Answer
$(A_1)$ holds, $(A_2 )$ fails
Work Step by Step
Set of natural numbers = $\mathbf{N}=\{1,2,...\}$
($A_1$) Let $n$ and $m$ are two natural numbers , i.e. $n,m\in\mathbf{N}$
and sum of two natural numbers is again a natural number , i.e. $n+m\in\mathbf{N}$ for all $n,m\in\mathbf{N}$
Therefore, Set of natural numbers $\mathbf{N}$ is closed under addition of vectors ( here natural numbers)
($A_2$) In particular , Let $x=1\in\mathbf{N}$
and we can choose scalars from set of real numbers
In particular take $k=-2$
Consider $\:k.x=1.(-2)=-2$
But $\;k.x=-2$ does not belong to $\mathbf{N}$
Therefore, $\mathbf{N}$ is not closed under scalar multiclipation.
