TITLE: $\forall m \exists n$, $mn = n$ True or False
QUESTION [7 upvotes]: Identify if the statement is true or false. If false, give a counterexample.
$\forall m \exists n$, $mn = n$, where $m$ and $n$ are integers.

I said that this statement was false; specifically, that it is false when $m$ is any integer other than $1$
Apparently this is incorrect; honestly though, I can't see how it is.

REPLY [6 votes]: As others have noted, the statement is true: 

$$\forall m \exists n, \; mn = n, \;\;\; m,\,n \in \mathbb Z \tag {1}$$

For all $m$, there exists an $n$ such that $mn = n$. To show this is true we need only to find the existence of such an $n$: and $n = 0:\;\; m\cdot 0 = 0 \forall m$.
Since there exists an $n$ ($n = 0$) such that for every $m$, $mn = n$, this is one case where one can switch the order of the quantifiers and preserve truth:

$$\exists n \forall m,\; mn = n,\;\;\;m, n\in \mathbb Z \tag{2}$$

And further more, this $n = 0$ is the unique $n$ satisfying $(1), (2)$. It is precisely the defining property of zero under multiplication, satisfied by and only by $0$.
The existence of a unique "something" is denoted: $\exists !$, giving us, the strongest (true) statement yet:

$$\exists! n\forall m,\;\; mn = n,\;\;\;m, n\in \mathbb Z\tag{3}$$