Zero Padded Basis Representation/Proof

Case 1
Let $n \in \Z_{> 0} : n < b^m$.

From Basis Representation Theorem there exists a sequence $\sequence {s_j}_{0 \mathop \le j \mathop \le k}$ such that:


 * $(a): \quad \ds n = \sum_{j \mathop = 0}^k s_j b^j$
 * $(b): \quad \forall j \in \closedint 0 k: s_j \in \N_b$
 * $(c): \quad s_k \ne 0$

First it is shown that:
 * $k \le m - 1$


 * $m - 1 < k$
 * $m - 1 < k$

Then:

This contradicts the premise:
 * $n < b^m$

It follows that:
 * $k \le m - 1$

Let $\sequence {r_j}_{0 \mathop \le j \mathop \le m - 1}$ be the sequence defined by:
 * $r_j = \begin {cases}

s_j & : \text{if } j \le k\\ 0 & : \text{if } k < j \le m - 1 \end {cases}$

Then:

To show that $\sequence {r_j}_{0 \mathop \le j \mathop \le m - 1}$ is unique, let $\sequence {r'_j}_{0 \mathop \le j \mathop \le m - 1}$ be a sequence such that:
 * $(1): \quad \ds n = \sum_{j \mathop = 0}^{m - 1} r'_j b^j$
 * $(2): \quad \forall j \in \closedint 0 {m - 1}: r'_j \in \N_b$

Since $n > 0$, then:
 * $\exists j \in \closedint 0 {m - 1} : r'_j > 0$

Let:
 * $k' = \max \set {j : j \in \closedint 0 {m - 1} \text{ and } r'_j > 0}$

Then:
 * $(1): \quad \ds n = \sum_{j \mathop = 0}^{k'} r'_j b^j$
 * $(2): \quad \ds \forall j \in \closedint 0 {k'}: r'_j \in \N_b$
 * $(3): \quad r_{k'} \ne 0$

and
 * $(4): \quad \ds \forall j \in \closedint {k + 1} {m - 1}: r'_j = 0$

By definition the sequence $\sequence {r_j}_{0 \mathop \le j \mathop \le k'}$ is a basis representation for $n$.

From Basis Representation Theorem the basis representation for $n$ is unique.

Then:
 * $\sequence {r'_j}_{0 \mathop \le j \mathop \le k'} = \sequence {s_j}_{0 \mathop \le j \mathop \le k} = \sequence {r_j}_{0 \mathop \le j \mathop \le k}$

That is:
 * $k' = k$

and
 * $\forall j \in \closedint 0 k :r'_j = r_j$

It follows that:
 * $\forall j \in \closedint {k + 1} {m - 1}: r'_j = 0 = r_j$

So:
 * $\forall j \in \closedint 0 {m - 1}: r'_j = r_j$

This establishes that the sequence $\sequence {r_j}_{0 \mathop \le j \mathop \le m - 1}$ is unique.

Case 2
Let $n = 0$.

Let $\sequence {r_j}_{0 \mathop \le j \mathop \le m - 1}$ be the sequence defined by:
 * $\forall j \in \closedint 0 {m - 1} : r_j = 0$.

Then:
 * $\ds \sum_{j \mathop = 0}^{m - 1} r_j b^j = \sum_{j \mathop = 0}^{m - 1} 0 \cdot b^j = 0$

This shows that $\sequence {r_j}_{0 \mathop \le j \mathop \le m - 1}$ is a sequence such that:
 * $(1): \quad \ds 0 = \sum_{j \mathop = 0}^{m - 1} r_j b^j$
 * $(2): \quad \forall j \in \closedint 0 {m - 1}: r_j \in \N_b$

Let $\sequence {r'_j}_{0 \mathop \le j \mathop \le m - 1}$ be a sequence such that:
 * $(1): \quad \ds 0 = \sum_{j \mathop = 0}^{m - 1} r'_j b^j$
 * $(2): \quad \forall j \in \closedint 0 {m - 1}: r'_j \in \N_b$

Suppose for some $i \in \closedint 0 {m - 1}$, $r'_i > 0$.

Then:
 * $\ds \sum_{j \mathop = 0}^{m - 1} r'_j b^j \ge r'_i b^i > 0$.

It follows that:
 * $\forall j \in \closedint 0 {m - 1} : r'_j = 0$.

This proves that $0$ has one and only one sequence $\sequence {r_j}_{0 \mathop \le j \mathop \le m - 1}$ such that:
 * $(1): \quad \ds 0 = \sum_{j \mathop = 0}^{m - 1} r_j b^j$
 * $(2): \quad \forall j \in \closedint 0 {m - 1}: r_j \in \N_b$