# $\forall m \exists n$, $mn = n$ 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.

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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:

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

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I Like the bit about changing the order of the quantifiers--I never thought of doing that. –  Mack Mar 14 at 12:58

For any $m$, take $n=0{}{}{}{}{}$.

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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.

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Hint: Consider $n=0$.${}{}{}{}$
Oh, yes. $0$ is in fact an integer! –  Mack Mar 13 at 19:18