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Note: my goal is maximum mathematical rigor.

The toy problem here is (from [{Spivak's Calculus book}'s 1st chapter]'s problems):

  • Prove that, if $b,d\ne 0$, then $\frac{a}{b} = \frac{c}{d}$ if and only if $ad = bc$.
  • Allowed axioms are only: additive association/inverse/identity, multiplicative association/inverse/identity, and distributive property of addition and multiplication.

Now, what puzzles me is: how would my proof be different if there was no "if and only if" phrase?

Below is my proving attempt:

First let's rewrite the problem in a language closer to the axioms:

  • Prove $ab^{-1} = cd^{-1}$ if and only if $ad = bc$ (of course, $b,d\ne 0$).

Proof--- By multiplying both sides by $bd$:

$$ ab^{-1}bd = cd^{-1}bd $$

By [axiom mul assotition]

$$ a(b^{-1}b)d = c(d^{-1}d)b $$

By [axiom mul inverse]:

$$ a(1)d = c(1)b $$

By [axiom mul identity]:

$$ ad = cb $$

$\blacksquare$

My questions

Is my proof above adequate to show "if" (way forward) and "and only if" (way back) properties?

If maximum rigor is to sought for my proof, should I add any extra wordings there to signify that my proof is actually proving the "iff" (both ways) and not only "if" (forward)? E.g. should I reverse the proof backwards to show that it's both ways?

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caveman
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1 Answers1

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Your proof assumes, from the start, that, given three real numbers $a$, $b$, and $c$, with $c\neq0$,$$a=b\iff ac=bc.$$Perhaps that you could prove it.

Besides, after multiplying both sides by $bd$, what you should get is $(ab^{-1})(bd)=(cd^{-1})(bd)$. Only then can you use associativity to prove what you want to prove. And note that you have used the commutativity of the multiplication, when you went from $cd^{-1}bd$ to $c(d^{-1}d)b$ .

José Carlos Santos
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  • Thanks. What do you mean by "perhaps that you could prove it"? Are you implying that I did not prove it, but I could? I'm also confused, are you also saying that my proof is perfect rigor, except for only one fault as I misapplied the multiplication? – caveman May 08 '19 at 23:47
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    I meant that your proof is incomplete and that, in order to be complete you should prove that statement. And, yes, there was that small fail that I mentioned. – José Carlos Santos May 09 '19 at 06:28
  • Thanks. Could you please also explain how is my proof assuming that $a = b \iff ac = bc$? To me it seems that I am just starting from the question ($ab^{-1} = cd^{-1}$) and playing with it until I get $ad=cb$. The only errors that I did that I understand is that misused multiplication and skipped commutativity. – caveman May 19 '19 at 21:40
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    At the first step, when you claim that $ab^{-1}=cd^{-1}\iff ab^{-1}bd=cd^{-1}bd$. – José Carlos Santos May 19 '19 at 22:00
  • Oh, you mean I should assume form those axioms that multiplying both sides of the equation by the same value (i.e. $bd$) would work, as this by itself needs a proof? – caveman May 21 '19 at 13:48
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    Almost. You should prove that $a=b\iff ac=bc$. – José Carlos Santos May 21 '19 at 13:50
  • Sorry, I meant that I should not assume (typo; can't edit now). Was your "Almost" only because of my typos (forgetting "not" there)? --- Is the proof only substituting $a$ into $b$, so $ac=bc$ becomes $ac=ac$? – caveman May 21 '19 at 14:05
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    Substituting $a$ into $b$ only proves that $a=b\implies ac=bc$. In order to prove that $ac=bc\implies a=b$, you should do$$ac=bc\implies(ac)c^{-1}=(bc)c^{-1}\iff a(cc^{-1})=b(cc^{-1})\iff a=b.$$ – José Carlos Santos May 21 '19 at 14:16
  • Isn't this cyclic? E.g. as you tried to prove the theorem: $a=b \implies ac=bc$, you had to use the theorem in step $ac=bc \implies (ac)c^{-1} = (bc)c^{-1}$ (which seems to me to at least assume that $a=b \to ac=bc$). No? Or am I totally missing the point of this whole discussion? :) – caveman May 21 '19 at 14:33
  • No. $a=b\implies ac=bc$ is trivial, in the sense that it works for each and every operation we can possibly imagine. But $ac=bc\implies a=b$ is not of the same nature. It works in goroups, not not for every conceivable operation. – José Carlos Santos May 21 '19 at 14:36