I'm reading "Understanding Analysis" by Abbott, and I'm confused about the density of $Q$ in $R$ and how that ties to the cardinality of rational vs irrational numbers.
First, on page 20, Theorem 1.4.3 "Density of $Q$ in $R$" Abbot states:
For every two real numbers a and b
with a < b, there exists a rational number r satisfying a < r < b.
For which he provides a proof.
Later, on page 22, in the section titled "Countable and Uncountable Sets" he states:
Mentally, there is a temptation to think of $Q$ and $I$ as being intricately mixed together in equal proportions, but this turns out not to be the case…the irrational numbers far outnumber the rational numbers in making
up the real line.
My question is: how are these two statements not in direct contradiction? Given any closed interval of irrational numbers of cardinality $X$, $A$, shouldn't be the case that we would have corresponding set of $X-1$ rational numbers, $B$, where each rational in $B$ falls "between" two other irrationals in $A$?
If this is not the case, how do we have so many more irrationals than rationals while still satisfying our theorem that between every two reals there is a rational number?
I know there are other questions similar to this, but I haven't found an answer that explains this very well, and none that address this (perceived) contradiction.
Best Answer
That will certainly be true if you change the word "interval" to "set" and stipulate that $X$ is a finite integer.
Consider a finite set $A$ containing $X$ distinct irrational numbers and nothing else, where $X \in \mathbb Z$. Then you can arrange the members of $A$ in increasing sequence, that is, write $A = {a_i}, 1 \leq i \leq X$ such that $a_i > a_{i-1}$ when $i > 1$. And then you can insert $X - 1$ rational numbers in the "gaps" between the consecutive members of ${a_i}$.
The problem with this in the more general case is that there are more than a finite number of irrational numbers in any closed interval in $\mathbb R$. In fact, there are more than a countable number of them. You can't just go and insert a rational number between each consecutive pair of irrational numbers, because there is no such thing as a consecutive pair of irrational numbers in an interval. In fact, take any two irrational numbers $r, s$ in the interval; there will be an uncountably infinite number of irrational numbers between $r$ and $s$.
We do indeed have a rational number $q$ that falls between $r$ and $s$, in fact a countably infinite set of such numbers; but we also have an uncountably infinite set of irrational numbers that fall between $r$ and $s$. There is no way to organize these numbers into an increasing sequence of alternating irrational and rational numbers, like this:
$$ r_1 < q_1 < r_2 < q_2 < r_3 < \cdots < r_{X-1} < q_{X-1} < r_X, $$
so any counting argument based on imagining such a sequence is incorrect.