The exercise:
$A = \{ n \in \mathbb{N} : n = 4 m ^{2} \text{ for certain } m \in \mathbb{N}\},$
$B = \{ n \in \mathbb{N} : n \text{ is even} \},$
$C = \{ n \in \mathbb{Z} : n = m ^{2} \text{ for certain } m \in \mathbb{Z}\}.$
now prove that
$A \subseteq B \cap C$
The proof:
If we have $n$ that is in $A$, it has divisor $2$ and must be in $B$ as well. If we state that for each $m$ $4m^{2}=n^{2}$ for $n=2m$ then $A$ must be in $C$ as well. If $n$ that is in $A$ is in both $B$ and $C$, than it is proven. How do I correctly write this down?