Krdytkyu

$mathcal{R} = {(1,2), (2,3), (3,4)} :: mathcal{A} = {1, 2, 3, 4}$

The definition of antisymmetric states if $(a,b)$ and $(b,a)$ are both elements of $mathcal{R} implies a = b$. However, $(b,a)$ is not an element and yet it is considered antisymmetrical. What is the reasoning behind this?

You are confused with the definition of symmetric relation here, for symmetric relation we need $(b,a)$ if we have $(a,b)$ in $R$. In other hand, we call a relation anti-symmetric, if it contains $(a,b)$ and $(b,a)$ then $a=b$. That means- if there exist some pair $(a,b)$ and also $(b,a)$ with $aneq b$, then the relation is not anti-symmetric. More simply,in anti-symmetric relation we can't have $(a,b)$ with unequal elements $a$ and $b$ such that both $(a,b),(b,a)in R$.

For visual aid, you can think of a $ntimes n$ table, with $n$ elements of the set(on which you are defining the relation) on top and $n$ in left. We will write $1$ in $(i,j)$ position if $i$ is related to $j$ and $0$ otherwise. Observe that, for anti-symmetric relation, if you have $1$ in $(h,k)$ for some $h$ and $k$ with $hneq k$ then you can't have $1$ at $(k,h)$(though entries at $(i,i)$ will be $1$). But for symmetric relation, if you have $1$ at $(a,b)$ then you must have $1$ at $(b,a)$.

For the given relation follow the tables below:
$$
begin{array}{c|lcr}
text{Fig I} & 1 & 2 & 3 & 4 \
hline
1 & 0 & 1 & 0 & 0 \
2 & 0 & 0 & 1 & 0 \
3 & 0 & 0 & 0 & 1\
4 & 0 & 0 & 0 & 0
end{array}~~~~~~begin{array}{c|lcr}
text{Fig II} & 1 & 2 & 3 & 4 \hline
1 & 0 & 1 & 0 & 0 \
2 & 1 & 0 & 1 & 0 \
3 & 0 & 1 & 0 & 1\
4 & 0 & 0 & 1 & 0
end{array}
$$

Observe that, $text{Fig I}$ doesn't contain the $1$ at the diagonally opposite position where $1$ is placed. Hence, is the anti-symmetric relation. $text{Fig II}$ contains $1$ for every diagonally opposite position where $1$ is placed above the diagonal-this type of table corresponds to symmetric relation.

Both may contain $1$ at the diagonal, means $(i,i)$ position for $iin{1,2,3,4}$.

Hope it will help to distinguish between different kind of relations. (Can you work out how the table will look like reflexive or say asymmetric?)

The definition of an anti-symmetric relation can be thought of as follows:
$$forall a, b in mathcal{A} ,, left[((a neq b) longrightarrow ((a,b) notin mathcal{R}) vee ((b,a) notin mathcal{R}))right].$$

So the implication is false ONLY when we have $a,b in mathcal{A}$ such that $a neq b$ and BOTH $(a,b)$ and $(b,a)$ are in the relation $mathcal{R}$.

Consider $a=1,b=2$. Here $a neq b$ is true, so the hypothesis of the implication is true. Now $(1,2) in mathcal{R}$ and $(2,1) notin mathcal{R}$. This means the conclusion of the (above) implication
$$((1,2) notin mathcal{R}) vee ((2,1) notin mathcal{R})$$
is true. So the whole implication is true. Now you can check the remaining pairs as well.

When the statement does not say anything about what happens when both are not elements of $R$ as in this case, we can say that this statement is true because for a statement $p$, $F implies p$ is always $T$ (For $p = F$, $F implies F = T$ and for $p = T$, $F implies T = T$).