5
$\begingroup$

I found a question in my textbook but I think the answer provided is wrong. The question says: Let $S$ be a relation defined over $\mathbb R$ such that $(a,b) \in S \iff ab≥0$. Is $S$ equivalence?
Book says that it is.
But let $x=4, y=0$ and $z=-6$ be three real numbers. Then clearly $(x,y),(y,z)\in S$ but $(x,z)\notin S$. Therefore $S$ is intransitive, hence not equivalence relation. Am I doing some conceptual blunder? Please help.

$\endgroup$
3
  • 1
    $\begingroup$ Yes you are right! It is supposed to be strict inequality. $\endgroup$ Commented Mar 4 at 9:14
  • 5
    $\begingroup$ @MadhanKumar A small correction: strict inequality would mean the relation is not reflexive (since $0 \cdot 0 \not > 0$), so it’s still not an equivalence relation. It would need to be restricted to $\mathbb{R}\setminus\{0\}$ (in which case it doesn’t matter if the inequality is strict or not), or the case of $0$ needs to be separately dealt with. $\endgroup$
    – David Gao
    Commented Mar 4 at 9:17
  • $\begingroup$ yeah right..... $\endgroup$ Commented Mar 4 at 9:19

0

You must log in to answer this question.

Browse other questions tagged .