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

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.

• 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. Feb 6, 2015 at 21:34

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

• 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. Feb 6, 2015 at 18:16
• @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. Feb 6, 2015 at 18:33
• @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. Feb 6, 2015 at 18:35
• @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. Feb 6, 2015 at 18:50
• @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." Feb 6, 2015 at 18:55

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 $x=1$.

• 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? Feb 6, 2015 at 18:44
• @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. Feb 6, 2015 at 20:17