Proving $(ab)^{-1} = a^{-1}b^{-1}$, if $a,b\ne 0$

Having only these axioms:

• mul associativity.
• mul identity.
• mul inverse.
• mul commutative.
• distributive.

Prove that $$(ab)^{-1} = a^{-1}b^{-1}$$, if $$a,b\ne 0$$

My attempt 1 (edit: false)

Using [mul identity]: $$\begin{split} (ab)^{-1} &=(ab)^{-1}\times 1 \times 1\\ \end{split}$$

Using [mul associativity]: $$\begin{split} (ab)^{-1}\times 1 \times 1 &= (a^{-1}1) \times (b^{-1}1)\\ \end{split}$$

Using [mul identity]: $$\begin{split} (a^{-1}1) \times (b^{-1}1) &= a^{-1} b^{-1}\\ \end{split}$$

$$\blacksquare$$

My attempt 2 (edit: false)

Using [mul associativity]: $$\begin{split} (ab)^{-1} &= a^{-1} b^{-1}\\ \end{split}$$

$$\blacksquare$$

My attempt 3 (I think I nailed it here)

If $$a$$ and $$b$$ are numbers, then their product, $$(ab)$$, is a number too.

By [mul inverse], we know that: $$(ab)(ab)^{-1}=1$$

We also know that the equality holds if we multiply both sides by the same numbers:

$$(ab)(ab)^{-1} a^{-1} b^{-1}=(1) a^{-1} b^{-1}$$

By [mul associativity] we know: $$(aa^{-1}) (bb^{-1}) (ab)^{-1} =(1) a^{-1} b^{-1}$$

By [mul inverse] we know:

$$(1) (1) (ab)^{-1} =(1) a^{-1} b^{-1}$$

By [mul identity] we know:

$$(ab)^{-1} = a^{-1} b^{-1}$$

$$\blacksquare$$

Questions

My goal is to achieve maximum rigor based on the axioms above.

1. What are the mistakes in my 1st attempt?
2. What are the mistakes in my 2nd attempt?
3. What is the best way of proving it?
• How would you show something is the inverse of $ab$? Feb 14, 2019 at 4:44
• If $(ab) \ne 0$, $(ab)$'s inverse $(ab)^{-1}$, by axiom [mul inverse]. So $a$'s inverse is $a^{-1}$. $b$'s is $b^{-1}$. By the [mul associativity] axiom, I know that $(ab)$ = $(a)(b)$. I also know that $(ab)(ab)^{-1} = (a)(a)^{-1}(b)(b)^{-1} = 1$. I can multiply all sides by $(ab)^{-1}$ or $a^{-1}b^{-1}$ to cancel out those $(ab)$ and $(a)(b)$, and get $(ab)^{-1} = a^{-1}b^{-1}$. Is this what you mean I should be doing? I don't see why I couldn't pop $(ab)^{-1}$ open directly by simply using [mul associativity] as in my 2nd attempt? I also don't see why my 1st attempt is wrong. Feb 14, 2019 at 4:52
• You can't "pop open" $(ab)^{-1}$ because the notation $^{-1}$ is not an operation and doesn't distribute and other than a axiom that for any $x$ a $y$ so the $xy=1$ exists, you don't have any axioms about inverses at all. Is $y=ab$ you know $y^{-1}$ exists, but you don't have *any* idea that if $y$ "breaks apart" to $a$ and $b$ you have no idea how $y^{-1}$ breaks apart if it does at all. Feb 14, 2019 at 5:05
• Bear in mind that although $a^k$ might be notation for $aaa*...*a$, that although the notation $a^{-1}$ looks similar it has nothing to do with exponentiation and means something entirely different. At least entirely different until you can prove there's a connection. Feb 14, 2019 at 5:09
• " I also know that $(ab)(ab)^{-1} = (a)(a)^{-1}(b)(b)^{-1} = 1$. " No! You know absolutely NOTHING even remotely close to that! You know that if $y=ab$ and $y$ factors to $a\times b$ you know that $y^{-1}$ exists and that $y*y^{-1}=1$ you have utterly no idea how $y^{-1}$ factors at all. ... your job is to prove $y*(a^{-1}b^{-1})=1$. Feb 14, 2019 at 5:21

My proof.

\begin{align*} &\quad (ab)(a^{-1}b^{-1})\\ &= (aba^{-1})b^{-1}\tag{mul asso}\\ &= (aa^{-1}b)b^{-1} \tag {mul comm}\\ &= (1b)b^{-1} \tag {mul inv}\\ &= 1(bb^{-1}) \tag {mul asso}\\ &= 1\cdot 1 \tag {mul inv}\\ &= 1. \tag {mul identity} \end{align*} By the definition of multiplicative inverse, $$(ab)^{-1} = a^{-1}b^{-1}$$.

• In fact, if you get rid of the commutative law of multiplication (but keep the other laws), you'll find via a similar proof that $(ab)^{-1}=b^{-1}a^{-1}$. Feb 14, 2019 at 7:24
• @RobertShore Yes, I know this. Thanks for the comment.
– xbh
Feb 14, 2019 at 7:26
• @caveman If you want to prove "$x$ is the inverse of $y$", what would you do?
– xbh
Feb 19, 2019 at 1:10
• @xbh Oh I see what did he do. Nice. Got it. Thanks. Feb 19, 2019 at 1:14
• @caveman You're welcome. Keep the definitions in mind.
– xbh
Feb 19, 2019 at 1:15