If R and S are both anti-symmetric, then R ∪ S is anti-symmetric The question states. Suppose R an S are both relations on  a non empty set A. Determine if the following is correct.
If R and S are both anti-symmetric, then R ∪ S is anti-symmetric.
Now I now that Binary relation R on a set A is antisymmetric if and only if
if a, b ∈ A and both (a, b) and (b, a) are in R, then a = b.
With this definition I believe that its true  R ∪ S is anti-symmetric but I'm having a hard time explaning why. If R and S are both antisemtric it means that they both equal a=b therefor the or statement of them must be anti-Symmetric?
 A: I think the answer is no. Take $(a,b) \in R$ and $(b,a) \in S$, then $(a,b),(b,a) \in R \cup S$ but it doesn't imply $(a,b)=(b,a)$.
The key here is that $\{(a,b),(b,a)\} \subset R \cup S$ but it is not the case that $\{(a,b),(b,a)\} \subset R$ nor $\{(a,b),(b,a)\} \subset S$ so you can't use the antisymmetry of $R$ or $S$ to conclude $(a,b)=(b,a)$
Added: If $R\cup S$ were antisymmetric, then the following should hold for any $a,b\in A$: $$(a,b),(b,a) \in R \cup S \implies (a,b)=(b,a)$$
For your example in the comment, let $A=\{1,2,3\}$, $R=\{(1,2),(2,2),(3,1)\}$, and $S=\{(1,2),(1,3)\}$.
Note both $R$ and $S$ are antisymmetric but $(1,3)$ and $(3,1)$ are in $R \cup S$ but $1 \not=3$, hence $R \cup S$ is not antisymmetric
A: This not not true.  Consider the set $(A,B)$ and the relation $(A,B)$ and $(B,A)$.  Both relations are anti symmetric but their union isn't anti symmetric because both $(A,B)$ and $(B,A)$ are members of the relation but $A \neq B$.
A: Let $R$ be a total order relation on the set $X$ (for instance the usual ordering $\leq$ on $X=\mathbb{N}$), so antisymmetric. Let
$$
S=R^{\mathrm{op}}=\{(x,y):(y,x)\in R\}
$$
which is obviously antisymmetric as well.
Then $R\cup S=X\times X$ which is only antisymmetric if $X$ has at most one element.
More generally, if $R$ is antisymmetric and there exist $x,y$ such that $(x,y)\in R$ but $x\ne y$, then $R^{\mathrm{op}}$ is also antisymmetric, but $R\cup R^{\mathrm{op}}$ isn't.
