Q:

Is it correct to say that $\mathbb{R}$ has fewer elements than $\mathbb{C}$ if both are infinite?

My math teacher said that. I disagreed, but he said that I was wrong. But I'm not convinced - is it really right? Please notice that I'm not talking about $\mathbb{R}$ $⊂$ $\mathbb{C}$,  but $\mathbb{R}$ $<$ $\mathbb{C}$.

A:

It really depends on what one means by "fewer elements".
The usual notion that most mathematicians use for this is "cardinality". Two sets $X$ and $Y$ have the same cardinality if and only if there is a bijection between them (a function $f\colon X\to Y$ that "pairs them up", so that every element of $Y$ corresponds, under $f$, to one and only one element of $X$). In the case of finite sets, this corresponds to our usual notion of "size", in that two finite sets have the same "cardinality" if and only if they have the same number of elements.
But for infinite sets, the notion introduces many counterintuitive consequences: the set of all even integers has "the same cardinality" as the set of all integers, even though it is a proper subset of it. In fact, one way to define "infinite set" is to say that a set is infinite if and only if it has the same cardinality as a proper subset.
We say a set $X$ has "strictly smaller cardinality" than a set $Y$ if and only if there is a one-to-one function $f\colon X\to Y$, but there is no onto function $g\colon X\to Y$. In particular, the fact that there is a one-to-one but not onto surjection from $\mathbb{R}$ to $\mathbb{C}$ shows that the cardinality of $\mathbb{R}$ is less than or equal to that of $\mathbb{C}$, but it does not show that they are not equal (just because this particular function is not onto does not mean there are no onto functions).
In that sense, $\mathbb{R}$ and $\mathbb{C}$ both have "the same size" because they both have the same cardinality: there is in fact a bijection between $\mathbb{R}$ and $\mathbb{R}^2$, and of course there is a bijection between $\mathbb{R}^2$ and $\mathbb{C}$.
The arithmetic of cardinalities is interesting, and not without its dangers (the answers may depend on exactly what set-theoretic axioms you accept or not accept, for example). 
On the other hand, you can try to measure "size" in other ways. There is a natural notion of "size" for subsets of the complex numbers: their area, or at least their outer measure, when viewed as subsets of the plane; if you use that notion of size, then the real numbers have "size" $0$, while the complex numbers have infinite size, so the complex numbers are "larger".  Added. Or as t.b. suggests in the comments, you could try to compare them in terms of their dimension as an $\mathbb{R}$-vector space ($\dim_{\mathbb{R}}(\mathbb{R}) = 1\lt 2 = \dim_{\mathbb{R}}(\mathbb{C})$. Or any other of several ways in which you could try to compare their "size".
I would argue, however, that it would not be very felicitous to call these other ways of comparing their size, "the reals have fewer elements than the complex numbers", because "fewer elements" seems to refer to a "count" of elements, and that leads you straight back to the notion of cardinality.

A:

$\mathbb R$ is smaller than $\mathbb C$ ... ??? In the sense of cardinality, NO.  (As explained in other answers.)  In the sense of Hausdorff dimension, YES.  What sense did your teacher mean?   
I'm sorry, but we often find when a student reports what a teacher says, some details may get distorted or even omitted.  Unless the teacher comes here and makes a defense, we cannot know what is going on.  Or, even better, if the student talks to the teacher (what an idea!) and finds out what was intended.

A:

No, it is not correct. Both the real numbers $\mathbb{R}$ and the complex numbers $\mathbb{C}$ have the same cardinality, $\mathfrak{c}$, which is the cardinality of the continuum. See this wikipedia page.
As a general rule, if $A\subseteq B$ then you can say that $|A|\leq|B|$, but if $A\subset B$ (that's a strict subset relation) you cannot conclude that $|A| < |B|$.

