How does one find the Zariski closure of a set? I've started to learn algebraic geometry this week (so I do not have much knowledge in the subjet) and, after reading the definition of the Zariski closure $V(I(S))$ of a set $S$, I've tried to do the following exercises without much success:

Find the Zariski closure of the following sets:
1) $\{(n^2,n^3): n \in \mathbb{N} \} \subset \mathbb{A}^2(\mathbb{Q})$
2) $\{(x,y): x^2+y^2 < 1 \} \subset \mathbb{A}^2(\mathbb{R})$

Any help with these? In general, how does one usually attacks this kind of problem?
Thank you!
 A: Since the first part has been dealt with in the comments: 
Assume $f(X,Y)=\sum_{i,j\ge 0}a_{i,j}X^iY^j\in\mathbb R[X,Y]$ is a polynomial with $f(x,y)=0$ whenever $x^2+y^2<1$.
Let $\alpha\in\mathbb R$ and consider the polynomial $g(T)=f(T,\alpha T)\in\mathbb R[T]$. Then $g(t)=0$ whenever $t^2(1+\alpha^2)<1$. As there are infinitely many such $t$, $g$ must be the zero polynomial. We conclude that each of its coefficients
$$\sum_{i+j=k}a_{i,j}\alpha^j $$
is zero - no matter what $\alpha\in\mathbb R$ we pick. Hence each of the polynomials $\sum_{i+j=k}a_{i,j}X^j$ has infinitely many roots, hence is the zero polynomials, hence all $a_{i,j}$ are zero. We conclude $f=0$.
A: It So question is “find Zariski closure for infinite set of 1-dimension objects (points) in two-dimension space”, and the answer is f=0. Isn’t that obvious? To some extent. We need projection to 1st dimension (x,  and to find “closure” there, it will be f(x, y) = g(y) space. Because non-trivial polynomial of (x) can have only finite number of roots. Then second projection f(x,y) = h(x). And intersection of these spaces is f(x,y) = 0.
