Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The motivation to this question is: Is this an equivalence relation (reflexivity, symmetry, transitivity) :

Let $\theta(s):\mathbb{C}\to \mathbb{R}$ be a well defined function. I define the following relation in $\mathbb{C}$.

$\forall s,q \in \mathbb{C}: s\mathbin{R}q\iff\theta(s)\ne 0 \pmod {2\pi}$ (and)

$\theta(q)\ne 0 \pmod {2\pi}$


My question is: Can I consider the case with equality as the inverse relation of this one and it is also an equivalence relation.

share|improve this question
1  
I simply added the relation to which you are referring to this post, to help "round out" your question (so it's self-contained). I left your link. –  amWhy Dec 21 '12 at 13:52

1 Answer 1

up vote 4 down vote accepted

First: just to be sure you understand: your relation $R$ is not an equivalence relation if there exists $x$ such that it is not the case that $x R x$: when, e.g., $\theta(x) = 0 \pmod {2\pi}$, since in this situation, it is NOT the case that $\theta(x) \ne 0 \pmod {2\pi}\;$ AND $\;\theta(x) \ne \pmod {2\pi}$

That is, the relation IS reflexive if and only if FOR ALL $x \in \mathbb{C}$, it is true that $\theta(x) \ne 0 \pmod {2\pi}\;$ AND $\;\theta(x) \ne 0 \pmod {2\pi}$.

You need reflexivity (as well symmetry and transitivity), so if $x$ satisfies $\theta(x) = 0 \pmod {2\pi} $, then you do not have $x R x$, hence it is not an equivalence relation. Not knowing exactly how $\theta(x)$ is defined makes it difficult to ascertain whether or not it can be the case that reflexivity fails. But if, say, $\theta: \mathbb{C} \to \mathbb{R}$ is onto, then reflexivity fails.


Your relation defines what it means for $s$ to be related to $q$: $$sRq \iff (\theta(s) \ne 0 \pmod {2\pi} \land \theta(q) \ne 0\pmod {2\pi}.$$ So it defines the set of all ordered pairs $(s, q)$, $s, q \in \mathbb{C}$, which satisfy the given relation. To obtain the inverse relation, you need to determine what relation defines how $q$ is related to $s$: what relation $R^{-1}$ defines the set of all ordered pairs $(q, s)$, when $(s, q)\in R$?

In this case, it turns out that the inverse relation $R^{-1}$ defines exactly the same relation as does $R$: simply commute the conditions: $s \ R^{-1} q \iff \theta(q) \ne 0 \pmod {2\pi} \land \theta(s) \ne 0\pmod {2\pi}$. Then you have $q\ R \ s $.

share|improve this answer
    
Yes, you make things very clear. Thank you very much. –  ZE1 Dec 21 '12 at 14:53

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.