I think I got the right answer for the case regarding the real numbers. The problem is to prove for the rational numbers.
Here is my proof for real numbers:
Given an arbitrary real number $h>0$, Archimedean property of the real-number system tells that there is a natural number $n$ so that $n(y-x)>h$ (Because $(y-x)$ and $h$ are positive).
This implies that $x+\frac{h}{n}<y$.
Since $\frac{h}{n}>0$, $x<x+\frac{h}{n}$, which gives us $x<x+\frac{h}{n}<y$.
$h$ is a arbitrary number, and we can choose infinitely many numbers, because the set of real numbers is not bounded above. Therefore, we have infinite real numbers greater than $x$ and smaller than $y$.
Is this correct?
Now I'm struggling to solve this for rational numbers. Can someone give me any hints or insights?
Best Answer
If you were interested only in proving that there are infinitely many reals intermediate between $x$ and $y$, a simple proof is to first note that $x < \frac{x+y}{2} < y$ (which is easily proven using the axioms provided in Apostol's book), and then simply repeat this averaging process to find $x < \frac{x + \frac{x+y}{2}}{2} < \frac{x+y}{2} < \frac{\frac{x+y}{2} + y}{2} < y$. Continuing this we have infinitely many reals between $x$ and $y$. (To formally prove this you could set up some sequence like $a_1 = \frac{x+y}{2}$, $a_{n+1} = \frac{x+a_n}{2}$ and then prove by induction that all $a_n$ satisfy the inequality $x < a_n < y$).
If you required infinitely many rationals (which implies the infinitely many reals case anyway), proceed as follows. Find a natural $n$ such that $\frac{1}{n} < y-x$ (which is easy to find using the Archimedean property since this is equivalent to $n(y-x) > 1$), and then set $m$ to be the smallest natural number such that $\frac{m}{n} > x$ (here, you would need to formally prove from the given axioms that such an $m$ exists). The idea here is that as $m$ increases, $\frac{m}{n}$ increases in increments of $\frac{1}{n}$, which is smaller than $y-x$, the "gap" between $x$ and $y$, and so the first $\frac{m}{n}$ to exceed $x$ should lie in between $x$ and $y$. To see that this is true, by definition of $m$, $x < \frac{m}{n}$. Further, since $m$ is the smallest such natural number, $\frac{m-1}{n} < x$, which upon rearrangement gives $\frac{m}{n} < x + \frac{1}{n} < x + (y-x) = y$. And this is what was required, $x < \frac{m}{n} < y$.
EDIT: Sorry, I didn't finish of the case of the rationals. After having found a rational, $\frac{m}{n}$ between any two reals $x$ and $y$, taking $x$ and $\frac{m}{n}$ as two reals, we can find a rational between them, and similarly we find a rational between $\frac{m}{n}$ and $y$. Repeating this, we may construct an infinite sequence of rationals between $x$ and $y$.