0

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?
caveman
  • 393
  • 1
  • 13
  • 1
    How would you show something is the inverse of $ab$? – J. W. Tanner Feb 14 '19 at 04:44
  • 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. – caveman Feb 14 '19 at 04:52
  • 2
    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. – fleablood Feb 14 '19 at 05:05
  • 2
    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. – fleablood Feb 14 '19 at 05:09
  • 3
    " 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 $yy^{-1}=1$ you have utterly no* idea how $y^{-1} $ factors at all. ... your job is to prove $y*(a^{-1}b^{-1})=1$. – fleablood Feb 14 '19 at 05:21
  • Thank you very much @fleablood. I added a 3rd attempt. Have I nailed it now? – caveman Feb 19 '19 at 00:32

1 Answers1

2

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

xbh
  • 8,835
  • 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 $(ab)^{-1}=b^{-1}a^{-1}$. – Robert Shore Feb 14 '19 at 07:24
  • @RobertShore Yes, I know this. Thanks for the comment. – xbh Feb 14 '19 at 07:26
  • I think your proof is wrong, because you started with $(a^{-1}b^{-1})$, which is really $a^{-1}b^{-1}$ as the brackets are only serving for association purposes. The question starts with $(ab)^{-1}$, and the goal is to prove that you can pop it out into $a^{-1}b^{-1}$. So all you did there is totally unrealted to the question. – caveman Feb 19 '19 at 00:28
  • 1
    @caveman If you want to prove "$x$ is the inverse of $y$", what would you do? – xbh Feb 19 '19 at 01:10
  • 1
    @xbh Oh I see what did he do. Nice. Got it. Thanks. – caveman Feb 19 '19 at 01:14
  • 1
    @caveman You're welcome. Keep the definitions in mind. – xbh Feb 19 '19 at 01:15