=== 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: :$(a): \quad \displaystyle n = \sum_{j \mathop = 0}^k s_j b^j$ :$(b): \quad \displaystyle \forall j \in \closedint 0 k: s_j \in \N_b$ :$(c): \quad s_k \ne 0$ First it is shown that: :$k \le m - 1$ {AimForCont}}: :$m - 1 < k$ Then: {{begin-eqn}} {{eqn | l = m       | o = \le       | r = k       | c = }} {{eqn | lo = \leadsto       | l = b^m       | o = \le       | r = b^k       | c = Power Function on Integer between Zero and One is Strictly Decreasing }} {{eqn | o = \le       | r = s_k b^k       | c = as $s_k > 0$ }} {{eqn | o = \le       | r = \displaystyle \sum_{j \mathop = 0}^k s_j b^j       | c = as $b \ge 0$ and $\forall j \in \closedint 0 {k - 1} : s_j \ge 0$ }} {{eqn | r = n       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^{m - 1} r_j b^j       | r = \sum_{j \mathop = 0}^k r_j b^j + \sum_{j \mathop = k + 1}^{m - 1} r_j b^j }} {{eqn | r = \sum_{j \mathop = 0}^k s_j b^j + \sum_{j \mathop = k + 1}^{m - 1} 0 \cdot b^j       | c = Definition of the Sequence $\sequence {r_j}_{0 \mathop \le j \mathop \le m - 1}$ }} {{eqn | r = \sum_{j \mathop = 0}^k s_j b^j }} {{eqn | r = n }} {{end-eqn}} 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: :$(1): \quad \displaystyle n = \sum_{j \mathop = 0}^{m - 1} r'_j b^j$ :$(2): \quad \displaystyle \forall j \in \closedint 0 {m - 1}: r'_j \in \N_b$ 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: :$(1): \quad \displaystyle n = \sum_{j \mathop = 0}^{k'} r'_j b^j$ :$(2): \quad \displaystyle \forall j \in \closedint 0 {k'}: r'_j \in \N_b$ :$(3): \quad r_{k'} \ne 0$ and :$(4): \quad \displaystyle \forall j \in \closedint {k + 1} {m - 1}: r'_j = 0$ 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. {{qed|lemma}} === 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: :$\sum_{j \mathop = 0}^{m - 1} r_j b^j = \sum_{j \mathop = 0}^{m - 1} 0 \cdot b^j = 0$ This shows that $\sequence {r_j}_{0 \mathop \le j \mathop \le m - 1}$ is a sequence such that: :$(1): \quad \displaystyle 0 = \sum_{j \mathop = 0}^{m - 1} r_j b^j$ :$(2): \quad \displaystyle \forall j \in \closedint 0 {m - 1}: r_j \in \N_b$ Let $\sequence {r'_j}_{0 \mathop \le j \mathop \le m - 1}$ be a sequence such that: :$(1): \quad \displaystyle 0 = \sum_{j \mathop = 0}^{m - 1} r'_j b^j$ :$(2): \quad \displaystyle \forall j \in \closedint 0 {m - 1}: r'_j \in \N_b$ Suppose for some $i \in \closedint 0 {m - 1}$, $r'_i > 0$. Then: :$\sum_{j \mathop = 0}^{m - 1} r'_j b^j \ge r'_i b^i > 0$. 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: : $(1): \quad \displaystyle 0 = \sum_{j \mathop = 0}^{m - 1} r_j b^j$ : $(2): \quad \displaystyle \forall j \in \closedint 0 {m - 1}: r_j \in \N_b$ {{qed}} Category:Number Bases 0b1nrl13pdfqe9umxl8r967fb5h0wm2