Consider the relation $R$ on $\Bbb R\times \Bbb R$ by $(a,b)R(c,d)\iff (ad-bc)=0$.

What are the equivalence classes?

I worked out that the relation is an equivalence relation.

But I cant find the equivalence classes

Can someone help please?

  • 1
    $\begingroup$ This is a particularly important relation. If this were defined on $\Bbb Z\times \Bbb Z^*$ instead, we use this as we rigorously define the set of rational numbers and this would be how we actually define equality between rational numbers. $\endgroup$ – JMoravitz Aug 27 '19 at 16:57

You should be careful of zero here. As the hint and my other comments allude to, this is very closely related to how we define equality of fractions.

However!!! You should be aware that division by zero causes a problem. And zero divided by zero is equally troubling.

Indeed, if you were to pay attention, you would notice that $(0,0)$ is related to everything.

$(1,1)$ is related to $(0,0)$ which is in turn related to $(2,1)$, however $(1,1)$ is not related to $(2,1)$.

As such, your relation is not transitive. This can be corrected by restricting your relation from being over all of $\Bbb R\times \Bbb R$ to instead being over $\Bbb R\times \Bbb R^*$, essentially forbidding zero from being a second entry from any ordered pair. Without doing this however, your relation then fails to be an equivalence relation.

If we were to make this change., then our set of equivalence classes will be those sets of ordered pairs of real numbers who when divided equal one another. For example, one of the equivalence classes will be $[(\pi,3)]$ Included in this would also be $(\pi,3),(2\pi,6),(3\pi,9)$ as well as $(\pi\times \sqrt{2},3\sqrt{2}), (\pi e, 3e)$ and so on.

A common choice of canonical element from an equivalence class here would be the one which has $1$ as the second element. The canonical element of $[(\pi, 3)]$ might be $(\pi/3,1)$ for instance.

The set of equivalence classes then might be written as $\{[(x,1)]~:~x\in\Bbb R\}$ where $[(x,1)] = \{(a,b)~:a,b\in\Bbb R,b\neq 0,~a/b = x\}$., i.e. those pairs $(a,b)$ who when you perform the division $a/b$ results in $x$.

  • $\begingroup$ I understand that this relation on $Z\times Z$ gives $Q\times Q$ but what about $R\times R$ $\endgroup$ – Math_Freak Aug 27 '19 at 17:11
  • $\begingroup$ I really dont understand how to write the classes $\endgroup$ – Math_Freak Aug 27 '19 at 17:23

Hint: if $ad - bc = 0$, and $b \neq 0 \neq d$, then $\frac{a}{b} = \frac{c}{d}$.

  • $\begingroup$ Can you explain a bit how to find the equivalence classes in $\Bbb R^2$ $\endgroup$ – Math_Freak Aug 27 '19 at 16:59
  • $\begingroup$ @Math_Freak You should be familiar with fractions of integers and that $\frac{3}{6} = \frac{1}{2}$. In exactly the same way and for the same reason, in your relation you have $(3,6)$ is related to $(1,2)$ $\endgroup$ – JMoravitz Aug 27 '19 at 17:01
  • $\begingroup$ @JMoravitz Here the set is $\Bbb R\times \Bbb R$ $\endgroup$ – Math_Freak Aug 27 '19 at 17:03
  • $\begingroup$ @Math_Freak And? $\frac{\pi}{3} = \frac{3\pi}{9}$, just as $(\pi,3)$ is related by your relation to $(3\pi,9)$ $\endgroup$ – JMoravitz Aug 27 '19 at 17:04
  • $\begingroup$ @JMoravitz so what should be the classes then? $\endgroup$ – Math_Freak Aug 27 '19 at 17:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.