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In the homework solution in the image the professor represents s as a ratio of integers, despite s itself being defined as an integer. Is this allowed? I proved the relation transitive without using this step.

Let $R$ be the relation on $mathbbZtimes(mathbbZ-0)$ defined by $(p,q)R(s,t)iff pt=qs$.
Show that $R$ is an equivalence relation.$\$
I disagreed with his proof of transitivity, which was:$\\$
Let $A=mathbbZtimes(mathbbZ-0)$. Suppose that $(p,q), (s,t), (x,y)in A$ satisfy $(p,q)R(s,t)$ and $(s,t)R(x,y)$ (thus $q,t,yne 0)$. Hence $pt=qs$ and $sy=tx$. Using $yne 0$, we can get $s=frac txy$. Substituting this into $pt=qs$, we obtain that $pt=qfractxy$. Since $tne 0$ we may multiply both sides by $fracyt$ and derive that $py=qz$, which implies that $(p,q)R(x,y)$.

Homework Problem

In the homework solution in the image the professor represents s as a ratio of integers, despite s itself being defined as an integer. Is this allowed? I proved the relation transitive without using this step.

Sure, the definition of the relationship guarantees that if $langle s,trangle$ and $langle x,yrangle$ are in $Bbb Ztimes(Bbb Zsetminus0)$, and it holds that $langle s,tranglemathrm Rlangle x,yrangle$, then the ratio of $tfracxty$ will be an integer.   It will equal $s$, which is an integer.

Indeed, we can represent the relation as $langle p,qranglemathrm Rlangle s,trangleiff tfrac pq=tfrac st$. [Since the pairs are from $Bbb Ztimes(Bbb Zsetminus0)$ , therefore $q,t$ are definitely non-zero.]

So we have that for any three pairs, $langle p,qrangle,langle s,trangle,langle x,yrangle$ taken from $Bbb Ztimes(Bbb Zsetminus0)$ ...

• 
  $langle p,qranglemathrm Rlangle s,trangle$ by assumption


• 
  $langle s,tranglemathrm Rlangle x,yrangle$ by assumption


• 
  $tfrac pq=tfrac st$ by definition of $mathrm R$
  


• 
  $tfrac st=tfrac xy$ by definition of $mathrm R$
  


• 
  $tfrac pq=tfrac xy$ by equality elimination


• $langle p,qranglemathrm Rlangle x,yrangle$

Hence $langle p,qranglemathrm Rlangle s,tranglelandlangle s,tranglemathrm Rlangle x,yrangletolangle p,qranglemathrm Rlangle x,yrangle$, and thus the relationship is transitive.