\begin{definition}[Definition:Basis Expansion/Positive Real Numbers]
Let $x \in \R$ be a real number such that $x \ge 0$.
Let $b \in \N: b \ge 2$.
Let us define the recursive sequence:
:$\forall n \in \N: n \ge 1: \sequence {f_n} = \begin {cases}
b \paren {x - \floor x} & : n = 1 \\
b \paren {f_{n - 1} - \floor {f_{n - 1} } } & : n > 1
\end{cases}$
Then we define:
:$\forall n \in \N: n \ge 1: \sequence {d_n} = \floor {f_n}$
It follows from the method of construction and the definition of the floor function that:
:$\forall n: 0 \le f_n < b$ and hence $\forall n: 0 \le d_n \le b - 1$
:$\forall n: f_n = 0 \implies f_{n + 1} = 0$ and hence $d_{n + 1} = 0$.
Hence we can express $x = \floor x + \displaystyle \sum_{j \mathop \ge 1} \frac {d_j} {b^j}$ as:
:$\sqbrk {s \cdotp d_1 d_2 d_3 \ldots}_b$
where:
:$s = \floor x$
:it is not the case that there exists $m \in \N$ such that $d_M = b - 1$ for all $M \ge m$.
(That is, the sequence of digits does not end with an infinite sequence of $b - 1$.)
This is called the '''expansion of $x$ in base $b$'''.
The generic term for such an expansion is a '''basis expansion'''.
It follows from the Division Theorem that for a given $b$ and $x$ this expansion is unique.
\end{definition}