Use the axioms for the field of real number to prove $1/(-a) = -(1/a)$

Question

How would you prove using the axioms for the real numbers that $1/(-a) = -(1/a)$?

I tried the following:

$1/(-a) = 1/(-a) + 0(1/a) = 1/(-a) + (1+(-1))(1/a) = 1/(-a)+1/(a) + (-(1/a))$

But I don't know if $1/(-a)+1/(a)=0$ or not?

Answer 1

$1/(-a)+1/(a)=0$ is equivalent to what you want to prove, so your argument is not an improvement.

Here is an argument:

$$ 1/(-a)=1/((-1)\cdot(a))=1/(-1) \cdot 1/a = (-1) (1/a) = -(1/a) $$

For that to work, you need to prove that

• $-a = (-1)(a)$
• $1/(ab)=(1/a) \cdot (1/b)$
• $1/(-1)=-1$

all of which are easy because they follow from uniqueness: $x=-a$ iff $x+a=0$ and $x=1/a$ iff $xa=1$.