Why are some sets neither symmetric or anti-symmetric? In this relation set R1 = {(2, 2),(2, 3),(2, 4),(3, 2),(3, 3),(3, 4)}, when finding its property of relation- antisymmetric, transitive, symmetric etc the answer states that its neither antisymmetric or symmetric. However as the set is not symmetric should it not be antisymmetric. Trying to grasp the concept any help would be greatly appreciated. 
 A: Let $R$ be a relation on a set $A = \{1,2,\ldots,n\}$, so that $R \subseteq A \times A$.   The relation $R$ can be represented by an $n$ by $n$ matrix $M$ of zeros and ones, where $M_{ij} = 1 $ whenever $(i,j) \in R$ and $M_{ij}=0$ whenever $(i,j) \notin R$.  
A relation $R$ is symmetric if $M$ is equal to its transpose, i.e. if $M_{ij} = M_{ji}$ for all $i, j \in A$.  Thus, whenever an off-diagonal element $(i,j)$ is in the relation, its mirror image $(j,i)$ must also be in the relation (here, the mirror image is taken to mean a reflection about the main diagonal of the matrix).  This must hold for every $i,j$.  So, if there is even one location $(i,j)$ where the property "$(i,j) \in R \implies (j,i) \in R$" is violated, then the relation is not symmetric. In other words, if there exists an $(i,j)$ such that $(i,j) \in R$ and $(j,i) \notin R$, then the relation is not symmetric. 
(To understand the definition of a relation being not symmetric, take the negation of the definition for a relation being symmetric: the negation of the statement $\forall x (P(x) \implies Q(x))$ is the statement $\exists x, P(x) \wedge \neg Q(x)$.)
A relation $R$ is antisymmetric if whenever $(i,j) \in R$ and $i \ne j$, then $(j,i) \notin R$. Thus, if any off-diagonal element is in the relation, because this relation is against symmetry ("anti" symmetric),  the mirror image must not be in the relation.  So, if there is even one violation to the property $(i,j) \in R, i \ne j \implies (j,i) \notin R$, then the relation is not antisymmetric. Thus, if there exists an $(i,j)$ such that both $(i,j)$ and its mirror image $(j,i)$ belong to the relation, then the relation is not antisymmetric.
It's possible to have both kinds of violations at the same time, but they would have to be in different locations of the matrix.  In your example, $(2,3),(3,2)$ both belong to $R$. This violates antisymmetry.  So the relation is not antisymmetric.  Also, $(2,4) \in R$ and $(4,2) \notin R$, and this gives a location where the symmetry property is not satisfied (this location where you see a violation of the symmetry property is different from the location of the violation of the antisymmetry property).
To understand this better, ask the question: when can a relation be both symmetric and antisymmetric. If any off-diagonal element $(i,j)$ is present in such a relation (recall off-diagonal means $i \ne j$), then symmetry of the relation requires that $(j,i)$ be in the relation, and antisymmetry requires $(j,i)$ not be in the relation.  The only way both these can be satisfied if there is no off-diagonal element $(i,j)$ in the relation in the first place to cause a violation of one of the two properties.  Thus, if a relation is both symmetric and antisymmetric, it must have all its elements on the diagonal - meaning, $M$ is a diagonal matrix, with all its $1$'s on the main diagonal. The number of such relations is $2^n$, because the number ways to choose a subset of positions for the $1$'s from the $n$ available positions on the main diagonal is $2^n$. 
A: One symmetric pair is enough to break anti-symmetry. $(2,3),(3,2) \in R_1$ are symmetric pairs in this case.
An element whose symmetric pair is not in the relation is enough to break symmetry. In this case $(3,4) \in R_1$ but its symmetric pair $(4,3) \notin R_1$.
Therefore this relation is neither symmetric nor anti-symmetric.
