Proof that $(-x)^{-1} = -(x^{-1})$

Question

How do I prove technically that

$$(-x)^{-1} = -(x^{-1})$$

with the field axioms? I tried the following but in the second step I lack an argument that $(a \cdot b)^{-1} = a^{-1} \cdot b^{-1}$.

$$(-x)^{-1} = (-1 \cdot x)^{-1} = (-1)^{-1} \cdot x^{-1} = -1 \cdot (x^{-1}) = -(x^{-1})$$

I feel like it must be really simple but I'm stuck right now .

Answer 1

With problems like these, the easiest way is just to use definition and uniqueness of multiplicative inverse: $$(-x)^{-1} = -x^{-1} \iff -x\cdot (-x^{-1}) = -x^{-1}\cdot(-x) = 1.$$

Now, this is almost direct, but you need to use that $$-(xy) = (-x)y = x(-y)$$ which can again be proven from definition and uniqueness of additive inverse:

$$-(xy) = (-x)y \iff xy + (-x)y = (-x)y + xy = 0$$ and similarly for the other one.

Can you prove these statements?

Answer 2

Well, due to associativity and commutativity, \begin{align}&(a^{-1}b^{-1})(ab)=((a^{-1}b^{-1})a)b=(a(a^{-1}b^{-1}))b=((aa^{-1})b^{-1})b=\\=&b^{-1}b=1\end{align}

For non-commutative fields, the formula is actually $(ab)^{-1}=b^{-1}a^{-1}$, for which \begin{align}(ab)(b^{-1}a^{-1})&=((ab)b^{-1})a^{-1}=(a(bb^{-1}))a^{-1}=aa^{-1}=1\\(b^{-1}a^{-1})(ab)&=b^{-1}(a^{-1}(ab))=b^{-1}((a^{-1}a)b)=b^{-1}b=1 \end{align}