The question is: "is $\mathbb{Q}_{>0}$ well ordered by the following operator?"
Operator '$($': Write inputs as fractions in lowest terms, $a=\frac{p_a}{q_a}, b=\frac{p_b}{q_b}$. Then, $a(b\iff p_aq_b<p_bq_a$.
I have been told by peers that no, $($ does not well-order the rationals. Still, I cannot figure out why not. My guess is that the operator, $($, does not respect transitivity. IE. if $a(b$ and $b(c$, then it is not necessarily true that $a(c$. However, I have yet to find an example of this. Any help is greatly appreciated.
A well-order is a linear order $<^*$ on a set $S$ such that every non-empty $T\subset S$ has a $<^*$-least member.
$a(b\iff p_aq_b<p_bq_a\iff p_a/q_a<p_b/q_b\iff a<b.$ But the usual order $<$ on $Q^+$ is not a well-order. E.g. $\{1/n: n\in N\}$ has no $<$-least member.
BTW it is preferable to not use $($ as a relation-symbol. If used in an expression involving brackets, it may be hard to read. Common options include $aRb$ and $a<^*b.$