Here’s a way to think about symmetry and antisymmetry that some people find helpful. A relation $R$ on a set $A$ has a directed graph (or digraph) $G_R$: the vertices of $G_R$ are the elements of $A$, and for any $a,b\in A$ there is an edge in $G_R$ from $a$ to $b$ if and only if $\langle a,b\rangle\in R$. Think of the edges of $G_R$ as streets. The properties of symmetry, antisymmetry, and reflexivity have very simple interpretations in these terms:
$R$ is reflexive if and only if there is a loop at every vertex. (A loop is an edge from some vertex to itself.)
$R$ is symmetric if and only if every edge in $G_R$ is a two-way street or a loop. Equivalently, $G_R$ has no one-way streets between distinct vertices.
$R$ is antisymmetric if and only every edge of $G_R$ is either a one-way street or a loop. Equivalently, $G_R$ has no two-way streets between distinct vertices.
This makes it clear that if $G_R$ has only loops, $R$ is both symmetric and antisymmetric: $R$ is symmetric because $G_R$ has no one-way streets between distinct vertices, and $R$ is antisymmetric because $G_R$ has no two-way streets between distinct vertices.
To make a relation that is neither symmetric nor antisymmetric, just find a digraph that has both a one-way street and a two-way street, like this one:
$$0\longrightarrow 1\longleftrightarrow 2$$
It corresponds to the relation $R=\{\langle 0,1\rangle,\langle 1,2\rangle,\langle 2,1\rangle\}$. on $A=\{0,1,2\}$.
Your first answer is correct for the reason that you give; your second is not. The relation $\le$ on $\Bbb Z^+$ is not symmetric, but it is antisymmetric: if $m\le n$ and $n\le m$, then $m=n$.
The easiest way to find a relation $R$ that is neither symmetric nor antisymmetric is to build one from scratch. To ensure that $R$ is not symmetric, we must put two distinct elements, say $0$ and $1$, into the underlying set $A$ and put exactly one of the ordered pairs $\langle 0,1\rangle$ and $\langle 1,0\rangle$ into $R$; I’ll put $\langle 0,1\rangle$ into $R$ and leave $\langle 1,0\rangle$ out. So far, then, we have $0,1\in A$ and $\langle 0,1\rangle\in R$.
To ensure that $R$ is not antisymmetric, we must have two elements of $A$ — call them $a$ and $b$ for a moment — such that $a\ne b$, but both of the ordered pairs $\langle a,b\rangle$ and $\langle b,a\rangle$ belong to $R$. We can’t use $0$ and $1$ for $a$ and $b$, since we’ve already required that $\langle 1,0\rangle\notin R$, but I can add $2$ to $A$ and use $0$ and $2$ for $a$ and $b$. That is, I’ll set $A=\{0,1,2\}$ and $R=\{\langle 0,1\rangle,\langle 0,2\rangle,\langle 2,0\rangle\}$; then
- $R$ is a relation on $A$,
- $R$ is not symmetric, because $\langle 0,1\rangle\in R$ but $\langle 1,0\rangle\notin R$, and
- $R$ is not antisymmetric, because $\langle 0,2\rangle,\langle 2,0\rangle\in R$, but $0\ne 2$.
Best Answer
If every pair satisfies $aRb\rightarrow bRa$ then the relation is symmetric. If there is at least one pair which fails to satisfy that then it is not symmetric.
Similarly if there is at least one pair which has $(aRb\rightarrow bRa)\land a\neq b$ then antisymmetry is also not satisfied.
We can therefore take the following relation: $\{a,b,c\}$ would be our universe and $R=\{\langle a,b\rangle,\langle b,a\rangle,\langle a,c\rangle\}$.
The fact that $aRc\land\lnot cRa$ shows that the relation is not symmetric, but $a\neq b$ and both $aRb$ and $bRa$ hold.