq$. Let $b=0$ and $a$ any integer. Then $a\cdot 0=0$ and therefore $a$ divides $0$. As to your worry about $0/0$, nowhere did we use that expression. But from the fact that $a\cdot 0=0$, we can conclude that there seems to be no pleasant way to assign a value to $0/0$. $\endgroup$ Commented Feb 5, 2015 at 3:32

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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 $0$.

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  • $\begingroup$ Thank you David! So, divisibility and and division are not the same. $\endgroup$
    – Johnathan
    Commented Feb 5, 2015 at 3:37
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    $\begingroup$ Definitely not the same. Division is something like $7/3$ which is a number. Divisibility is something like $7\mid 3$ which is not a number but a statement. It has no numerical value but only a logical value, true or false (in this case false). $\endgroup$
    – David
    Commented Feb 5, 2015 at 3:41

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