Basis Representation Theorem

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

For every $$n \in \Z_+$$, there exists one and only one sequence $$\left \langle {r_k} \right \rangle_{0 \le k \le m}$$ such that:


 * 1) $$n = \sum_{j = 0}^k r_j b^j$$;
 * 2) $$\forall j \in \left[{0 \, . \, . \, k}\right]: r_j \in \N_b$$;
 * 3) $$r_k \ne 0$$.

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 $$s_b \left({n}\right)$$ be the number of ways of representing $$n$$ to the base $$b$$.

We need to show that $$s_b \left({n}\right) = 1$$ 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:

$$ $$ $$

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

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 $$s_b \left({n}\right) \le s_b \left({n - 1}\right)$$.

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

So this inequality implies the following:


 * $$\forall m, n: s_b \left({m}\right) \le s_b \left({m - 1}\right) \le \ldots \le s_b \left({n + 1}\right) \le s_b \left({n}\right)$$

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


 * $$1 \le s_b \left({b^n}\right) \le s_b \left({n}\right) \le s_b \left({1}\right) = 1$$

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

So $$s_b \left({n}\right) = 1$$ and the theorem has been proved.

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


 * $$n = \sum_{j = 0}^k {r_j b^j}: r_0, r_1, \ldots, r_k \in \left\{{0, 1, \ldots, b-1}\right\}$$

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


 * $$\left[{r_m r_{m-1} \ldots r_2 r_1 r_0}\right]_b$$