We have set $X = \mathbb{C},z_1\sim z_2 \leftrightarrow$ Im $z_1$ $\leq$ Im $z_2$.

How to determine if this relation is:

  • reflexive

  • symmetric

  • antisymmetric

  • transitive?

  • $\begingroup$ To take one thing at a time: You are asking, how to determine whether the relation is reflexive? You want to know whether $Im\ z_1\le Im\ z_1$ is always true? Is that where you're stuck? $\endgroup$
    – bof
    Commented Mar 20, 2017 at 10:06
  • $\begingroup$ Maybe it will help to look at a concrete example. Suppose $z_1=2+3i.$ That $Im\ z_1=3,$ right? Well, is $3\le3$? Can you generalize that? $\endgroup$
    – bof
    Commented Mar 20, 2017 at 10:07

1 Answer 1


Use the following definitions:


$z_1\sim z_1$


$z_1\sim z_2 \implies z_2 \sim z_1$ for $z_1\neq z_2$


$z_1\sim z_2 \text{ and }z_1\neq z_2 \implies z_2\not\sim z_1$


$z_1\sim z_2 \text{ and } z_2\sim z_3 \implies z_1 \sim z_3$

  • 1
    $\begingroup$ The normal equality(=) on real numbers is both symmetric and anti-symmetric. Also the given relation is neither symmetric nor anti-symmetric. $\endgroup$ Commented Mar 20, 2017 at 10:21
  • $\begingroup$ @UdayanJoshi My bad, I've removed that part. I've never come across antisymmetric before, I just Googled that defintion (I knew the others) so at a first glance it appeared they were oppositves of each other. Still can't quite see it, but I'll take your word for it $\endgroup$
    – lioness99a
    Commented Mar 20, 2017 at 10:31
  • $\begingroup$ By symmetric the clause "for $z_1\neq z_2$" is redundant. By antisymmetric I would choose for (the more symmetric formulation) $z_1\sim z_2\text{ and }z_2\sim z_1\implies z_1=z_2$. $\endgroup$
    – drhab
    Commented Mar 20, 2017 at 10:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .