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The relation given is $ R = \{(a,b); 1ab>0; a,b ∈ R \} $

I clearly understand that this is symmetric since $a*b = b*a$ but I'm not able to understand that why is this reflexive also and not at all transitive. (Would be great if you can explain with the roster form of the given relation in set-builder form.)

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    $\begingroup$ This is incorrect. $(0,0)$ is not in the relation. $\endgroup$ – vadim123 Jul 29 '15 at 16:43
  • $\begingroup$ On the other hand, it is transitive. $\endgroup$ – Henning Makholm Jul 29 '15 at 16:44
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It's not reflexive. $0\times 0\not\gt 0$.

It IS transitive. If $ab \gt 0$ then $a,b$ have the same sign. If $bc\gt 0$ then $b,c$ have the same sign and thus $a,c$ have the same sign and $ac\gt0$.

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  • $\begingroup$ So, that means that it is NOT reflexive and IS transitive. If that is the case, then the book that I'm consulting is probably not a good one. :-/ $\endgroup$ – Anoneemus Jul 29 '15 at 16:48

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