I need to prove that $\frac{a}{b}=\frac{c}{d}$ iff $\frac{d}{c}=\frac{b}{a}$ using this definition (Eudoxus Def.):
$\frac{a}{b}=\frac{c}{d}$ iff for any positive integers $p$ and $q$ it holds
(1) $pa>qb$ iff $pc>qd$
(2) $pa=qb$ iff $pc=qd$
(3) $pa<qb$ iff $pc<qd$
My approach has been using algebra to obtain each of the same conditions for $\frac{d}{c}=\frac{b}{a}$ without luck, but I've been using algebra for (2) like this:
If $pa=qb$ then $\frac{p}{q}=\frac{b}{a}$ iff $pc=qd$ then $\frac{p}{q}=\frac{d}{c}$. Therefor $\frac{d}{c}=\frac{b}{a}$.
This is a right approach? How should I faced the other two conditions?
Assume $\frac{a}{b}=\frac{c}{d}$. Then for any positive integers k and h
(1) ka>hb iff kc>hd
(2) ka=hb iff kc=hd and
(3) ka$<$hb iff kc$<$hd
Flipping the iff statements we get,
(1) kc>hd iff ka>hb
(2) kc=hd iff ka=hb and
(3) kc$<$hd iff ka$<$hb
Flipping the order of the inequalities we get
(1) hd>kc iff hb>ka
(2) hd=kc iff hb=ka and
(3) hd$<$kc iff hb$<$ka
h and k are arbitrary integers so the criteria are true for all integers, so $\frac{d}{c}=\frac{b}{a}$