How to prove that $f(x) = (x-1)/ (x+1)$ and $g(x) = x$ are ordinally equivalent on domain $[0, ∞)$?

Question

I can see that for all $x$ and $y$, $f(x)≥f(y)$ iff $g(x)≥g(y)$. But what would the formal proof look like? Would it be by induction?

Answer 1

Well, for all $x\in[0,\infty),$ we have $x+1>0,$ so if $x,y\in[0,\infty),$ we have $$f(x)\ge f(y)\quad\iff\quad (x-1)(y+1)\ge(y-1)(x+1).$$ The idea from here is to keep manipulating through a chain of equivalent inequalities until you end up with $g(x)\ge g(y).$ Can you justify the above equivalence and take it from there?