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I want to prove that the inverse of a product equals the product of inverses, but I’m not quite sure if to I am using what I want to demonstrate. If so, can you explain me a way to avoid this kind of mistakes?

Thanks for the advice!

Let $a$ and $b$ be nonzero real numbers. Then

\begin{align}(ab)(ab)^{-1}=1&\implies(b^{-1}a^{-1})[(ab)(ab)^{-1}]= (b^{-1}a^{-1})[1]\\ &\implies[b^{-1}(a^{-1}a)b](ab)^{-1}]= (b^{-1}a^{-1})\\ &\implies[b^{-1}(1)b](ab)^{-1}]= b^{-1}a^{-1}\\ &\implies[b^{-1}b](ab)^{-1}]= a^{-1}b^{-1}\\ &\implies[1](ab)^{-1}]= a^{-1}b^{-1}\\ &\implies(ab)^{-1}]= a^{-1}b^{-1}\hspace{1.5mm}._{QED}\end{align}

Another User
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Aya Monge
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  • Looks good to me. There’s an easier way, though. – While I Am Jun 25 '22 at 18:39
  • I suggest you give at least a brief justification of each step. That will help you (and your Readers) be sure you didn't use circular arguments. – hardmath Jun 25 '22 at 18:57
  • Welcome to Math.SE! <> Your instincts are good: As written, this proof uses what you want to prove (so is not OK), and arrives at a conclusion we already know. Instead, assume $a^{-1}$ and $b^{-1}$ denote the inverses of $a$ and $b$, and use this together with associativity to prove $(ab)(b^{-1}a^{-1}) = 1$ (and similarly with the factors swapped if your operation is not commutative). This simultaneously shows $ab$ is invertible, and $(ab)^{-1} = b^{-1}a^{-1}$. – Andrew D. Hwang Jun 25 '22 at 20:23
  • Your proof is fine. But have you proven that inverses are unique yet? That is $ac = 1 \iff c = a^{-1}$? If so this becomes much simpler. Just show that $(ab)[a^{-1}b^{-1}] =1$ and then because inverses are unique that means $(ab)^{-1}$ must equal $a^{-1}b^{-1}$. So just show $(ab)(a^{-1}b^{-1})=(ab)(b^{-1}a^{-1})$ (by commutivity) $= a(bb^{-1})a^{-1}$ (by associativity) $=a\cdot 1 \cdot a^{-1}$ (def of inv)$=a\cdot a^{-1} = 1$. That's all. – fleablood Jun 25 '22 at 21:48
  • @AndrewD.Hwang " As written, this proof uses what you want to prove" Are you sure of that? Can you point out where. I seriously do not think it does.... The OP denotes the inverse of $(ab)$ as $(ab)^{-1}$ and notes by definition of inverses $(ab)(ab)^{-1}=1$. Then they multiply both sides but the same value. And from then on out it is simple and legitimate simplification of the RHS to get the final result. The only assumptions are 1)$(ab)$ has an inverse and 2) all the axioms of arithmetic (commutivity, associativity, etc) – fleablood Jun 25 '22 at 21:51
  • FWIW and BTW. To prove inverses are unique: Suppose $a$ has $c$ and $d$ as inverses-- that is to say suppose $ac =1$ and $ad = 1$. Take $ac =1$ and multiply both sides by $d$ to get $dac = d\cdot 1=d$. A little commutivity and associativity finese and $dac = (da)c =(ad)c =1\cdot c = c$ and therefore it must be true that $c =d$ and $a$ has only one unique inverse..... for what its worth and by the way.... – fleablood Jun 25 '22 at 22:00
  • @fleablood Admittedly my comment was hasty. (Particularly, I misread the final line.) The proof does tacitly assume $ab$ is invertible, however, which I construed as part of the conclusion. – Andrew D. Hwang Jun 26 '22 at 00:59
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    It does tacitly assume $ab$ is invertible but as $a$ and $b$ are presumably invertable, they are non zero so $ab$ is non-zero so $(ab)$ is invertable.... presumably. We do need to know which axioms and proposition the OP has at their finger tips. – fleablood Jun 26 '22 at 03:48

1 Answers1

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You are not using what you want to show.

If $a,b \ne 0$ then $ab \ne 0$ and so $(ab)^{-1}$ exists and is equal to something. Also $a^{-1},b^{-1}$ and therefore $a^{-1}b^{-1}$ exists as well.

By definition of inverses $(ab)(ab)^{-1} = 1$.

So you haven't assumed anything about what you want to prove yet.

Now you are just multiplying both sides by something. Nothing wrong with that. You can multiply both sides by anything you want! So sure you can multiply both sides by $a^{-1}b^{-1}$. Why not?

So $(b^{-1}a^{-1})(ab)(ab)^{-1} = (b^{-1}a^{-1})\cdot 1 = (b^{-1}a^{-1})$. There's nothing disputable about that.

And then all you have done is simplify the RHS by simple rules and the result is that the RHS simplifies to $(ab)^{-1}$. You are done and you did not assume what you wanted to show.

Good job!

......

But there is an easier shorter way.

Note, if $(ab)\cdot M = 1$ then by definition $M$ is $(ab)^{-1}$.

So just show that $(ab)(b^{-1}a^{-1}) = 1$ (which you are obviously capable of doing).

That means that by definition $(b^{-1}a^{-1})= (ab)^{-1}$.

fleablood
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  • Yeah, I was thinking of the exercise this way. I know that for any non-zero real number $x$, there exists an inverse such that $x \cdot x^{-1}=1$. I hadn’t shown yet that inverses are unique, but I’m already added to it. Thanks for your help! – Aya Monge Jun 26 '22 at 12:44
  • A lot matters an what axioms and definitions of numbers you are using and what order they were presented in. – fleablood Jun 26 '22 at 16:01