Existence of Base-N Representation

Theorem
Given a number $$x \in \left[{0 \,. \, . \ 1}\right) \ $$, there exists a representation of that number in a base-$p$ positional system.

Specifically, there exists a sequence $$\left \langle{a_n}\right \rangle$$ such that:


 * $$0 \le a_n < p \ $$, and


 * $$\sum_{n=1}^\infty \frac{a_n} {p^n} \ $$ converges to $x \ $.

Unless $$\left \langle{a_n}\right \rangle$$ terminates (i.e. $$a_n = 0 \ $$ for all sufficiently large $$n \ $$), then this representation is unique.

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

Existence of Representation
Define $$a_j = \left \lfloor{\left({ x-\sum_{i=1}^{j-1} \frac{a_i}{p^i} }\right) p^j} \right \rfloor \ $$, where we accept the abuse of notation $$\sum_{i=1}^0 a_ip^{-i} =0 \ $$. 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:
 * $$a_n = \left \lfloor{\left({x - \sum_{i=1}^{n-1} \frac{a_i}{p^i}}\right) p^n}\right \rfloor \ge p \ $$


 * But then we can pull out the final term of the sum and divide by $$p$$ to get:


 * $$\left({x - \sum_{i=1}^{n-2} \frac{a_i}{p^i}}\right) p^{n-1} \ge 1 + a_{n-1} \ $$


 * This left-hand side is of course just:


 * $$a_{n-1} + \text{something in} \ \left[{0 \, . \, . \ 1}\right) \ge 1 + a_{n-1} \ $$


 * which is impossible.

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

Since $$\forall i \in \N: a_i, p^{-i} > 0 \ $$, this series 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:
 * $$s_1 = a_1 p^{-1} = \left \lfloor{x p}\right \rfloor p^{-1} \le x p p^{-1} = x \ $$


 * Suppose we have $$s_j<x \ $$ for some $$j$$.
 * By definition:
 * $$s_{j+1} - s_j = a_{j+1} p^{-1-j} \ $$


 * But:
 * $$a_{j+1}p^{-1-j} = \left \lfloor{ \left({x - s_j}\right) p^{1+j}}\right \rfloor p^{-1-j} \le (x-s_j) $$


 * 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:
 * $$a_{j+1} p^{-1-j} = \left \lfloor{\left({x-s_j}\right) p^{1+j}}\right \rfloor p^{-1-j} = 0 \implies s_{j+1} = s_j = x \ $$


 * This completes the induction proof.

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

Observe that in the sum $$s_{k-1} + a_k p^{-k} = s_k \ $$, we have defined $$a_k = \left \lfloor{\left({x - \sum_{i=1}^{k-1} \frac{a_i}{p^i}}\right) p^k}\right \rfloor\ $$ 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 $$\left \langle {s_k}\right\rangle \ $$ is increasing, bounded above by $$x \ $$, and comes arbitrarily close to $$x$$, we have $$\left \langle{s_n}\right \rangle \to x \ $$.

Uniqueness of Representation
Let $$\left \langle {a_n}\right \rangle \ $$ be the sequence defined in the definition of the theorem.

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

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 $$\left \langle {b_n}\right \rangle \ $$ which does not agree with $$\left \langle {a_n}\right \rangle \ $$.

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

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 $$\left \langle {t_n}\right \rangle \ $$ 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: $$\exists N \in \N : \left({\forall n \ge N: \ a_n = 0}\right) \iff \left({x = q p^{1-N}}\right) \ $$


 * Proof:
 * ($$\implies$$)


 * Suppose $$\exists N : \forall n \ge N: a_n = 0 \ $$.


 * Then $$x = \sum_{n=1}^\infty a_n p^{-n} = \sum_{n=1}^{N-1} a_n p^{-n}$$.


 * 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 $$q \in \N \ $$.


 * ($$\ \Longleftarrow \ $$)
 * Suppose $$x = q p^{1-N}$$.


 * Observe that since $$p^{1-N} \backslash s_{N-1} \ $$ (where $$\backslash$$ indicates divides), we must have $$s_{N-1} = x \ $$.


 * Since $$\left \langle{s_n}\right \rangle \to x \ $$ and is strictly increasing, we must have all successive terms equal to zero.

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 $$\left \langle{s_n}\right \rangle = \sum_{n=1}^\infty \frac{a_n} {p^n} \ $$ as defined above;
 * another series we describe now.

Consider the sequence $$\left \langle{a_n}\right \rangle \ $$ 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

$$ $$ $$ $$ $$ $$ $$ $$ $$

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 $$p - 1 \ $$.

Also see

 * Basis Expansion for how this applies to the representation of a real number


 * Basis Representation Theorem for an equivalent proof for integers.