Basis Expansion of Irrational Number

Theorem

A basis expansion of an irrational number never terminates and does not recur.

Proof

• Suppose $x \in \R$ were to terminate in some number base $b$.

Then (using the notation of that definition):

$\exists k \in \N: f_k = 0$

and so we can express $x$ precisely as:

$x = \left[{s . d_1 d_2 d_3 \ldots d_{k-1}}\right]_b$

This means:

$\displaystyle x = s + \frac {d_1}{b} + \frac {d_2}{b^2} + \frac {d_3}{b^3} + \cdots + \frac {d_{k-1}}{b^{k-1}}$

This is the same as:

$\displaystyle x = \frac {s b^{k-1} + d_1 b^{k-2} + d_2 b^{k-3} + \cdots + d_{k-1}} {b^{k-1}}$

Both top and bottom are integers and so $x$ is a rational number.

• Now suppose $x \in \R$ were to recur in some number base $b$.