Why is reflexive property necessary? equivalence relations

[hocuspocus102]

Homework Statement

Provide an example that shows why the reflexive property is not redundant in determining whether a relation is an equivalence relation or not. For example, why can't you just say, "If xRy then yRx by symmetric property, and then using transitive property you get xRx." Give a counterexample to that statement.

Homework Equations

The Attempt at a Solution

I know the reflexive property is necessary but I can't find a good example of why. The TA for my class said "for all integers greater than 0, aRb if and only if ab is odd" is an example and said if a is 2 then aRa doesn't hold because aa would be even and not odd as the relation requires it to be. But in my mind, you can't even use a = 2 to begin with because 2 times any number is even so no matter what b is it wouldn't be odd anyway. So I don't understand if his example is correct or if I'm just misinterpreting it. Thanks.

 

[jbunniii]

Provide an example that shows why the reflexive property is not redundant in determining whether a relation is an equivalence relation or not. For example, why can't you just say, "If xRy then yRx by symmetric property, and then using transitive property you get xRx." Give a counterexample to that statement.

This argument implicitly assumes that $x \neq y$, i.e., that the equivalence class of $x$ contains at least two elements. That isn't necessarily the case.

 

[hocuspocus102]

ok, yeah I see that and understand why that makes sense. I just don't see how the example works if it says aRb iff ab odd. it'd be reflexive for say a=3 because 3R3 iff 3*3 odd which it is. but it's not even a relation for a=2 because the iff doesn't hold because 2*2 is even which the relation clearly says it's only a relation if it's odd.

 

[jbunniii]

P.S. Your TA's example is a good one. It is an example of a relation that satisfies symmetry and transitivity, but not reflexivity. I don't understand your argument against it: by definition, the relation is defined on the positive integers, so it is certainly applicable when a = 2. You are correct that 2 times any number is even, but all that means is that aRb is false for all b when a = 2. In other words, the "equivalence class" for 2 is empty - it doesn't even contain 2!

 

[hocuspocus102]

oh so if the equivalence class of it is empty, it can't be a relation because every equivalence class is supposed to have at least one element? and if it were defined over the odd integers for example it would hold?

 

[jbunniii]

oh so if the equivalence class of it is empty, it can't be a relation because every equivalence class is supposed to have at least one element? and if it were defined over the odd integers for example it would hold?

It's a relation, but not an equivalence relation.

A relation R is simply a rule that defines whether aRb is true or false for each pair a,b.

So your TA's rule defines a relation: aRb is defined to be true if ab is odd, and it is defined to be false if ab is even.

An equivalence relation is a special type of relation. In addition to requiring you to define the rule for when aRb is true or false, it imposes three additional constraints: reflexivity, symmetry, and transitivity.

Your TA's example happens to satisfy the latter two, but not reflexivity. Thus it is a relation but not an equivalence relation. The example shows that without reflexivity, it's possible to have a situation where aRb is false for a given a and every b. Since in an equivalence relation, "R" is supposed to represent "is equivalent to", it's not very cool that you can have values that aren't equivalent to anything, not even to themselves. We want to disallow this, and that's why we need reflexivity.

 

[hocuspocus102]

oh yeah sorry I meant "not an equivalence relation" but thank you very much, I think I get it now!

 

[jbunniii]

P.S. In answer to your last question, if your TA's rule R was defined the same way, but the "input" set was limited to the odd integers, then yes, it would be an equivalence relation because then all three requirements would be satisfied.

However, it would not be a very interesting equivalence relation. Can you see why? (Hint: what's a complete list of all the elements that are equivalent to 1 in that case?)

 

[NameNvrMattrd]

Why is reflexive property necessary has to come with a "w.r.t". In mathematics abstract structures with minimal properties have been created so that they conform to certain properties.Concept of equivalence relation is used in many fields and i would like to quote one example to mention the importance of the reflexive property:

Every equivalence property on a set defines a partition (comprised of equivalence sets of the elements) of a set :

If reflexive property wasn't true, this above mentioned property would fail. As rightly shown by example of R: aRb iff ab is odd. This satisfies transitive and symmetric properties. But not reflexive property . If this relation was to be equivalence relation then 2 or any even number for that matter would not belong to any of the equivalence sets implying : it does not give a partition on the set.

 

[LCKurtz]

Just to add my 2 cents worth, it might help the OP to think of the relation as a subset of NxN:

It isn't an equivalence relation on N because, as has been pointed out, (2,2) isn't in the subset.