Are there more rectangles than squares? My son just asked me a question that I thought was pretty good.  He asked, "Are there more rectangles than squares?"  
He's 13 and very good at math.  For instance, we've talked about how there are more real numbers than integers and seems to understand.  
My question is the reverse of this question, I think:
Do the real numbers and the complex numbers have the same cardinality?
So it would seem that there is the same number of rectangles as squares.  But it's very strange, since we know that squares are a subset of rectangles.
Any thoughts?
 A: There are the same number of rectangles as squares for the reason you mention, even though the set of squares is a proper subset of the set of rectangles.  It's no stranger than the fact that there are the same number of even integers as there are integers.  Every infinite set can be placed in one-to-one correspondence with a proper subset of itself.  
A: It depends. Two answers have pointed out that, as far as cardinality goes, the set of squares has the same cardinality as the set of rectangles. But there are other ways to look at the question. 
For example, if you know where two opposite vertices of the square are, then you know the square. But a rectangle is not determined by two vertices – you have to know three vertices to determine a rectangle. So in a sense (and it can be made a quite precise sense), the set of all squares is a two-dimensional object, while the set of all rectangles is a three-dimensional object, and in that sense the set of all rectangles is a bigger object. It's like the two-dimensional surface of a three-dimensional ball; surface and ball have the same cardinality, but ball has the bigger dimension. 
A: We can define the set of rectangles in $\mathbb{R}^2$ (disregarding translations by sending the bottom-left corner to the origin) to be the set $\mathcal{R}=\{[0,a]\times[0,b]:a,b\in\mathbb{R}\}$ and the set of squares $\mathcal{S}=\{[0,a]\times[0,a]:a\in\mathbb{R}\}$. There is a bijection $\mathcal{R}\to\mathbb{R}^2$ given by $[0,a]\times[0,b]\mapsto(a,b)$, and a bijection $\mathcal{S}\to\mathbb{R}$ given by $[0,a]\times[0,a]\mapsto a$. Since $|\mathbb{R}^2|=|\mathbb{R}|$, there exists a bijection $\mathcal{R}\to\mathcal{S}$ and so $|\mathcal{R}|=|\mathcal{S}|$, i.e. there are "the same number" of rectangles as there are squares.
A: It depends on what you mean by "more", a "square", and a "rectangle".  For the sake of this discussion, I am going to assume that a square is a regular quadrilateral, and that a rectangle is a equiangular quadrilateral.  I will be working in Euclidean space, which is coordinate-free (i.e. I do not impose a Cartesian coordinate system upon this space; the answer of both Gerry Myerson seems to implicitly impose this condition, while csch2's answer does not, though it still seems to think of rectangles and squares as living in Cartesian space, rather than Euclidean space).  In this setting, a square corresponds to a single parameter (its side-length), and a rectangle corresponds to two parameters (the lengths of two adjacent sides).
Cardinality
Now comes the problem of understanding "more".  When working with finite sets, it easy relatively simple to understand what "more" means—we just count the number of objects, and use the ordering of the natural numbers.  Stated somewhat formally, a finite set $A$ contains "more" elements than a finite set $B$ if

*

*there is a bijection between $A$ and $\{1,2,\dotsc,m\} \subseteq \mathbb{N}$,


*there is a bijection between $B$ and $\{1,2,\dotsc,n\} \subseteq \mathbb{N}$, and


*$m > n$.
This notion of "more" corresponds to cardinality.  By definition, two sets have the same cardinality if there is a bijection between them.  For example, $\mathbb{N}$ and $\mathbb{Z}$ have the same cardinality, because the map
$$ n \mapsto \begin{cases}
k-1 & \text{if $n = 2k$ is even, and} \\
-k & \text{if $n=2k+1$ is odd}
\end{cases}
$$
is bijective (here $\mathbb{N} = \{1,2,3,\dotsc\}$; the bijectivity of the map can be verified).  With respect to this notion of "more", there are not more squares than rectangles.  Let $\mathscr{S}$ denote the set of all squares, and let $\mathscr{R}$ denote the set of all rectangles.  As noted above, every square is entirely described by a single positive real number.  It is also not too hard to see that every positive real number corresponds to a square.  Thus there is a bijection
$$a : \mathscr{S} \to (0,\infty).$$
Similarly, there is a bijection
$$b : \mathscr{R} \to (0,\infty)^2.$$
Technically, the rectangle corresponding to $(a,b)$ is the same as the rectangle corresponding to $(b,a)$, as these two rectangles are congruent.  However, this distinction is inessential throughout the following discussion, and leaving things as described above avoids some annoying details.  Finally, there is a bijection
$$c : (0,\infty) \to (0,\infty)^2;$$
I am not going to describe this bijection explicitly, but it can be shown that if $A$ is an infinite set, then the cardinality of $A$ is the same as the cardinality of $A\times A$.  By composing these bijections, we obtain a bijection
$$b^{-1} \circ c \circ a : \mathscr{S} \to \mathscr{R}.$$
This is all very pedantic, but it shows that the set of squares is in one-to-one correspondence with the set of rectangles, thus these sets have the same cardinality.  In this sense, there are not "more" rectangles than squares, nor "more" squares than rectangles.
Measure
Another way to quantify "more" is in terms of measures.  I am not going to go into tremendous detail on the underlying theory, but the essential idea is that we can generate consistent ways of measuring the size of sets (even infinite sets).  This theory is usually used in conjunction with notions of integration, but it makes sense here, too.
The applicable measure here is the Lebesgue measure on $\mathbb{R}^2$.  Roughly speaking the measure of subset of $\mathbb{R}^2$ is obtained by approximating that set with sets of the form $(a_1, b_1) \times (a_2, b_2)$, where $(a_j, b_j)$ denotes an open interval with endpoints $a_1 < b_1$.  Such a set (called a measurable rectangle; this is a distinct notion of rectangle) has measure
$$ (b_1 - a_1) (b_2 - a_2). $$
Given any set $E \subseteq \mathbb{R}^2$, we can cover that set with measurable rectangles, then add up the measures of those measurable rectangles to get an approximation of the measure of $E$.  The measure of $E$ is the infimum (e.g. the minimum) of all possible approximations.  For a more rigorous discussion, see (for example) the Wikipedia article on Lebesgue measure.
As noted above, the set of rectangles is in one-to-one correspondence with the set $Q_I := (0,\infty)^2$, which is the first quadrant in $\mathbb{R}^2$.  The set of squares, seen as a subset of the set of rectangles, is in one-to-one correspondence with the set
$$
\Delta = \{ (x,x) \mid x \in (0,\infty) \} \subseteq Q_I.
$$
Here, $\Delta$, the capital Greek letter Delta, denotes the "diagonal".  The set $Q_I$ has infinite Lebesgue measure:  one property of the Lebesgue measure is that if $A \subseteq B$, then $m(A) \le m(B)$, where $m$ denotes the measure.  But
$$ (0,b)^2 \subseteq Q_I\  \forall b\in (0,\infty)
\qquad\text{and}\qquad
m\bigl( (0,b)^2 \bigr) = b^2. $$
Hence $m(Q_I) > b^2$ for any $b\in(0,\infty)$, which implies that this measure is infinite.  On the other hand, it can be shown that
$$ m(\Delta) = 0. $$
A rigorous proof is a bit tedious, but the basic idea is that for any $\varepsilon > 0$, we can cover $\Delta$ by a sequence of measurable rectangles with rational vertices such that the total measure of all of those rectangles is smaller than $\varepsilon$.
Therefore
$$ 0 = m(\Delta) < m(Q_I) = \infty. $$
This implies that the set of squares (which corresponds to $\Delta$) is much, much smaller than the set of rectangles (which corresponds to $Q_I$).  In this sense, there are "more" rectangles than squares.  Or, perhaps, this is better stated as "the set of rectangles is bigger than the set of squares."
Dimension
As Gerry Myerson notes, there is also a dimensional argument.  Given that dimension theory is very near and dear to my heart, I'd like to expand on this a bit.  The notion of dimension which is relevant here is, more or less, the dimension of a vector space.  Very roughly speaking, a vector space $V$ can be described in terms of a basis.  A basis is a collection of vectors $\{v_j\}_{j=1}^{n}$ such that

*

*the vectors $\{v_j\}$ span $V$, in the sense that if $u \in V$, then there is some set of scalars $\{a_j\}$ such that $u = \sum a_j v_j$, and


*the vectors $\{v_j\}$ are linearly independent, in the sense that if $\sum a_j v_j = 0$, then $a_j = 0$ for all $j$.
It turns out that a vector space may have many bases ("bases" is the plural of "basis"), but all of these bases will have the same cardinality.  Thus a basis can be thought of as the number of parameters which you need in order to specify an element of the vector space:  each vector in $u \in V$ may be uniquely specified by listing the coefficients $\{a_j\}$ such that
$$ u = \sum_{j=1}^{n} a_j v_j. $$
The dimension of a space is the size of any basis (since they are all the same size, this notion is well-defined).
In the context of squares and rectangles, these sets are not quite vector spaces—it isn't quite clear how we "add" squares, for example, and side lengths must be positive, so there are no "negative" vectors, which is a problem.  However, if we think in terms of the number of parameters required to specify a square or rectangle, a notion of dimension makes sense.  Specifically, we think of each parameter as corresponding to some kind of "basis" object.
It takes only one parameter to specify a square (a single side length), whereas it takes two parameters to specify a rectangle (two side lengths).  Thus the space of squares is 1-dimensional, while the space of rectangles is 2-dimensional.  In this sense, the space of rectangles is larger than the space of squares, and so there are "more" rectangles.
