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I tried to prove that:

$ \forall a \in \mathbb{R}: (-(-a)) = a $

using the following field axioms:
(1) Commutative law (addition)
(2) Associative law (addition)
(3) Existence and uniqueness of the number zero (addition)
(4) Commutative law (multiplication)
(5) Associative law (multiplication)
(6) Distributive law
(7) Existence and uniqueness of the number 1 (multiplication)

I know that,
$ \forall a \in \mathbb{R}: a + (-a) = 0 $

The proof:
Let $a \in \mathbb{R}$.
$(-(-a))$
$= (-1)(-1)a $. using (6) and (7)
$= (-1+1-1)(-1+1-1)a $. using (3)
$= ((-1)(-1)+(-1)+(-1)(-1)+(-1)+1+(-1)+(-1)(-1)+(-1)+(-1)(-1))a $. using (6)
$= ((-(-1))+(-1)+(-(-1))+(-1)+(-(-1))+(-1)+(-(-1))+(-1)+1)a$. using (6) and (1)
$= ((-1)+(-(-1))+(-1)+(-(-1))+(-1)+(-(-1))+(-1)+(-(-1))+1)a$. using (1)
$= (4(-1)+4(-(-1))+1)a$. using (6)
$= (4(-1)-4(-1)+1)a$. using (2)
$= (0 + 1)a$
$= a$. using (6) and (7)

Is the proof correct? I'm almost sure there is a simpler way using the third axiom.

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    $\begingroup$ You've used existence of a $1$, but that wasn't listed as one of your axioms. (And indeed, it is not needed.) $\endgroup$
    – Randall
    Nov 28, 2022 at 20:39
  • $\begingroup$ Thank you. I've added the axiom to the list. I know there is an easier (or maybe I should say more elegant) way to prove this by starting from a = a. Is my proof correct still? $\endgroup$
    – tfb
    Nov 28, 2022 at 20:45
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    $\begingroup$ After the 7th axiom, you've included an equation for the existence of the additive inverse. Are you including that as an axiom? If you can use the "existence and uniqueness of additive inverse" axiom, then I think it can be proved simpler $\endgroup$
    – Tony Mathew
    Nov 28, 2022 at 20:55
  • $\begingroup$ Yes. In my book it immediately follows from the axiom "unambiguous solvability of the equation a + x = 0". $\endgroup$
    – tfb
    Nov 28, 2022 at 21:09
  • $\begingroup$ Can we use the fact that $0x = 0$ for all $x$ ? This is usually not given as an axiom, but follows from earlier axioms. $\endgroup$
    – Adam Rubinson
    Nov 28, 2022 at 21:11

1 Answer 1

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We have:

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

both by definition. Then:

$$a + (-a) = (-a) + -(-a)$$

$$\implies a + a + (-a) = a + (-a) + -(-a)$$

$$ \implies a + 0 = 0 + -(-a)$$

$$\implies a = -(-a)$$

where I have only used $(1),(2),(3)$, and as Tony Mathew remarked, assumed that inverses do exist. This is the standard argument which shows that inverses are unique in any group (if they are not already assumed to be). If it is not applicable for some reason I have missed, please let me know and I will delete this.

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