I am a sophomore in college and just started an introductory real analysis course. Our professor decided to start with the construction of real and rational numbers. I feel I have a good intuitive understanding of what this statement means, but cannot seem to reach conclusive results while creating a proof. Any help is greatly appreciated.
Given the relation $\sim$ defined on $\mathbb{Z}\times\mathbb{Z}\setminus\{0\}$ with (a,b) $\sim$ (c,d) when ad = bc.
Prove the following proposition
Proposition. Let (a,b), (c,d), (e,f), (g,h) $\in\mathbb{Z}\times\mathbb{Z}\setminus\{0\}$ with (a,b) $\sim$ (e,f) and (c,d) $\sim$ (g,h). If (a,b) < (c,d), then (e,f) < (g,h).
For clarification by (a,b) < (c,d) we can assume (cb - ad,db) > $\mathbb{O}$, where $\mathbb{O}$ is the additive inverse defined as (0,1). We can also assume that if (a,b) < (c,d), then (cb - ad)(db) > 0.
Here is what I know and where I am struggling:
I know that since (a,b) $\sim$ (e,f) that af = be and since (c,d) $\sim$ (g,h) that ch = dg. I also know since (a,b) < (c,d) that (cb - ad)(bd) > 0. To show that (e,f) < (g,h) I must show that (gf - eh)(hf) > 0. Initially I though to multiply af = be and ch = dg and subtract the equations to get (cb - ad)(bd) = (gf - eh)(hf), but after half an hour I wanted to try something else. A friend suggested showing that (cb - ad, db) $\sim$ (gf - eh)(hf), but neither of us could find a way. I have a strong feeling the latter is correct but need some guidance.
Let $a,b,c,d,e,f,g\in\Bbb Z$ with $b,d,f,h\ne0.$ If $$af=be\quad\text{and}\quad ch=dg,$$ then $$(gf-eh)bd=dgfb-behd=chfb-afhd=(cb-ad)hf $$ hence if moreover $$(cb-ad)bd>0,$$ then $$(gf-eh)hf(bd)^2=(cb-ad)bd(hf)^2>0$$ which implies $$(gf-eh)hf>0.$$