Proving $(ab)^{-1}=a^{-1}b^{-1}$ where $F$ is a field and $a,b\in F$.

Question

Proving $(ab)^{-1}=a^{-1}b^{-1}$ where $F$ is a field and $a,b\in F$.

One thing to note is $a^{-1}\ne \large\frac{1}{a}$ (same goes for $b$) in this instance as there could be fields where this isn't true.

Just before this I did a similar problem where I needed to prove $(-a)^{-1}=-(a^{-1})$. What I did there was:

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

So $-(a^{-1})$ is the multiplicative inverse of $-a$.

The problem I am having is finding an appropriate way similar to the last problem that doesn't involve using $\large\frac{1}{a}$ that I'm not allowed to use.

Answer 1

I assume you mean $a,b$ are nonzero.

It's important to remember that $(ab)^{-1}$ is by definition the unique element $c$ such that $c(ab)=(ab)c=1$. So to show $(ab)^{-1}=b^{-1}a^{-1}$, you just need to show $b^{-1}a^{-1}$ satisfies the role of the element $c$ in the previous sentence.

This follows from associativity, $$ (ab)(b^{-1}a^{-1})=a(bb^{-1})a^{-1}=aa^{-1}=1 $$ and similarly $(b^{-1}a^{-1})(ab)=1$. So by definition, $b^{-1}a^{-1}=(ab)^{-1}$.

Answer 2

$$(ab)^{-1}(ab)=e \\ \implies (ab)^{-1}ab=e \\ \implies (ab)^{-1}abb^{-1}=b^{-1}\\ \implies (ab)^{-1}a=b^{-1} \\ \implies (ab)^{-1}=b^{-1}a^{-1}$$