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For each of the following sets A and binary relations ~, decide whether ~ defines an equivalence relation on $A$

a) Set $A=\mathbb{R}$ Relation: $x\sim y$ if $x=ay$ for some $a \in \mathbb{Q}$

shaylene
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    “Reflexive”, “symmetric”, and “transitive” are usually properties of (binary) relations. While binary relations are “sets”, you want to specify that you are talking about relations, and not sets in general. – Arturo Magidin May 16 '21 at 22:41
  • You should not use the letter $a$ to mean different things. Instead, you can use another letter $b$ to also be a rational number, for example "$x = ay$ for some $a\in \mathbb{Q}\setminus{0}$ then $y=bx$ where $b=1/a\in \mathbb{Q}"$ –  May 16 '21 at 22:50
  • Remember also that examples can not be proofs. In your symmetry argument you rely on the fact that if $a\in \mathbb{Q}\setminus{0}$ then $1/a\in \mathbb{Q}$ which does not account for the case $a=0$, does $a=0$ break the symmetry or is there a way to get around it? –  May 16 '21 at 22:54
  • Proof for (b) is the same as in the proposed dupe since $\Bbb Q$ is a multiplicative subgroup of $\Bbb R$, and the counterexample for (a) comes from looking at where the proof breaks down when you adjoin the noninvertible element $0$, destroying the group structure (as hinted in Andre's comment). – Bill Dubuque May 16 '21 at 23:04
  • For $y \ne 0$ in part a, $0 \sim y$ but $y \not\sim 0$. – Geoffrey Trang May 17 '21 at 04:29
  • This post seems rather similar to one part of your question: Prove that $E$ is an equivalence relation, where $E$ is given by $qr=s$ for some $q\in\mathbb Q^*$ – Martin Sleziak May 17 '21 at 10:05

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If you include the number $0$ you get $\;0\sim y\;$ with $\;0=0\cdot y\;$ for any $\;y\in\mathbb{R}\;$. Though, if $\;y\neq 0\;$, you never get $\;y\sim 0\;$, as $\;a\cdot 0=0\neq y\;$ for all $\;a\in\mathbb{Q}\;$. What follows, is that under the inclusion of $0$, the symmetry won't hold.

If you remove the $0$, as shown in case (b), you maintain the symmetry, given that for all $\;a\in\mathbb{Q}\;$ where $\;a\neq 0\;$, there exists a multiplicative inverse $\;a^{-1}\;$, such that $\;x=ay\iff y=a^{-1}x\;$, or $\;x\sim y\implies y\sim x\;$. The above exception won't be a problem anymore, since $\;0\sim y\iff 0=ay\iff y=0\;$.

SimonK
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