$\mathcal{R} = \{(1,2), (2,3), (3,4)\} \:\: \mathcal{A} = \{1, 2, 3, 4\}$
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?
On
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 $a\neq b$, 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 $n\times n$ 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 $h\neq k$ 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 $i\in\{1,2,3,4\}$.
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?)
On
The definition of an anti-symmetric relation can be thought of as follows: $$\forall a, b \in \mathcal{A} \,\, \left[((a \neq b) \longrightarrow ((a,b) \not\in \mathcal{R}) \vee ((b,a) \not\in \mathcal{R}))\right].$$
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 $(2,1) \not\in \mathcal{R}$. This means the conclusion of the (above) implication $$((1,2) \not\in \mathcal{R}) \vee ((2,1) \not\in \mathcal{R})$$ 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 $p = T$, $F \implies T = T$).