How would I prove $-a=(-1)a$ for any integer $a$ using Peano axioms?

Question

How can I prove that for any integer $a$ it holds that $-a=(-1)a$ using Peano axioms? I think that it can be partially proven with the distributive property, multiplicative identity, and additive inverses property, but I think it's lacking and I'm not sure if it is correct. Any help is appreciated.

Answer 1

Peano axioms are about natural number: $-a$ is not defined.

You have to use the Axioms for rings and the existence of inverse elements:

$a + (-a)=0$.

Proof sketch:

$1+(-1)=0$, thus: $(1+(-1)) \times a = 0 \times a =0$. By distributivity : $(1 \times a)+ ((-1) \times a)=0$.

Finally, "comparing" $a+((−1)×a)=0$ with $a+(−a)=0$, gives us the desired result.