# Basis Representation Theorem

## Contents

## Theorem

Let $b \in \Z: b > 1$.

For every $n \in \Z_{> 0}$, there exists one and only one sequence $\sequence {r_k}_{0 \le k \le m}$ such that:

- $(1): \quad \displaystyle n = \sum_{j \mathop = 0}^k r_j b^j$
- $(2): \quad \displaystyle \forall j \in \closedint 0 k: r_j \in \N_b$
- $(3): \quad r_k \ne 0$

This unique sequence is called the **representation of $n$ to the base $b$**, or, informally, we can say **$n$ is (written) in base $b$**.

## Proof

Let $\map {s_b} n$ be the number of ways of representing $n$ to the base $b$.

We need to show that $\map {s_b} n = 1$ always.

Now, it is possible that some of the $r_i = 0$ in a particular representation.

So we may exclude these terms, and it won't affect the representation.

So, suppose:

- $n = r_k b^k + r_{k - 1} b^{k - 1} + \cdots + r_t b^t$

where $r_k \ne 0, r_t \ne 0$.

Then:

\(\displaystyle n - 1\) | \(=\) | \(\displaystyle r_k b^k + r_{k - 1} b^{k - 1} + \cdots + r_t b^t - 1\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle r_k b^k + r_{k - 1} b^{k - 1} + \cdots + \paren {r_t - 1} b^t + b^t - 1\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle r_k b^k + r_{k - 1} b^{k - 1} + \cdots + \paren {r_t - 1} b^t + \sum_{j \mathop = 0}^{t - 1} {\paren {b - 1} b^j}\) | $\quad$ Sum of Geometric Progression | $\quad$ |

from the identity $\displaystyle \sum_{j \mathop = 0}^{n - 1} x^j = {\frac {x^n - 1} {x - 1} }, x \ne 1$.

Note that we have already specified that $b > 1$.

So for each representation of $n$ to the base $b$, we can find a representation of $n-1$.

If $n$ has another representation to the base $b$, then the same procedure will generate a new representation of $n - 1$.

Thus:

- $\map {s_b} n \le \map {s_b} {n - 1}$

Note that this holds even if $n$ has no representation at all, because if this is the case, then $\map {s_b} n = 0 \le \map {s_b} {n - 1}$.

So this inequality implies the following:

- $\forall m, n: \map {s_b} m \le \map {s_b} {m - 1} \le \ldots \le \map {s_b} {n + 1} \le \map {s_b} n$

given that $m \ge n$.

From N less than M to the N and the fact that $b^n$ has at least one representation (itself), we see:

- $1 \le \map {s_b} {b^n} \le \map {s_b} n \le \map {s_b} 1 = 1$

The entries at either end of this inequality are $1$, so all the intermediate entries must also be $1$.

So $\map {s_b} n = 1$ and the theorem has been proved.

$\blacksquare$

## Comment

So, once we have chosen a base $b > 1$, we can express any positive integer $n$ uniquely as:

- $\displaystyle n = \sum_{j \mathop = 0}^k {r_j b^j}: r_0, r_1, \ldots, r_k \in \set {0, 1, \ldots, b - 1}$

Then we can write $\displaystyle n = \sum_{j \mathop = 0}^m {r_j b^j}$ as:

- $\sqbrk {r_m r_{m - 1} \ldots r_2 r_1 r_0}_b$

## Also see

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 24$: Theorem $24.2$ - 1971: George E. Andrews:
*Number Theory*... (previous) ... (next): $\S 1.2$: Theorem $1.3$ - 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $1$: Properties of the Natural Numbers: $\S 21 \alpha$