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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5You 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 08 '12 at 22:00
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$$a+(-1)a=(1)a+(-1)a=(1+(-1))a=0\cdot a=0$$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: $$x=x+0=x+(a+y)=(x+a)+y=0+y=y$$
Daniel Montealegre
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You have to remember $-a$ is by definition the element you add to $a$ to obtain zero. – Daniel Montealegre Mar 08 '12 at 21:59
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Hint $\ $ Both are inverses of $\rm\,a\,$ so they are equal by uniqueness of inverses, i.e.
$$\rm \begin{array}{}\rm a'+a\ =\, 0\\ \rm \, a+a'' =\,\! 0\end{array}\quad\Rightarrow\quad a'\, =\ a' + (a + a'')\ =\ (a' + a) + a'' =\, a''\qquad$$ 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.
Bill Dubuque
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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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