# Existence of Base-N Representation

## Theorem

Given a number $x \in \hointr 0 1$, there exists a representation of that number in a base-$p$ positional system.

Specifically, there exists a sequence $\sequence {a_n}$ such that:

$0 \le a_n < p$, and
$\displaystyle \sum_{n \mathop = 1}^\infty \frac {a_n} {p^n}$ converges to $x$.

Unless $\sequence {a_n}$ terminates (that is $a_n = 0$ for all sufficiently large $n$), then this representation is unique.

If $\sequence {a_n}$ does terminate, then there is exactly one other sequence which satisfies the criteria of the theorem.

## Proof

### Existence of Representation

Define:

$\displaystyle a_j = \floor {\paren {x - \sum_{i \mathop = 1}^{j - 1} \frac {a_i} {p^i} } p^j}$

where we accept the vacuous summation $\displaystyle \sum_{i \mathop = 1}^0 a_i p^{-i} = 0$.

This recursive definition allows for all $a_n$ to be computed.

Lemma: This will always be less than $p$.
Proof: Suppose to the contrary $\exists n$ such that $a_n\geq p$.
Then:
$\displaystyle a_n = \floor {\paren {x - \sum_{i \mathop = 1}^{n - 1} \frac {a_i} {p^i} } p^n} \ge p$
But then we can pull out the final term of the sum and divide by $p$ to get:
$\displaystyle \paren {x - \sum_{i \mathop = 1}^{n - 2} \frac {a_i} {p^i} } p^{n - 1} \ge 1 + a_{n - 1}$
This left-hand side is of course just:
$a_{n - 1} + \text{something in } \hointr 0 1 \ge 1 + a_{n - 1}$
which is impossible.

$\Box$

Define $\displaystyle s_n = \sum_{i \mathop = 1}^n a_i p^{-i}$.

Since $\forall i \in \N: a_i, p^{-i} > 0$, this sequence of partial sums is increasing.

It is also bounded above by $x$ by construction: at every point in the series, we add precisely as many $p^{-n - 1}$ as will fit in $x-s_n$ without going over $x$:

Lemma: $\forall n \in \N: s_n \le x$
Proof: We have:
$s_1 = a_1 p^{-1} = \floor {x p} p^{-1} \le x p p^{-1} = x$
Suppose we have $s_j<x$ for some $j$.
By definition:
$s_{j + 1} - s_j = a_{j + 1} p^{-1 - j}$
But:
$a_{j + 1}p^{-1 - j} = \floor {\paren {x - s_j} p^{1 + j} } p^{-1 - j} \le \paren {x - s_j}$
So:
$s_{j + 1} - s_j \le x - s_j \implies s_{j + 1} \le x$
Now suppose we have instead $s_j = x$.
Again we have:
$s_{j + 1} - s_j = a_{j + 1} p^{-1 - j}$
But now:
$a_{j + 1} p^{-1 - j} = \floor {\paren {x - s_j} p^{1 + j} } p^{-1 - j} = 0 \implies s_{j + 1} = s_j = x$
This completes the induction proof.

$\Box$

It remains to be shown this series converges to $x$.

Observe that in the sum $\displaystyle s_{k - 1} + a_k p^{-k} = s_k$, we have defined $\displaystyle a_k = \floor {\paren {x - \sum_{i \mathop = 1}^{k - 1} \frac {a_i} {p^i} } p^k}$ to count precisely how many $p^{-k}$ will fit in $x-s_{k - 1}$.

We could never have $x - s_k \ge p^{-k}$ because that would mean $a_k$ had undercounted by $1$.

Therefore:

$x - s_k < p^{-k}$.

Let $\epsilon >0$.

Then set $z = -\log_p \epsilon$.

Then:

$N > z \implies x - s_N < p^{-N} < p^{\log_p} \epsilon = \epsilon$

Since $\sequence {s_k}$ is increasing, bounded above by $x$, and comes arbitrarily close to $x$, we have $\sequence {s_n} \to x$.

$\Box$

### Uniqueness of Representation

Let $\sequence {a_n}$ be the sequence defined in the definition of the theorem.

Let $\sequence {b_n}$ be some sequence of integers $0 \le b_n < p$ such that $\sequence {t_n} \to x$ where $\displaystyle t_n = \sum_{i \mathop = 1}^n b_i p^{-i}$.

We wish to show that $a_n = b_n \forall n$, unless $x = q p^{-k}$ for some $k \in \N$.

Assume to the contrary that there are terms which do not agree and let $b_m$ be the first term of $\sequence {b_n}$ which does not agree with $\sequence {a_n}$.

Then $b_m > a_m \lor b_m < a_m$.

Let us consider the first case.

We know that $a_m$ counts precisely how many $p^{-m}$ can be added to $s_{m-1}$ without exceeding $x$.

So we can be sure that if $b_m > a_m$, then $s_{m- 1 } + b_m p^{-m} = t_{m - 1} + b_m p^{-m} = t_m > x$.

Since $\sequence {t_n}$ is always increasing, it can never converge to $x$.

Now consider the second case, $b_m < a_m$.

First, we will need a lemma:

Lemma: $\exists N \in \N : \paren {\forall n \ge N: a_n = 0} \iff \paren {x = q p^{1 - N} }$
Proof:
($\implies$)
Suppose $\exists N : \forall n \ge N: a_n = 0$.
Then:
$\displaystyle x = \sum_{n \mathop = 1}^\infty a_n p^{-n} = \sum_{n \mathop = 1}^{N - 1} a_n p^{-n}$
But:
$a_n p^{-n} = a_n p^{N - 1 - n} p^{1 - N}$
Since $x$ is a sum of these terms of $p^{N - 1}$, we must have $x = q p^{N - 1}$ for some $q \in \N$.

($\impliedby$)
Suppose $x = q p^{1 - N}$.
Observe that since $p^{1 - N} \divides s_{N - 1}$ (where $\divides$ indicates divisibility), we must have $s_{N - 1} = x$.
Since $\sequence {s_n} \to x$ and is strictly increasing, we must have all successive terms equal to zero.

$\Box$

Now suppose that $x = q p^{-k}$ for some $k \in \N$.

We wish to show that there are only two series which converge to $x$:

the series $\displaystyle \sequence {s_n} = \sum_{n \mathop = 1}^\infty \frac {a_n} {p^n}$ as defined above
another series we describe now.

Consider the sequence $\sequence {a_n}$ when $x = q p^{-k}$.

Now we define:

$b_n = \begin {cases} a_n & : n < k \\ a_n - 1 & : n = k \\ p - 1 & : n > k \end{cases}$

Then we see that:

 $\ds \sum_{j \mathop = 1}^\infty b_j p^{-j}$ $=$ $\ds \paren {\sum_{j \mathop = 1}^{k - 1} b_j p^{-j} } + \paren {a_k - 1} p^{-k} + \sum_{j \mathop = k + 1}^\infty \paren {p - 1} p^{-j}$ $\ds$ $=$ $\ds \paren {\sum_{j \mathop = 1}^{k - 1} a_j p^{-j} } + a_k p^{-k} - p^{-k} + \sum_{j \mathop = k + 1}^\infty \paren {p - 1} p^{-j}$ $\ds$ $=$ $\ds \paren {\sum_{j \mathop = 1}^k a_j p^{-j} } - p^{-k} + \sum_{j \mathop = k + 1}^\infty \paren {p^{1 - j} - p^{-j} }$ $\ds$ $=$ $\ds x - p^{-k} + \sum_{j \mathop = k + 1}^\infty \paren {p^{1 - j} - p^{-j} }$ $\ds$ $=$ $\ds x - p^{-k} + p^{-k} \sum_{j \mathop \ge 0} \paren {p^{-j} } - p^{-k - 1} \sum_{j \mathop \ge 0} \paren {p^{-j} }$ $\ds$ $=$ $\ds x - p^{-k} + p^{-k} \paren {\frac 1 {1 - p^{-1} } } - p^{-k - 1} \paren {\frac 1 {1 - p^{-1} } }$ Sum of Infinite Geometric Sequence $\ds$ $=$ $\ds x - p^{-k} + \frac {p^{-k} - p^{-k - 1} } {1 - p^{-1} }$ $\ds$ $=$ $\ds x - p^{-k} + p^{-k} \frac {1 - p^{-1} } {1 - p^{-1} }$ $\ds$ $=$ $\ds x$

So, this series converges to $x$ as well.

Let us suppose, finally, that $x \ne q p^{-k}$ for any $k \in \N$.

We have already shown that if the first differing term of another series $b_n$ is greater than the corresponding term $a_n$, the sum series cannot converge to $x$.

Now we examine the case $b_m < a_m$ at the first differing term.

As we saw above, if the first term to differ is only one less, ie, $b_m = a_m - 1$, then it is necessary for every other term afterwards to be increased from $0$ to $p - 1$ in order to make up for this deficit.

The remaining terms of course, cannot be increased more than this, or they would violate the condition that all terms be less than $p$.

Since in the case $x \ne q p^{-k}$, there are no infinite strings of zeroes, we cannot decrease any one term and increase the succeeding terms by $p - 1$.

$\blacksquare$