# Between two Real Numbers exists Rational Number

## Theorem

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

Then:

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

### Corollary: Real Numbers are Close Packed

- $\forall a, b \in \R: a < b \implies \paren {\exists c \in \R: a < c < b}$

## Proof 1

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.

$\blacksquare$

## Proof 2

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 $M := \set {x \in \Z: x > a n}$.

By Set of Integers Bounded Below has Smallest Element, there exists $m \in \Z$ 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 $r = \dfrac m n$.

$\blacksquare$