Consider a relation $R = \{(x,y) \in (\mathbb{N}\times\mathbb{N}): 2y\leq x \leq 3y\}$
Is this relation antisymmetric? I can't even find any $(x,y)$ such that $(x,y)\in R \land (y,x)\in R$ , (note -$0\notin \mathbb{N}$)
or the relation
$R = \{(x,y) \in (\mathbb{N}\times\mathbb{N}): 2x \mid y\}$
In this case also there isn't any $(x,y)$ such that $(x,y)\in R \land (y,x)\in R$
does that make the relation antisymmetric?
On
Yes, these relations are antisymmetric. To show a relation is antisymmetric, you need to show that if $(x,y) \in R$ and $(y,x) \in R$, then $x=y$. This is a statement of the form "if $P$, then $Q$." As you noted, $P$ is always false - there is no pair $(x,y)$ such that $(x,y) \in R$ and $(y,x) \in R$. This makes the implication "if $P$, then $Q$" vacuously true.
Another way to think about it is that "if $(x,y) \in R$ and $(y,x) \in R$, then $x=y$" is the same as saying "there is no pair of distinct elements $x \neq y$ such that both $(x,y) \in R$ and $(y,x) \in R$." Certainly this is true, because for any pair $(x,y)$ you come up with such that $(x,y) \in R$, you know that $(y,x) \notin R$.
By definition a relation $R$ on a set $X$ is antisymmetric if $$(x,y),(y,x)\in R\ \Rightarrow\ x=y.$$ So both the relations you give are antisymmetric; I believe you are able prove the implication for both relations.
I'm not sure what you mean by your question title; what does it mean for a relation 'not to apply'? If there are no $x,y\in X$ such that $(x,y),(y,x)\in R$ then the implication above is (voidly) valid.