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I am learning proofs, and I am stuck with this proposition:

Let $x \in\mathbb{Z}$. If $x$ has the property that for all $m \in\mathbb Z$, $mx = m$, then $x = 1$.

I want to use the additive identity to get $mx = m \cdot 1$ to introduce the 1. I am tempted to simply cancel the $m$, but I am supposed to use axioms. Any idea? If $m$ would be any integer except 0, I could use the cancellation axiom. However, $m$ accounts for all integers.

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  • $\begingroup$ Please reconsider how you think about "cancelling". There's no such thing as "cancelling" in mathematics. In this case, you can divide both sides by $m$. But thinking in terms of "cancel this" and "drop that" lead to sloppy math and ultimately bad habits. $\endgroup$
    – corsiKa
    Feb 6, 2015 at 21:34

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Use $m = 1$. Then $1\cdot x = 1$ so that $x = 1$.

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    $\begingroup$ To add some context for the OP, the hypothesis says that it holds for all such $m$. That means, you can pick the $m$ you want! Choosing $m=1$ is convenient because it has the additional property of being the multiplicative identity. $\endgroup$
    – Emily
    Feb 6, 2015 at 18:16
  • $\begingroup$ @Arkamis Thank you for your answer! However, I am a bit confused because if I had chosen 0, which is also an integer, x could have been anything, no? I was always under the impression that "for all m" meant all m, and "there exists an integer m" meant you could pick the integer, no? Sorry for being confused. $\endgroup$
    – Johnathan
    Feb 6, 2015 at 18:33
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    $\begingroup$ @user2472704 You could have chosen $m=0$, but it would not be a useful choice. Note that $0\cdot x = 0$ is true for all (finite) $x$, so yes, choosing $m=0$ means that $x$ could be anything. However, choosing $m=1$ eliminates that problem entirely. $\endgroup$
    – Emily
    Feb 6, 2015 at 18:35
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    $\begingroup$ @Arkamis I think I understand now my problem: 1 is the unique solution that is valid for all m. However, this does not imply that some integers (i.e. 0) only have one solution. This is why choosing 1 is valid. $\endgroup$
    – Johnathan
    Feb 6, 2015 at 18:50
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    $\begingroup$ @Johnathan Yes, the key is that the property holds for every possible value of $m$. That is not the same as saying "every possible value of $m$ is useful." $\endgroup$
    – Emily
    Feb 6, 2015 at 18:55
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Assuming you have the field axioms, you could also say

$$mx = m \\ mx - m = m-m \\ mx-m = 0\\ m(x-1) = 0.$$

Since this must hold for all $m$, including when $m \neq 0$, we must have $x-1 = 0$, so $x=1$.

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  • $\begingroup$ Thank you! Actually, I wrote this on my board last night but got stuck at how to interpret the last line. Actually, there was another proposition that I had to prove: For all m that belong to the set of integers, m x 0 = 0 = 0 x m. Hence, this could explain why (x - 1) has to be equal to 0 for all m, right? $\endgroup$
    – Johnathan
    Feb 6, 2015 at 18:44
  • $\begingroup$ @Johnathan $m\cdot 0=m\cdot (1-1)=m\cdot 1-m\cdot 1=0, \forall m\in\mathbb Z$. Same goes for $0\cdot m=0, \forall m\in\mathbb Z$. But this is not really something you need to prove for this: what you have to prove is that $ab=0\implies a=0\lor b=0$, the contrapositive of which is 'if $a,b\neq 0$, then $ab\neq 0$', which is simple to prove if you accept the existence of multiplicative inverses of non-zero integers, so that the assumption of $ab=0$ leads to a contradiction: $ab\cdot a^{-1}b^{-1}=0\cdot a^{-1}b^{-1}\iff 1=0$, a contradiction. $\endgroup$
    – user26486
    Feb 6, 2015 at 20:17

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