0
$\begingroup$

Having only these axioms:

  • add associativity.
  • add identity.
  • add inverse.
  • add commutative.
  • 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?
$\endgroup$
6
  • 1
    $\begingroup$ How would you show something is the inverse of $ab$? $\endgroup$ Feb 14 '19 at 4:44
  • $\begingroup$ 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. $\endgroup$
    – caveman
    Feb 14 '19 at 4:52
  • 2
    $\begingroup$ 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. $\endgroup$
    – fleablood
    Feb 14 '19 at 5:05
  • 2
    $\begingroup$ 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. $\endgroup$
    – fleablood
    Feb 14 '19 at 5:09
  • 3
    $\begingroup$ " 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 $y*(a^{-1}b^{-1})=1$. $\endgroup$
    – fleablood
    Feb 14 '19 at 5:21
2
$\begingroup$

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

$\endgroup$
6
  • $\begingroup$ 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 $(ab)^{-1}=b^{-1}a^{-1}$. $\endgroup$ Feb 14 '19 at 7:24
  • $\begingroup$ @RobertShore Yes, I know this. Thanks for the comment. $\endgroup$
    – xbh
    Feb 14 '19 at 7:26
  • 1
    $\begingroup$ @caveman If you want to prove "$x$ is the inverse of $y$", what would you do? $\endgroup$
    – xbh
    Feb 19 '19 at 1:10
  • 1
    $\begingroup$ @xbh Oh I see what did he do. Nice. Got it. Thanks. $\endgroup$
    – caveman
    Feb 19 '19 at 1:14
  • 1
    $\begingroup$ @caveman You're welcome. Keep the definitions in mind. $\endgroup$
    – xbh
    Feb 19 '19 at 1:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.