# Fractional and Non-Integer Number Bases

Recall that a number in base \( b\) is an expression

\[a_ka_{k-1}\ldots a_1a_0a_{-1}a_{-2}\ldots,\]

where the \( a_i \) are nonnegative integers less than \( b\). So, for instance, the number \( 25 = 2\cdot 3^2 + 2\cdot 3 + 1\cdot 3^0 \) is represented as \( 221\) in base \( 3 \).

Although most applications of bases restrict \( b \) to be a positive integer, it is worth considering what happens when \( b\) is, more generally, a positive real number.

Let \( b > 1 \) be a real number. The

\(b\)-expansionof a nonnegative real number \( x \) is a representation\[x = a_k b^k +a_{k-1}b^{k-1} + \cdots + a_0 + a_{-1}b^{-1} + \cdots,\]

where the \( a_i \) are nonnegative integers less than \( b \).

Note that \( b > 1 \) is required because there need to be some positive integers less than \( b\) (otherwise there are no nonzero choices for the \( a_i \)).

From this definition, it is straightforward to deduce the following fact:

For any \( b > 1\), every nonnegative real number has a \( b \)-expansion.

The algorithm to generate a \( b\)-expansion of a real number \( x \) is the familiar one: first let \( k \) be the integer satisfying \( b^k \le x < b^{k+1} \), and then let \( a_k =\left \lfloor\frac{ x}{b^k}\right\rfloor. \) Then \( a_k < b \) and \( x-a_kb^k < b^k \), so the process can be iterated: let \( x' = x-a_kb^k \) and let \( a_{k-1} = \left\lfloor \frac{x'}{b^{k-1}}\right \rfloor,\) and so on.

Let \( \phi \) be the golden ratio \( \frac{1+\sqrt{5}}2 \). Find a \(\phi\)-expansion of \( 6\).

First, \( \phi^3 \le 6 < \phi^4 \), so the expansion starts \( 6 = \phi^3+(6-\phi^3) \). Now \( 6-\phi^3 = 6-(2\phi+1) = 5-2\phi \approx 1.8 \), so the expansion starts \( 6 = \phi^3 + \phi + (5-3\phi) \). Since \( 5-3\phi \approx 0.15 \), this will need to be expressed using negative coefficients. In fact, \( \phi^{-2} = 2-\phi \) so \( \phi^{-4} = 4-4\phi+\phi^2 = 5-3\phi \), so the \( \phi\)-expansion of \( 6 \) is \( 6 = \phi^3+\phi+\phi^{-4} \). In base notation, this is

\[6_{10} = 1010.0001_{\phi}.\]

Note that this representation is not unique: for example, since \(\phi = 1 +\phi^{-1} \), it is also true that \( 6_{10} = 1001.1001_{\phi}. \) \(_\square\)

This illustrates a situation that is different for non-integer bases: finite (terminating) integer-base expansions are always unique \((\)but integer-base expansions are not unique, e.g. \( 1 = 0.9999\ldots \) in base \( 10 \), as discussed here\().\) However, finite base-\(b\) expansions may not be unique for algebraic integers \( b \).

The golden ratio \( \phi \) is the larger positive root to \( x^2 = x + 1 \).

We can calculate that

\[ \begin{array} { l l } \phi^0 = 1, & \\ \phi^1 = \phi, &\phi^{-1} = \phi - 1, \\ \phi^2 = \phi + 1, & \phi^{-2} = -\phi + 2, \\ \phi^3 = 2\phi + 1, & \phi^{-3} = 2\phi -3 . \\ \end{array} \]

This allows us to write numbers in base \(\phi \), where each place value is a non-negative integer less than \( \phi \). For example,

\[ \begin{array} { l l l l l l }

1 & = & 1 & = & 1 _\phi \\
2 & = & \phi + (-\phi + 2) & = & 10.01_\phi \\

3 & = & (\phi+1) + (-\phi + 2) & = & 100.01 _\phi.
\end{array} \]

Give the finite decimal representation of 5 in base \( \phi \) that doesn't use a consecutive pair of 1's.

(You may assume the fact that a finite decimal representation with no consecutive pair of 1's is unique. This is known as the standard form for base \( \phi\).)

**Cite as:**Fractional and Non-Integer Number Bases.

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