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Let $R$ be a relation on $\Bbb Q$ defined by \begin{equation*} aRb \iff a^3 + b^2 + a^2 b = b^3 + a^2 + ab^2, \end{equation*} for all $a,b \in \Bbb Q$.

I've shown that this relation $R$ is a equivalence relation. But, I got confused to determine the equivalence classes since it was on $\Bbb Q$. For the integers part I've understood, but not yet for the fractions part. Any ideas? Thanks in advanced.

lap lapan
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    I don't see why this would be transitive. (-1,1) and (1,0) but no relation -1,0 – Will Jagy Jul 31 '21 at 00:55
  • by factoring, the three ways to get $aRb$ are $a=b , ; ;$ $a+b=0, ; ; $ $a+b=1, ; ; $ – Will Jagy Jul 31 '21 at 01:08
  • @WillJagy But in the question is to showing that the given relation is a equivalence relation and finding its classes. How? – lap lapan Jul 31 '21 at 02:41
  • Equivalence relations are transitive by definition. What you want is ${b: aRb}$ for the NON-equivalence relation $R$ – DanielWainfleet Jul 31 '21 at 03:45
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    @DanielWainfleet Since $R$ is not an equivalence relation, then its not have an equivalence classes, right? So, I answering the question by showing the counter-ex that $R$ is not an equivalence relation. – lap lapan Jul 31 '21 at 03:56

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We have \begin{equation*} aRb \iff a^3 + b^2 + a^2 b - (b^3 + a^2 + ab^2)=0\iff (a-b)(a+b)(a+b-1)=0 \end{equation*} So set of all numbers related to a given number $a$ is $\{a,-a,1-a\}$. So for example, $\frac12R(-\frac12)$, $(-\frac12)R\frac32$ but $\frac12\not R\frac32$.

Martund
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  • And hence, the relation is not a equivalence relation? Oh my bad. I didn''t see that there exists a counter-ex. Thanks! – lap lapan Jul 31 '21 at 01:00