Question

Comment on the strength of the following argument and show that it is false by providing a counter example: "The reflexive property in the definition of an equivalence relation is redundant. Equivalence relations are symmetric so that if x ~ y, then y ~ x. Using the transitive property, we can deduce that x ~ x."

Answer 1

I think I have a counterexample. Let there be a set \(S=\{1,2,3\}\) and define a relation \(R\) on \(S\) such that \((2,3),(3,2)\in R\). Clearly, this is symmetric and transitive, but it is not reflexive because \((1,1)\notin R\).

Answer 2

@helder_edwin that's the symmetric property, and not reflexive.

Answer 3

reflexive means that it goes back to itself right?

Answer 4

Yes u r right. sorry. but

in the example @KingGeorge gave you
\[ \large R=\{(2,3),(3,2),(2,2),(3,3)\} \]
is symetric but not reflexive

Answer 5

Reflexive is \((x,x)\in R\,\,\forall x\in S\)
Symmetric is \((x,y)\in R \implies (y,x)\in R\)
Transitive is \((x,y),(y,z)\in R\implies (x,z)\in R\)

Answer 6

so how do we see transitivity in ur example?

Answer 7

In my example I should have said let \(R\) be a symmetric and transitive relation such that \((2,3)\in R\). That would force \(R=\{(2,3),(3,3),(2,3),(3,2)\}\)

Answer 8

shld be (2,2) the third element in the set?

Answer 9

It should :/

Answer 10

okkkk gotcha. Thanks :)

Answer 11

you're welcome.