If $m+x=m$ for all integers $m$, then $x=0$

Question

I am learning proofs and, have the following statement:

Let $x$ belong to the set of integers. If $x$ has the property that for each integer $m$, $m+x=m$, then $x=0$.

Here is my strategy:

\begin{align*} m + x &= m \\ (m + x) &= (m) \\ (-m) + (m + x) &= (-m) + (m) \\ ((-m) + m) + x &= 0 \\ (0) + x &= 0 \\ x &= 0 \\ \end{align*}

Would that be good? This would say that there is only one unique solution for any $m$ that belongs to $\mathbb Z$ and it's $0$.

Answer 1

Your algebra is right, though at this level you might be expected to explain it better. But your conclusion could be misinterpreted: it is not enough to say that for any $m$ there is a unique solution (i.e. $x$), since this means that $x$ might depend on $m$. What is true is that if $x$ that satisfies the equation for each $m$, then $x=0$. Your words are not exactly the former, but they could be reasonably read in that way.

As Joel mentions, your proof is a bit longer than strictly necessary to solve this problem. But as Zubin mentions, your work shows something stronger, which is that if there is any $m$ such that $m+x=m$, then $x=0$.