# 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.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/fractional-number-base/