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?

  • $\begingroup$ If $x$ was rational, the same proof would work for rationals. So all you really need to do is find one rational $z$ between $x$ and $y$, and use the same proof with $x$ replaced by $z$. Now, what definition of the reals are you using? $\endgroup$ Aug 31, 2012 at 2:09
  • $\begingroup$ I'm using the axioms for real numbers presented on Apostol's book on Calculus. $\endgroup$
    – Thiago
    Aug 31, 2012 at 2:17
  • $\begingroup$ Alternatively, prove that if $y = x + \delta$ with $\delta > 0$ and $m$ and $n$ are positive integers with $m/n < \delta$, then there are at least $m-1$ integers $k$ with $x < k/n < y$. $\endgroup$ Aug 31, 2012 at 2:21

1 Answer 1


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 $y$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .