# Between two Real Numbers exists Rational Number/Proof 1

## Theorem

Let $a, b \in \R$ be real numbers such that $a < b$.

Then:

$\exists r \in \Q: a < r < b$

## Proof

Suppose that $a \ge 0$.

As $a < b$ it follows that $a \ne b$ and so $b - a \ne 0$.

Thus:

$\dfrac 1 {b - a} \in \R$

By the Archimedean Principle:

$\exists n \in \N: n > \dfrac 1 {b - a}$

Let $M := \set {x \in \N: \dfrac x n > a}$.

By the Well-Ordering Principle, there exists $m \in \N$ such that $m$ is the smallest element of $M$.

That is:

$m > a n$

and, by definition of smallest element:

$m - 1 \le a n$

As $n > \dfrac 1 {b - a}$, it follows from Ordering of Reciprocals that:

$\dfrac 1 n < b - a$

Thus:

 $\displaystyle m - 1$ $\le$ $\displaystyle a n$ $\displaystyle \leadsto \ \$ $\displaystyle m$ $\le$ $\displaystyle a n + 1$ $\displaystyle \leadsto \ \$ $\displaystyle \frac m n$ $\le$ $\displaystyle a + \frac 1 n$ $\displaystyle$ $<$ $\displaystyle a + \paren {b - a}$ $\displaystyle$ $=$ $\displaystyle b$

Thus we have shown that $a < \dfrac m n < b$.

That is:

$\exists r \in \Q: a < r < b$

such that $r = \dfrac m n$.

Now suppose $a < 0$.

If $b > 0$ then $0 = r$ is a rational number such that $a < r < b$.

Otherwise we have $a < b \le 0$.

Then $0 \le -b < -a$ and there exists $r \in \Q$ such that:

$-b < r < -a$

where $r$ can be found as above.

That is:

$a < -r < b$

All cases have been covered, and the result follows.

$\blacksquare$

## Historical Note

When this proof was first published in 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces, there was a mistake in it:

This was corrected in the second printing, along with an apologetic note:

Preface to reprinted edition
I am grateful to all who have pointed out errors in the first printing
(even to those who mentioned that the proof of Corollary 1.1.7
purported to establish the existence of a positive rational number
between any two real numbers). In particular ...