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Isn't the following relation supposed to be neither?

I believe for reflexive $x = ax$ holds true for only $a = 1$. But it isn't so for any other rational number. Isn't it that if we can prove any one case where the relation doesn't hold true, it isn't of that type?

I understand it isn't symmetric but how is it supposed to be transitive? Can someone please help me out with an example?

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    Maybe the idea was that $x \sim y$ if there exists rational $a$ s.t. $x = ay$? – mihaild Sep 20 '23 at 09:21
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    @mihaild If so then someone (either OP or the original problem creator) has badly misstated the problem. As stated the wording is kind of vague anyway, but I would certainly interpret it as asking about a separate relation for each $a$. I would probably read it as saying “let $a$ be a rational number. Which values of $a$ make the relation Symmetric? Reflexive? Transitive?” Also if this were the exact phrasing of the question I’d request a refund for textbook. – M W Sep 20 '23 at 10:30
  • You already concluded that the relation is reflexive iff $a=1$. For it to be symmetric, if $x=ay$ then $y=ax$, whence $x=a^2x$, or $a^2=1$. For transitivity, if $x=ay$ and $y=az$, then $x=a^2z$, so $(x,z) \in R$ iff $a=a^2$ iff $a=1$. Is this what you're asking? – amrsa Sep 20 '23 at 10:53
  • I myself am confused. This question came on my school test and this was the exact wording. And some people were saying that it is reflexive by the logic that we can assume a to be 1. Same eeith transitive. But how does that make sense cause its not actually stated in the question. My question was what would've been your answer had you been given this problem cause I chose its neither reflexive, symmetric or transitive. – Betelgeuse2051 Sep 20 '23 at 14:13
  • On what set is this relation being applied? – paw88789 Sep 20 '23 at 14:45

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If it is transitive, then we can use $x=ay$,$y=az$ to deduce $x=az$, which implies $a^2=a$, which indicates $a=0$ or $a=1$. But if $a=0$, then the relation is not reflexive.

yi li
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