Is a set Anti-Symmetric if it's all reflexive? Lets say I have a set of integers of $-7\ to \ 7$
And $ R = {(a,b)|a^3=b^3} $
I don't understand anti-symmetry fully and I'm wondering if a set is only reflexive then it must be anti-symmetric?
Also is it:
Reflexive?
Symmetric?
Anti-symmetric?
Transitive?
 A: Yes, if a relation consists of all and only ordered pairs relating each and element of the set with itself, then by definition it is reflexive, and it is trivially antisymmetric, symmetric, and transitive.
Your relation $R$, that is, consists of all and only those ordered pairs such that $\forall x$ in $S = \{-7, -6, \ldots 0, \ldots 6, 7\},\; (x, x)\in R$, with $(x, y) \notin R whenever $x, y \in S$ and x \neq y$.
Put differently, a reflexive set such as this is trivially antisymmetric, symmetric, and transitive because the relation cannot fail to be symmetric, cannot fail to be antisymmetric, and cannot fail to be transitive.
Recall the definitions of the properties of a relation, for example, symmetry: "Whenever it is the case that $x R y$...then...$y R x$". This allows us to only test pairs of elements that DO satisfy "$x R y ...$." In the case at hand, the only pairs that satisfy the relation are pairs of the form $(x, x)$. And given all and only such pairs, 


*

*it is always true that for all $x$ in your set, $x R x$, since $x^3 = x^3$ for all $x\in S$ (reflexive).

*It is always true that IF $x R x$, then $x R x$ (symmetric)

*It is always true that IF $x R x$ and $x R x$, then $x = x$ (antisymmetric)

*It is always true that IF $x R x$ and $x R x$, then $x R x$ (transitive)


Some equivalence relations can be antisymmetric, as in this case.
A: Use the squares, not cubes in your example. This equivalence relation is reflexive, but is't antisymmetric: $(-6)R(6) ~\&~ (6)R(-6)~ \& ~(-6 \ne 6  )$
