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?
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 $c(d^{-1}d)b$ .