Is equality under the integers {...-2,-1,0,1,2,...} symmetric and antisymmetric? I am reading a scripture from my university in discrete mathematics. It says that equality under the integers is reflexive, symmetric and transitive. However, isn’t it also antisymmetric? Since if (a,b) are equal and (b,a) are equal, it always follows that a=b because if we take any integer in the relation = in the set of integers, it can only be equal to itself and thus equality on the integers is both symmetric and antisymmetric.
 A: P be an antisymmetric realation iff "(a,b)∈P and (b,a)∈P " implies "a=b",(this is the definition of antisymmetric relation).
A relation P is symmetric iff "(a,b)∈P implies (b,a)∈P ".
What have you asked that is symmetric relation can be a antisymmetric? It is not possible if there be an element (a,b)∈P with a not equal b. This proves the necessary part for being together symmetricity and antisymmetricity of a relation.
For sufficient conditions, if we take all the elements of the relation P as in the form (a,a), then it's automatically provides a relation which is together symmetric and antisymmetric.
So, antisymmetricity and symmetricity together for a relation is possible iff a=b for all (a,b)∈P.
A: Equivalence under equality is such that, for some relation $\mathscr{R} \subseteq S \times S$ for a set $S$ then $\mathscr{R}$ is symmetric when the statement
$$(x, y) \in \mathscr{R} \implies (y, x) \in \mathscr{R}$$
is true.
Antisymmetry occurs if and only if
$$(x, y) \in \mathscr{R} \wedge x \neq y \implies (y, x) \notin \mathscr{R}$$
or equivalently
$$(x, y) \in \mathscr{R} \wedge (y, x) \implies  x=y$$
