The set of rationals has the same cardinality as the set of integers The set of rationals $\mathbb{Q}$ has the same cardinality as
the set of integers $\mathbb{Z}$. True or false?
This was a question on an old exam for our class. The correct answer is true. However, I did some additional reading and came across Cantor's transfinite numbers. In the book I'm reading, it says that "there are more real numbers (which include rational and irrational numbers) than there are integers". So can it also be said that there are more rational numbers than integers? And so can we say that the above statement is false?
 A: Just because the integers are a proper subset of the rationals doesn't mean that the rationals have a higher cardinality than the integers.  Actually, there is a theorem that says that a set is infinite if and only if it has the same cardinality to a proper subset of itself (so your logic would only apply to a finite set).  
A set is countable, or has the same cardinality as the integers, if you can count the elements.  In other words, you can label each element by a unique positive integer.  We can see from the diagonals argument (see this image on Wikipedia for a good illustration) that this holds for that rational numbers.  Once you get the hang of it, you can see that a lot of sets that seem to be a lot bigger than the set of integers are in fact the same "size" (have the same cardinality) as the integers.  See Hilbert's Paradax of the Grand Hotel for a good example of this.
A: No. The statement is still true.
The cardinality of the natural number set is the same as the cardinality of the rational number set. In fact, this cardinality is the first transfinite number denoted by $\aleph_0$ i.e. $|\mathbb{N}| = |\mathbb{Q}| = \aleph_0$. By first I mean the "smallest" infinity.
The cardinality of the set of real numbers is typically denoted by $\mathfrak{c}$. We have $\mathfrak{c}  > \aleph_0$, since we can set up a bijection from $\mathbb{R}$ to the power set of the natural numbers and by Cantor's theorem, for any set $X$, we have $|X| < |2^{X}|$. So we have $|\mathbb{R}| = |2^{\mathbb{N}}| > |\mathbb{N}|$. So what this essentially says is that "there are more real numbers (which include rational and irrational numbers) than there are integers" in some sense.
The continuum hypothesis states that "there is no set whose cardinality is strictly between that of the natural numbers and that of the real numbers" which essentially means real numbers form the second "smallest" infinity.
A: You can say whatever you want. However, if you want to be correct, and you are talking about cardinalities, then you cannot say that there are more rationals than integers.
The reason we can say that there are more reals than rational numbers is because they actually have greater cardinality, not just because the set of reals properly contains the natural numbers. While we can find a bijection between $\mathbb{N}$ and $\mathbb{Q}$, no function from $\mathbb{N}$ to $\mathbb{R}$ is surjective (and, naturally, no function from $\mathbb{Q}$ to $\mathbb{R}$ can be surjective either). 
