$\forall m \exists n$, $mn = n$ True or False 
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.
 A: Note that $mn=n$ if and only if $mn-n=0$ if and only if $(m-1)n=0$. Using the zero product property, we see that there is exactly one $n$ such that this holds for all integers $m$, and that $n$ happens to be an integer.
A: Hint: Consider $n=0$.${}{}{}{}$
A: 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:

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

A: For any $m$, take $n=0{}{}{}{}{}$.
A: so this stattment  says,that for all m  ,there exist such n  so that m*n=n,if n is equal to zero,then it would  be  identity,so true,if we  say that  there exist  n  except n=0,then maybe it would not be correct if n is prime,so  answer for this case because it is not  constraint  n not to be zero,then it is true 
