Prove $-(-a)=a$ using only ordered field axioms

Question

I need to prove for all real numbers $a$, $-(-a) = a$ using only the following properties:

I tried by starting with P12 written using symbolic logic:

$\begin{align} \forall a,b \in P \to & a \cdot b \in P \Leftrightarrow True \\ \forall a \in P \land \forall b \in P \to & a \cdot b \in P \Leftrightarrow True \\ \lnot \left[ \left( \forall a \in P \right) \land \left( \forall b \in P \right) \right] \lor & a \cdot b \in P \Leftrightarrow True \\ \exists a \notin P \lor \exists b \notin P \lor & a \cdot b \in P \Leftrightarrow True \\ \end{align} \\ $

and played with evaluating it with $b = -1$, $-a$ and so on. But, felt this wasn't really a solution because the last symbolic logic statement above is ambiguous.

Not to mention--other than P12--I haven't used any of the axioms listed above whatsoever. ( And I've not bothered to prove any of the theorems of symbolic logic I have used. )

Can anyone give me a start on this problem?

Many thanks in advance.

Edit

Thank you, everyone! I wrote up the proof I came up with using all of your help here.

Best regards!

Answer 1

First you have to prove that additive inverses are unique: That is, if there is a $b$ such that $a+b=0$ and a $c$ such that $a+c=0$, then $b=c$ (this is really why it makes sense to give the inverse a name, like $-a$). After you have done this, think about what the equation $$-a+a=0$$ says about the additive inverse of $-a$.

Answer 2

Try to prove the following statements:

$$0\cdot a=0\text{ for all $a\in\mathbb{R}$}$$

$$(-1)\cdot a=-a\text{ for all $a\in\mathbb{R}$}$$

$$\text{The additive inverse is unique.}$$

Now prove $$(-a)+(-(-a))=0\text{ for all $a\in\mathbb{R}$.}$$

Answer 3

Since the additive inverse of $a$ is $-a$, $P3$ implies that additive inverse of $-a$ is $a$, but we denote it by $-(-a)$; so $a=-(-a)$ (prove that additive inverse is unique.)

Answer 4

If you are able to prove (and the proof is relatively standard) that:

$$ 0 \cdot a=0 $$

Then you can finish it off by substitution on both sides

$$[1+(-1)]\cdot a=a+(-a)$$ $$1a+(-1)a=a+(-a)$$ $$a+(-1)a=a+(-a)$$ Which will get the required result: $$(-1)a=(-a)$$

I'll leave you to observe what axioms I used. You may also wish to explore the fact that if $$a+b=a+c$$ then $$b=c$$

Or similarly; if $$a=b$$ then $$a+c=b+c$$ (When I studied this in linear algebra, I was told that this is a well assumed result and doesn't require justification) So you can also add the additive inverse to both sides in the above solution provided that you are allowed to assume this.

Answer 5

It's barely a proof. It's almost a definition.

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

So $a $ is an additive inverse of $-a $.

For an additive inverse of $-a $ we call it $-(-a) $.

And that's it.

BUT there is one subtlety that everyone else has mentioned. The proposition seems to assume (at least in its language) that there is only one unique additive inverse. (Was this one of the axioms?) If this isn't so we can't really say that $-a $ is one specific number. It might represent many possible numbers. I.E. if $a + b =b+a =0$ and $a+c = c+a=0$ and $c\ne b $ then $-a $ can represent either $b $ or $c $.

If so $a $ is one of the values of $-(-a) $. There might be others. We have to show this can't happen by showing $a+b=b+a=0$ and $a+c=c+a=0$ and $b\ne c $ is impossible.

Then we will know $-a $ can only have one value and $-(-a) = a $; the one and only inverse of the one and only inverse of $a $ is $a $.

If $a+c=0;c+a=0$ and $a+b=0;b+a=0$ then $a+b=a+c $ . So $b+a+b=b+a+c $ so $0+b=0+c $ and $b=c $.

So additive inverses are unique. $-a $ represents one specific number only. And $-(-a)=a $ and only $a $ (as, $-a +a=a+(-a)=0$).