# Between two Real Numbers exists Irrational Number

Jump to navigation
Jump to search

## Theorem

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

Then there exists an irrational number $\xi \in \R \setminus \Q$ such that:

- $a < \xi < b$

## Proof

From Number of Type Rational r plus s Root 2 is Irrational we have that real numbers of the form $r \sqrt 2$, where $r \ne 0$ is rational, is irrational.

From Between two Real Numbers exists Rational Number there exists a rational number $r$ such that:

- $\dfrac a {\sqrt 2} < r < \dfrac b {\sqrt 2}$

and so:

- $a < r \sqrt 2 < b$

giving us $\xi = r \sqrt 2$ as the irrational number we seek.

If $a < 0$ and $b > 0$, it may be that the $r$ that we find happens to be zero

Then we locate a second rational number $r$ such that:

- $0 < s < \dfrac b {\sqrt 2}$

which leads us to:

- $a < 0 < s \sqrt 2 < b$

and so in this case $\xi = s \sqrt 2$ is the irrational number we seek.

$\blacksquare$

## Sources

- 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $1$: Review of some real analysis: $\S 1.1$: Real Numbers: Corollary $1.1.7$: Remark - 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 3$: Natural Numbers: Exercise $\S 3.6 \ (6)$