Set-builder notation proof

Question

$S=\{x^2:x \in \mathbb R\}$, $T=\{y^3:y \in \mathbb Z\}$, $V=\{z^{12}:z \in \mathbb Z\}$

Prove that $V \subsetneq (S \cap T)$.

Having some difficulties with figuring out how to start with these kind of questions.

Answer 1

Let $a \in V$, then $a = z^{12}$ for some $z \in \mathbb{z}$. But then $a^{1/3} = z^{4}$ is an integer such that cubing gives $a$, thus $a \in T$. Likewise $a^{1/2} = z^6$ is a real number such that squaring gives $a$ so that $a \in S$. This means $a \in V$ implies $a \in S$ and $a \in T$ thus $a \in S \cap T$ which means $V \subseteq S \cap T$. Now we need to show that the sets are not equal. $(1/2)^2 = .25$ is in $S$ but clearly not in $V$ for the integers are multiplicatively closed.

Answer 2

$S$ is the set of squared reals, $T$ the set of cubed integers, and $V$ the set of twelve-powered integers.

Now, $V\subsetneq S\cap T$ iff $(\exists s\in S\cap T.s\notin V)~\wedge~(\forall v\in V.v\in S\cap T)$.

Which is to say $(\exists s\,. s\in S\wedge s\in T\wedge s\notin V)\wedge (\forall v\,.v\in V\to v\in S\wedge v\in T)$

Notice that $s\in S$ iff $s^{1/2}\in \Bbb R$, and such.

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So is it so that?: There is some number which is in both a squared real and a cubed integer yet not a twelve-powered integer, but all twelve-powered integers are both a squared real and a cubed integer .