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As the title reveals, I want to prove (based on the axioms of field) that $$-a=(-1)\cdot a$$ I've been trying for a while now, but couldn't think of a way to do it and it got me thinking that maybe its does not require a proof, but it doesn't feel right. Any ideas?

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You don't need the full strength of the field axioms (in particular you don't need division). Just the ring axioms will do. –  Qiaochu Yuan Mar 8 '12 at 22:00
    
By the way, I removed the tag "calculus" and added "abstract algebra" instead. And as Qiaochu said you only need the ring axioms. –  Daniel Montealegre Mar 8 '12 at 22:01

3 Answers 3

up vote 8 down vote accepted

Hint $\ $ Both are inverses of $\rm\:a\:$ so they are equal by uniqueness of inverses, viz.

$$\rm \begin{array}{} a'+a\ =\: 0\\ \rm \: a+a'' =\:\! 0\end{array}\quad\Rightarrow\quad a'\: =\ a' + (a + a'')\ =\ (a' + a) + a'' =\: a''$$ Using that, it suffices to prove that $\rm\:\ (-1)\:a + a = 0,\ $ i.e. $\rm\:(-1)\:a\:$ is an inverse of $\rm\:a\:.$ This follows easily by the ring axioms (hint: scale $\rm\: -1\: +\: 1\: =\: 0\:$ by $\rm\:a\:$).

It is useful to abstract out the lemma on uniqueness of inverses since it is ubiquitous in algebra.

Also, as I often emphasize, uniqueness theorems provide powerful tools for proving equalities.

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@downvoter why? –  Bill Dubuque Mar 9 '12 at 16:26

$$a+(-1)a=(1)a+(-1)a=(1+(-1))a=0\cdot a=0$$Hence by definition we have that $-a=(-1)a$

Added: We also want to show that the additive inverse of $a$ has no ambiguities, i.e., if $x$ and $y$ are the additive inverses of $a$ then $x=y$ and hence we can just denote this element by $-a$. Proof: $$x=x+0=x+(a+y)=(x+a)+y=0+y=y$$

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You have to remember $-a$ is by definition the element you add to $a$ to obtain zero. –  Daniel Montealegre Mar 8 '12 at 21:59
    
+1. Next proof $(-1)(-1)=1$. –  NikolajK Mar 8 '12 at 22:01
    
@Nick: What do you mean "next proof"? –  Daniel Montealegre Mar 8 '12 at 22:05
    
Of course, for this to work, you need to prove that $-a$ is the only element you can add to $a$ to obtain zero. –  Chris Eagle Mar 8 '12 at 22:11
    
@Chris: Thank you! I edited my post. –  Daniel Montealegre Mar 8 '12 at 22:17

Well you have $$a+(-1)a=(1)a+(-1)a=(1+(-1))a=0a=0$$

To prove that $0a=0$ note that $0a=(0+0)a=0a+0a$ add $-0a$ in both sides

$0=0a+(-0a)=0a+0a+(-0a)=0a$.

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