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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?

  • 2
    It looks to me like you have written it down already2012-09-03

3 Answers 3