令
D
{\displaystyle D}
整環
⊂
C
{\displaystyle \subset \mathbb {C} }
|
⋅
|
{\displaystyle |\cdot |}
（阿基米德）絕對賦值 。
數
X
∈
D
{\displaystyle X\in D}
進位制 系統中可以表示為：
X
=
±
∑
ν
x
ν
ρ
ν
,
{\displaystyle X=\pm \sum _{\nu }^{}x_{\nu }\rho ^{\nu },}
其中
ρ
∈
D
{\displaystyle \rho \in D}
為底數 ，並滿足
|
ρ
|
>
1
{\displaystyle |\rho |>1}
ν
∈
Z
{\displaystyle \nu \in \mathbb {Z} }
是指數 ，代表各個位數，
x
ν
{\displaystyle x_{\nu }}
是進制中每個位數，來自有限的位數數碼集合
Z
⊂
D
{\displaystyle Z\subset D}
|
x
ν
|
<
|
ρ
|
.
{\displaystyle |x_{\nu }|<|\rho |.}
其勢
R
:=
|
Z
|
{\displaystyle R:=|Z|}
進位制 系統或編碼系統 是一對二元組：
⟨
ρ
,
Z
⟩
{\displaystyle \left\langle \rho ,Z\right\rangle }
包括了其底數
ρ
{\displaystyle \rho }
Z
{\displaystyle Z}
R
{\displaystyle R}
Z
R
:=
{
0
,
1
,
2
,
…
,
R
−
1
}
.
{\displaystyle Z_{R}:=\{0,1,2,\dotsc ,{R-1}\}.}
理想的進位制 系統或編碼系統 具有以下特性：
任何在環
D
{\displaystyle D}
Z
{\displaystyle \mathbb {Z} }
高斯整數
Z
[
i
]
{\displaystyle \mathbb {Z} [\mathrm {i} ]}
Z
[
−
1
+
i
7
2
]
{\displaystyle \mathbb {Z} \left[{\tfrac {-1+\mathrm {i} {\sqrt {7}}}{2}}\right]}
正負號 ±。
任何在分式環
K
:=
Quot
(
D
)
{\displaystyle K:=\operatorname {Quot} (D)}
完備化 （
|
⋅
|
{\displaystyle |\cdot |}
度量 的意義下）所得的
K
:=
R
{\displaystyle K:=\mathbb {R} }
K
:=
C
{\displaystyle K:=\mathbb {C} }
|
⋅
|
{\displaystyle |\cdot |}
ν
→
−
∞
{\displaystyle \nu \to -\infty }
X
{\displaystyle X}
測度 為0。後者要求集合
Z
{\displaystyle Z}
R
=
|
ρ
|
{\displaystyle R=|\rho |}
R
=
|
ρ
|
2
{\displaystyle R=|\rho |^{2}}
在這種表示法中，一般常見的標準十進制 表示為：
⟨
10
,
Z
10
⟩
,
{\displaystyle \left\langle 10,Z_{10}\right\rangle ,}
標準二進制 系統表示為：
⟨
2
,
Z
2
⟩
,
{\displaystyle \left\langle 2,Z_{2}\right\rangle ,}
負二進制系統表示為：
⟨
−
2
,
Z
2
⟩
,
{\displaystyle \left\langle -2,Z_{2}\right\rangle ,}
平衡三進位 系統表示為[ 2]
⟨
3
,
{
−
1
,
0
,
1
}
⟩
.
{\displaystyle \left\langle 3,\{-1,0,1\}\right\rangle .}
上述這幾個進位制 系統在
Z
{\displaystyle \mathbb {Z} }
R
{\displaystyle \mathbb {R} }
較廣為人知的複底數進位制 系統包括下列幾個進位制系統（其中
i
{\displaystyle \mathrm {i} }
虛數單位 ）：
⟨
R
,
Z
R
⟩
{\displaystyle \left\langle {\sqrt {R}},Z_{R}\right\rangle }
⟨
±
i
2
,
Z
2
⟩
{\displaystyle \left\langle \pm \mathrm {i} {\sqrt {2}},Z_{2}\right\rangle }
[ 1]
i
2
{\displaystyle \mathrm {i} {\sqrt {2}}}
⟨
±
2
i
,
Z
4
⟩
{\displaystyle \left\langle \pm 2\mathrm {i} ,Z_{4}\right\rangle }
[ 2] 2i進制 ，由高德納 於1995年提出。
⟨
2
e
±
π
2
i
=
±
i
2
,
Z
2
⟩
{\displaystyle \left\langle {\sqrt {2}}e^{\pm {\tfrac {\pi }{2}}\mathrm {i} }=\pm \mathrm {i} {\sqrt {2}},Z_{2}\right\rangle }
⟨
2
e
±
3
π
4
i
=
−
1
±
i
,
Z
2
⟩
{\displaystyle \left\langle {\sqrt {2}}e^{\pm {\tfrac {3\pi }{4}}\mathrm {i} }=-1\pm \mathrm {i} ,Z_{2}\right\rangle }
[ 3] [ 5] −1 ± i
⟨
R
e
i
φ
,
Z
R
⟩
{\displaystyle \left\langle {\sqrt {R}}e^{\mathrm {i} \varphi },Z_{R}\right\rangle }
φ
=
±
arccos
(
−
β
/
(
2
R
)
)
{\displaystyle \varphi =\pm \arccos {(-\beta /(2{\sqrt {R}}))}}
β
<
min
(
R
,
2
R
)
{\displaystyle \beta <\min(R,2{\sqrt {R}})}
β
{\displaystyle \beta _{}^{}}
R
{\displaystyle R}
[ 7]
β
=
1
{\displaystyle \beta =1}
R
=
2
{\displaystyle R=2}
⟨
−
1
+
i
7
2
,
Z
2
⟩
{\displaystyle \left\langle {\tfrac {-1+\mathrm {i} {\sqrt {7}}}{2}},Z_{2}\right\rangle }
−
1
+
i
7
2
{\displaystyle {\tfrac {-1+\mathrm {i} {\sqrt {7}}}{2}}}
⟨
2
e
π
3
i
,
A
4
:=
{
0
,
1
,
e
2
π
3
i
,
e
−
2
π
3
i
}
⟩
{\displaystyle \left\langle 2e^{{\tfrac {\pi }{3}}\mathrm {i} },A_{4}:=\left\{0,1,e^{{\tfrac {2\pi }{3}}\mathrm {i} },e^{-{\tfrac {2\pi }{3}}\mathrm {i} }\right\}\right\rangle }
[ 8]
⟨
−
R
,
A
R
2
⟩
{\displaystyle \left\langle -R,A_{R}^{2}\right\rangle }
A
R
2
{\displaystyle A_{R}^{2}}
r
ν
=
α
ν
1
+
α
ν
2
i
{\displaystyle r_{\nu }=\alpha _{\nu }^{1}+\alpha _{\nu }^{2}\mathrm {i} }
α
ν
∈
Z
R
{\displaystyle \alpha _{\nu }^{}\in Z_{R}}
⟨
−
2
,
{
0
,
1
,
i
,
1
+
i
}
⟩
{\displaystyle \left\langle -2,\{0,1,\mathrm {i} ,1+\mathrm {i} \}\right\rangle }
[ 8]
⟨
ρ
=
ρ
2
,
Z
2
⟩
{\displaystyle \left\langle \rho =\rho _{2},Z_{2}\right\rangle }
ρ
2
=
{
(
−
2
)
ν
2
if
ν
even,
(
−
2
)
ν
−
1
2
i
if
ν
odd.
{\displaystyle \rho _{2}={\begin{cases}(-2)^{\tfrac {\nu }{2}}&{\text{if }}\nu {\text{ even,}}\\(-2)^{\tfrac {\nu -1}{2}}\mathrm {i} &{\text{if }}\nu {\text{ odd.}}\end{cases}}}
[ 9]
若要將一高斯整數
z
{\displaystyle z}
b
{\displaystyle b}
底數 的進位制
⟨
b
,
Z
R
⟩
{\displaystyle \left\langle b,Z_{R}\right\rangle }
數 分成一個可被底數整除的高斯整數和一個位於位數數碼集合內的數，並將可被底數整除的高斯整數部分除以底數當作商，位於位數數碼集合內的數當作餘數，並用商數繼續計算，並重複以上步驟，直到商為零，一系列的餘數部分即為轉換完成的結果。[ 11] :41
z
=
q
1
b
+
a
0
{\displaystyle z_{\,}=q_{1}b+a_{0}}
q
1
=
q
2
b
+
a
1
{\displaystyle q_{1}=q_{2}b+a_{1}}
q
2
=
q
3
b
+
a
2
{\displaystyle q_{2}=q_{3}b+a_{2}}
⋮
{\displaystyle \quad \quad \vdots }
q
t
=
0
b
+
a
t
{\displaystyle q_{t}=0_{\,}b+a_{t}}
其中，
q
1
{\displaystyle q_{1}}
q
2
{\displaystyle q_{2}}
q
3
{\displaystyle q_{3}}
q
t
{\displaystyle q_{t}}
a
1
{\displaystyle a_{1}}
a
2
{\displaystyle a_{2}}
a
3
{\displaystyle a_{3}}
a
t
{\displaystyle a_{t}}
則
z
=
(
a
t
⋯
a
2
a
1
a
0
)
b
{\displaystyle z=\left(a_{t}\cdots a_{2}a_{1}a_{0}\right)_{b}}
以5+12i轉換成-2+i進制（
⟨
−
2
+
i
,
{
0
,
1
,
2
,
3
,
4
}
⟩
{\displaystyle \left\langle -2+\mathrm {i} ,\{0,1,2,3,4\}\right\rangle }
[ 11] :42
5
+
12
i
{\displaystyle 5+12\mathrm {i} }
=
{\displaystyle =}
(
2
−
5
i
)
(
−
2
+
i
)
+
4
{\displaystyle \left(2-5\mathrm {i} \right)\left(-2+\mathrm {i} \right)+4}
2
−
5
i
{\displaystyle 2-5\mathrm {i} }
=
{\displaystyle =}
(
−
1
+
2
i
)
(
−
2
+
i
)
+
2
{\displaystyle \left(-1+2\mathrm {i} \right)\left(-2+\mathrm {i} \right)+2}
−
1
+
2
i
{\displaystyle -1+2\mathrm {i} }
=
{\displaystyle =}
(
2
)
(
−
2
+
i
)
+
3
{\displaystyle \left(2\right)\left(-2+\mathrm {i} \right)+3}
2
{\displaystyle 2}
=
{\displaystyle =}
(
0
)
(
−
2
+
i
)
+
2
{\displaystyle \left(0\right)\left(-2+\mathrm {i} \right)+2}
故5+12i(10) 轉換成-2+i進制為2324(−2+i) 。
在−1+i 較常被討論的複底數進制是2i進制 和−1 ± i −1 + i −1 − i 高斯整數 。
−1 ± i [ 3] [ 4] [ 6]
−1±i
十進制
二進制
2i進制
−1+i −1−i
0
0
0
0
0
1
1
1
1
1
2
10
2
1100
1100
−1
−1
103
11101
11101
−2
−10
102
11100
11100
i i 10.2
11
111
2i 10i 10
1110100
100
3i 11i 20.2
1110111
110011
−i −i 0.2
111
11
−2i −10i 1030
100
1110100
−3i −11i 1030.2
110011
1110111
1+i 1+i 11.2
1110
111010
1−i 1−i 1.2
111010
1110
−1+i −1+i 113.2
10
110
−1−i −1−i 103.2
110
10
整數的捨入區域——即在這系統表達之下，共用整數部分的複數（非整數）集合
S
{\displaystyle S}
分形 ：twindragon。根據定義，集合
S
{\displaystyle S}
∑
k
≥
1
x
k
(
i
−
1
)
−
k
{\displaystyle \textstyle \sum _{k\geq 1}x_{k}(\mathrm {i} -1)^{-k}}
x
k
∈
Z
2
{\displaystyle x_{k}\in Z_{2}}
S
{\displaystyle S}
1
4
S
{\displaystyle {\tfrac {1}{4}}S}
S
{\displaystyle S}
1
2
S
{\displaystyle {\tfrac {1}{\sqrt {2}}}S}
(
i
−
1
)
S
=
S
∪
(
S
+
1
)
{\displaystyle (\mathrm {i} -1)S=S\cup (S+1)}
2
15
←
0.
00001100
¯
{\displaystyle {\tfrac {2}{15}}\gets 0.{\overline {00001100}}}
1
15
i
←
0.
00000011
¯
{\displaystyle {\tfrac {1}{15}}\mathrm {i} \gets 0.{\overline {00000011}}}
−
8
15
←
0.
11000000
¯
{\displaystyle -{\tfrac {8}{15}}\gets 0.{\overline {11000000}}}
−
4
15
i
←
0.
00110000
¯
{\displaystyle -{\tfrac {4}{15}}\mathrm {i} \gets 0.{\overline {00110000}}}
1 / 15 [ 2] :206 。
由此，複矩形
[
−
8
15
,
2
15
]
×
[
−
4
15
,
1
15
]
i
{\displaystyle [-{\tfrac {8}{15}},{\tfrac {2}{15}}]\times [-{\tfrac {4}{15}},{\tfrac {1}{15}}]\mathrm {i} }
透過單射
∑
k
≥
1
x
k
(
i
−
1
)
−
k
↦
∑
k
≥
1
x
k
b
−
k
{\displaystyle \textstyle \sum _{k\geq 1}x_{k}(\mathrm {i} -1)^{-k}\mapsto \sum _{k\geq 1}x_{k}b^{-k}}
映入實數區間
[
0
,
1
)
{\displaystyle [0,1)}
b
>
2
{\displaystyle b>2}
[ 註 2]
此外，還有兩個映射
Z
2
N
→
S
(
x
k
)
k
∈
N
↦
∑
k
≥
1
x
k
(
i
−
1
)
−
k
{\displaystyle {\begin{array}{lll}Z_{2}^{\mathbb {N} }&\to &S\\\left(x_{k}\right)_{k\in \mathbb {N} }&\mapsto &\sum _{k\geq 1}x_{k}(\mathrm {i} -1)^{-k}\end{array}}}
和
Z
2
N
→
[
0
,
1
)
(
x
k
)
k
∈
N
↦
∑
k
≥
1
x
k
2
−
k
{\displaystyle {\begin{array}{lll}Z_{2}^{\mathbb {N} }&\to &[0,1)\\\left(x_{k}\right)_{k\in \mathbb {N} }&\mapsto &\sum _{k\geq 1}x_{k}2^{-k}\end{array}}}
兩者皆滿射 ，也就是產生了一個滿射（空間填充）的映射
[
0
,
1
)
→
S
{\displaystyle [0,1)\qquad \to \qquad S}
然而，其並不連續，因此不是空間填充曲線。但是一個類似的曲線——戴維斯-高德納龍（Davis-Knuth dragon），是連續的空間填充曲線。
^ 1.0 1.1 無盡不循環小數
^ 不能取底數
b
=
2
{\displaystyle b=2}
2
−
1
=
0.1
bin
=
0.5
dec
{\displaystyle \textstyle 2^{-1}=0.1_{\text{bin}}=0.5_{\text{dec}}}
∑
k
≥
2
2
−
k
=
0.0
1
¯
bin
=
0.1
bin
=
0.5
dec
{\displaystyle \textstyle \sum _{k\geq 2}2^{-k}=0.0{\overline {1}}_{\text{bin}}=0.1_{\text{bin}}=0.5_{\text{dec}}}
(
i
−
1
)
−
1
=
−
0.1
bin
−
0.1
bin
i
=
−
0.5
dec
−
0.5
dec
i
{\displaystyle \textstyle (\mathrm {i} -1)^{-1}=-0.1_{\text{bin}}-0.1_{\text{bin}}\mathrm {i} =-0.5_{\text{dec}}-0.5_{\text{dec}}\mathrm {i} }
∑
k
≥
2
(
i
−
1
)
−
k
=
0.1
dec
+
0.3
dec
i
{\displaystyle \textstyle \sum _{k\geq 2}(\mathrm {i} -1)^{-k}=0.1_{\text{dec}}+0.3_{\text{dec}}\mathrm {i} }
^ 1.0 1.1 Knuth, D.E. An Imaginary Number System. Communications of the ACM. 1960, 3 (4): 245–247. S2CID 16513137 doi:10.1145/367177.367233 ^ 2.0 2.1 2.2 2.3 Knuth, Donald . Positional Number Systems. The art of computer programming 2 3rd. Boston: Addison-Wesley. 1998: 205. ISBN 0-201-89684-2OCLC 48246681 ^ 3.0 3.1 3.2 Khmelnik, S.I. Specialized digital computer for operations with complex numbers. Questions of Radio Electronics (In Russian). 1964, XII (2). ^ 4.0 4.1 W. Penney, A "binary" system for complex numbers, JACM 12 (1965) 247-248.
^ 5.0 5.1 Jamil, T. The complex binary number system. IEEE Potentials. 2002, 20 (5): 39–41. doi:10.1109/45.983342 ^ 6.0 6.1 Duda, Jarek. Complex base numeral systems. 2008-02-24. arXiv:0712.1309 math.DS ]. ^ Khmelnik, S.I. Positional coding of complex numbers. Questions of Radio Electronics (In Russian). 1966, XII (9). ^ 8.0 8.1 Khmelnik, S.I. Coding of Complex Numbers and Vectors (in Russian) (PDF) . Israel: Mathematics in Computer. 2004 [2022-11-03 ] . ISBN 978-0-557-74692-7存檔 (PDF) 於2022-11-10）. ^ 9.0 9.1 Khmelnik, S.I. Method and system for processing complex numbers . Patent USA, US2003154226 (A1). 2001 [2022-11-03 ] . （原始內容存檔 於2023-01-09）. ^ William J. Gilbert. Arithmetic in Complex Bases (PDF) . Mathematics Magazine. 1984-03,. Vol. 57 (No. 2) [2022-11-03 ] . （原始內容存檔 (PDF) 於2022-11-03）. ^ 11.0 11.1 Piché, Daniel Guy, Complex Bases, Number Systems and Their Application to Fractal-Wavelet Image Coding (PDF) , University of Waterloo, 2002 [2022-11-03 ] , （原始內容存檔 (PDF) 於2022-11-10）
![{\displaystyle \mathbb {Z} [\mathrm {i} ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b2f7e30b2083c5a1ffb637bd8c9ec111a486cd06)
![{\displaystyle \mathbb {Z} \left[{\tfrac {-1+\mathrm {i} {\sqrt {7}}}{2}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bce5a43e3c8fb5fdabc8d12b6c20e560b8bb9f48)
![{\displaystyle [-{\tfrac {8}{15}},{\tfrac {2}{15}}]\times [-{\tfrac {4}{15}},{\tfrac {1}{15}}]\mathrm {i} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1169f246ca710e557a9ff51c42d43f199b1100f)
[math.DS].
