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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?

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    On the set of natural numbers, the relations $\gt$ and $\lt$ are antisymmetric. There union is the relation $\not=$ which is not antisymmetric. – 2017-02-18
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    The antisymmetry condition is more intuitive in contrapositive form: If $a \ne b$ then you cannot have both $(a,b)$ and $(b,a)$ in $R$. – 2017-02-18

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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

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    But it says that both R and S are anti symmetric. Does that not mean that (a,b)∈R and (b,a)∈R, then a=b (a,b)∈S and (b,a)∈S, then a=b Therefor (a=b) = (a=b)? – 2017-02-18
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    Yes, but I didn't assume $(b,a) \in R$ nor $(a,b) \in S$, just $(a,b) \in R$ and $(b,a) \in S,$ so you cannot conclude $(a,b)=(b,a)$ by using the anti symmetric properties of $R$ or $S$ – 2017-02-18
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    Im confused on the definition of what anti-symmetry is then. The question says to prove it wrong then use the set A={1,2,3}. How would you do that? (1,2,3)∈R and (3,2,1) ∈S ?? – 2017-02-18
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    Your definition, just as you wrote it above is correct. I will edit my answer in order to make it clearer – 2017-02-18
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    What I dont get is does the question states that both R and S are anti-semmtric towards the non empty set A. Meaning they arent anti-symmetric towards each other. So would they both not be (a,b)∈R and (b,a)∈R, (a,b)∈S and (b,a)∈S ? – 2017-02-18
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    What do you mean by "being antisymmetric towards a set"? Both relations R and S aredefined over the set A, this means their elements are ordered pairs of elements of A (a relation over a set A is just a subset of the cartesian product of A with itself). There is no assumption concerning the relation betwen R and S, is it clearer now? – 2017-02-18
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    Yes! Thank you!!! – 2017-02-18
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    No problem, glad to help! – 2017-02-18
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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 $A \neq B$.

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    By sets $A,B$ do you mean the sets $\{A\},\{B\}$? – 2017-02-18
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    Oops. That should be the set $(A,B)$. – 2017-02-18
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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.