0
$\begingroup$

I'm having trouble understanding whether or not this relation would be considered antisymmetric and transitive. The a relation R on the set of real numbers by (x,y) ϵ R if and only if x-y=0. If I am understanding this correctly the this set would be correct:

R= { (1,1) (2,2) (3,3) (4,4) (5,5) }

It is clearly reflexive but I'm having trouble understanding whether or not it is antisymmetric and transitive.

$\endgroup$
0
$\begingroup$

Edit: Im sorry, i thought you defined the relation with the set you wrote. I am now looking into it further (with your full definition of $R$).

Edit 2: Well, What i wrote still holds for the arithmetic definition of your relation $R$. Try show transitivty with the definition of $R$ and the axioms of The field $\mathbb R$ . (Hint: $(x,y) \in R $ iff $ x=y$)

Using negation is always a useful tool. This relation is transitive because it's not not-transitive.

Formally speaking: $(a,b),(b,c) \in R $ yields $(a,c) \in R$ Which is clearly the case since the negation is Not true.

Try the same in order to understand if it is anti-symmetric

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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