Struggling with a proof that $-m=(-1)m$

Question

Prove that

For all integers $m$, $-m=(-1)m$. Any help would be greatly appreciated.

Answer 1

The integer $-m$ is the opposite of $m$ that's the integer such that $$(-m)+m=0$$ and notice that the opposite is unique (the proof is pretty easy).

Now to prove the desired result:

$$(-1)m+m=\left[(-1)+1\right]m=0m=0$$ where we used the distribituve law and the fact that $$0\times a=0\quad \forall a$$

so by unicity $$-m=(-1)m$$

Answer 2

Lemma: for all $m$, $$0\cdot m=0$$ Proof: $$\begin{align}0\cdot m+0\cdot m&=(0+0)\cdot m&&\mbox{by the distributive axiom}\\ &=0\cdot m &&\mbox{by the defining property of $0$}\\ (0\cdot m+0\cdot m)+-(0\cdot m)&=0\cdot m+-(0\cdot m)&&\mbox{adding the same quantity to both sides}\\ 0\cdot m+(0\cdot m+-(0\cdot m))&=0\cdot m+-(0\cdot m)&&\mbox{associativity of addition}\\ 0\cdot m+0&=0&&\mbox{defining property of negatives}\\ 0\cdot m&=0&&\mbox{defining property of $0$}\\ \end{align}$$

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Now $$\begin{align} (-1)\cdot m + m &=(-1)\cdot m + 1\cdot m &&\mbox{by the defining property of $1$}\\ &=(-1+1)\cdot m&&\mbox{by the distributive axiom}\\ &=0\cdot m&&\mbox{defining property of negatives}\\ &=0&&\mbox{by the lemma}\\ ((-1)\cdot m + m)+-m&=0+-m&&\mbox{adding the same quantity to both sides}\\ (-1)\cdot m +( m+-m)&=0+-m&&\mbox{associativity of addition}\\ (-1)\cdot m +0&=0+-m&&\mbox{defining property of negatives}\\ (-1)\cdot m &=-m&&\mbox{defining property of $0$}\\ \end{align}$$