Between two Real Numbers exists Irrational Number

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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 a real number 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 $s$ 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

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