The set of rationals has the same cardinality as the set of integers

Solution 1:

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.

Solution 2:

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).