# The consequence of divisibility definition in integer

1. Feb 15, 2014

### Seydlitz

So I think I've just proven a preposition, where $0$ is divisible by every integer. I prove it from the accepted result that $a \cdot 0 = 0$ for every $a \in \mathbb{Z}$. From then, we can just multiply the result by the inverse of $a$, to show that the statement holds for $0$. That is to say, there exist an integer $0$, such that $a^{-1} \cdot 0 = 0$.

But then there's another preposition, if $a \in \mathbb{Z}$ and $a \neq 0$, then $a$ is not divisible by $0$. Okay we can also use the fact that $a \cdot 0 = 0$. So far so good. But then I realize that the preposition seems to imply that if $a=0$ then $a$ is divisible by $0$. The first preposition where $0$ is divisible by every integer also points to the same result because $0 \in \mathbb{Z}$.

But we know isn't it, that we cannot divide any number by $0$, any operation that involves division by $0$ is automatically a no-no in math. It just doesn't sound right. (The preposition comes from a book and I don't propose that myself) Does it mean that technically (according to the definition of divisibility) $0$ is also divisible by $0$, but it's not a legal operation in cancellation, say when, $a \cdot 0$ = $b \cdot 0$. We cannot cancel the $0$ in this case. But still again, $0$ is divisible $0$.

2. Feb 16, 2014

### jbriggs444

What definition of "is divisible by" are you and your book using? Is it that "a is divisible by b iff a/b is an integer"? Or is it that "a is divisible by b iff there exists an integer c such that a = bc"?

If it is the former, then "zero is divisible by zero" is neither true nor false -- it is meaningless. If it is the latter then zero is divisible by zero and no contradiction ensues since the definition does not involve division by zero.

3. Feb 16, 2014

### Seydlitz

The book uses the latter version, a is divisible by b iff there exists an integer c such that a = bc.