Quaterimaginary base
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The quaterimaginary numeral system is a numeral system, first proposed by Donald Knuth in 1960. Unlike standard numeral systems, which use an integer (such as 10 in decimal, or 2 in binary) as their bases, it uses the imaginary number 2i (equivalent to [math]\displaystyle{ \sqrt{4} }[/math]) as its base. It is able to (almost) uniquely represent every complex number using only the digits 0, 1, 2, and 3.^{[1]} Numbers less than zero, which are ordinarily represented with a minus sign, are representable as digit strings in quaterimaginary; for example, the number −1 is represented as "103" in quaterimaginary notation.
Decomposing the quaterimaginary
In a positional system with base [math]\displaystyle{ b }[/math],
[math]\displaystyle{ \ldots d_3d_2d_1d_0.d_{1}d_{2}d_{3}\ldots }[/math]
represents[math]\displaystyle{ \dots + d_3\cdot b^3+d_2\cdot b^2+d_1\cdot b+d_0+d_{1}\cdot b^{1}+d_{2}\cdot b^{2}+d_{3}\cdot b^{3}\dots }[/math]
In this numeral system, [math]\displaystyle{ b = 2i }[/math],
and because [math]\displaystyle{ (2i)^2=4 }[/math], the entire series of powers can be separated into two different series, so that it simplifies to [math]\displaystyle{ \begin{align} &{} [\dots+d_4\cdot(4)^2 +d_2\cdot(4)^1+d_0+d_{2}\cdot(4)^{1}+\dots ] \end{align} }[/math] for evennumbered digits (digits that simplify to the value of the digit times a power of 4), and [math]\displaystyle{ \begin{align} 2i\cdot[\dots+d_3\cdot(4)^1+d_1+d_{1}\cdot (4)^{1}+d_{3}\cdot (4)^{2} + \dots] \end{align} }[/math] for those digits that still have an imaginary factor. Adding these two series together then gives the total value of the number.
Because of the separation of these two series, the real and imaginary parts of complex numbers are readily expressed in base −4 as [math]\displaystyle{ \ldots d_4d_2d_0.d_{2}\ldots }[/math] and [math]\displaystyle{ 2\cdot(\ldots d_5d_3d_1.d_{1}d_{3}\ldots) }[/math] respectively.
Converting from quaterimaginary
k  (2i)^{k} 

−5  −i/32 
−4  1/16 
−3  i/8 
−2  −1/4 
−1  −i/2 
0  1 
1  2i 
2  −4 
3  −8i 
4  16 
5  32i 
6  −64 
7  −128i 
8  256 
To convert a digit string from the quaterimaginary system to the decimal system, the standard formula for positional number systems can be used. This says that a digit string [math]\displaystyle{ \ldots d_3d_2d_1d_0 }[/math] in base b can be converted to a decimal number using the formula
 [math]\displaystyle{ \cdots + d_3\cdot b^3+d_2\cdot b^2+d_1\cdot b+d_0 }[/math]
For the quaterimaginary system, [math]\displaystyle{ b = 2i }[/math].
Additionally, for a given string [math]\displaystyle{ d }[/math] in the form [math]\displaystyle{ d_{w1}, d_{w2},\dots d_0 }[/math], the formula below can be used for a given string length [math]\displaystyle{ w }[/math] in base [math]\displaystyle{ b }[/math]
 [math]\displaystyle{ Q2D_w \vec d \equiv \sum_{k=0}^{w1} d_k \cdot b^k }[/math]
Example
To convert the string [math]\displaystyle{ 1101_{2i} }[/math] to a decimal number, fill in the formula above:
 [math]\displaystyle{ 1\cdot(2i)^3 + 1\cdot(2i)^2 + 0\cdot(2i)^1 + 1\cdot(2i)^0 = 8i  4 + 0 + 1 = 3  8i }[/math]
Another, longer example: [math]\displaystyle{ 1030003_{2i} }[/math] in base 10 is
 [math]\displaystyle{ 1\cdot(2i)^6 + 3\cdot(2i)^4 + 3\cdot(2i)^0 = 64 + 3\cdot 16 + 3 = 13 }[/math]
Converting into quaterimaginary
It is also possible to convert a decimal number to a number in the quaterimaginary system. Every complex number (every number of the form a+bi) has a quaterimaginary representation. Most numbers have a unique quaterimaginary representation, but just as 1 has the two representations 1 = 0.9 in decimal notation, so, because of 0.0001_{2i} = 1/15, the number 1/5 has the two quaterimaginary representations 0.0003_{2i} = 3·1/15 = 1/5 = 1 + 3·–4/15 = 1.0300_{2i}.
To convert an arbitrary complex number to quaterimaginary, it is sufficient to split the number into its real and imaginary components, convert each of those separately, and then add the results by interleaving the digits. For example, since −1+4i is equal to −1 plus 4i, the quaterimaginary representation of −1+4i is the quaterimaginary representation of −1 (namely, 103) plus the quaterimaginary representation of 4i (namely, 20), which gives a final result of −1+4i = 123_{2i}.
To find the quaterimaginary representation of the imaginary component, it suffices to multiply that component by 2i, which gives a real number; then find the quaterimaginary representation of that real number, and finally shift the representation by one place to the right (thus dividing by 2i). For example, the quaterimaginary representation of 6i is calculated by multiplying 6i × 2i = −12, which is expressed as 300_{2i}, and then shifting by one place to the right, yielding: 6i = 30_{2i}.
Finding the quaterimaginary representation of an arbitrary real integer number can be done manually by solving a system of simultaneous equations, as shown below, but there are faster methods for both real and imaginary integers, as shown in the negative base article.
Example: Real number
As an example of an integer number we can try to find the quaterimaginary counterpart of the decimal number 7 (or 7_{10} since the base of the decimal system is 10). Since it is hard to predict exactly how long the digit string will be for a given decimal number, it is safe to assume a fairly large string. In this case, a string of six digits can be chosen. When an initial guess at the size of the string eventually turns out to be insufficient, a larger string can be used.
To find the representation, first write out the general formula, and group terms:
 [math]\displaystyle{ \begin{align} 7_{10}& = d_{0}+d_{1}\cdot b+d_{2}\cdot b^{2}+d_{3}\cdot b^{3}+d_{4}\cdot b^{4}+d_{5}\cdot b^{5} \\ & = d_{0}+2id_{1}4d_{2}8id_{3}+16d_{4}+32id_{5} \\ & = d_{0}4d_{2}+16d_{4}+i(2d_{1}8d_{3}+32d_{5}) \\ \end{align} }[/math]
Since 7 is a real number, it is allowed to conclude that d_{1}, d_{3} and d_{5} should be zero. Now the value of the coefficients d_{0}, d_{2} and d_{4}, must be found. Because d_{0} − 4 d_{2} + 16 d_{4} = 7 and because—by the nature of the quaterimaginary system—the coefficients can only be 0, 1, 2 or 3 the value of the coefficients can be found. A possible configuration could be: d_{0} = 3, d_{2} = 3 and d_{4} = 1. This configuration gives the resulting digit string for 7_{10}.
 [math]\displaystyle{ 7_{10} = 010303_{2i} = 10303_{2i}. }[/math]
Example: Imaginary number
Finding a quaterimaginary representation of a purely imaginary integer number ∈ iZ is analogous to the method described above for a real number. For example, to find the representation of 6i, it is possible to use the general formula. Then all coefficients of the real part have to be zero and the complex part should make 6. However, for 6i it is easily seen by looking at the formula that if d_{1} = 3 and all other coefficients are zero, we get the desired string for 6i. That is:
 [math]\displaystyle{ \begin{align}6i_{10} = 30_{2i}\end{align} }[/math]
Another conversion method
For real numbers the quaterimaginary representation is the same as negative quaternary (base −4). A complex number x+iy can be converted to quaterimaginary by converting x and y/2 separately to negative quaternary. If both x and y are finite binary fractions we can use the following algorithm using repeated Euclidean division:
For example: 35+23i=121003.2_{2i}
35 23i/2i=11.5 11=12−0.5 35÷(−4)=−8, remainder 3 12/(−4)=−3, remainder 0 (−0.5)×(−4)=2 −8÷(−4)= 2, remainder 0 −3/(−4)= 1, remainder 1 2÷(−4)= 0, remainder 2 1/(−4)= 0, remainder 1 20003 + 101000 + 0.2 = 121003.2 32i+16×2−8i−4×0+2i×0+1×3−2×i/2=35+23i
Radix point "."
A radix point in the decimal system is the usual . (dot) which marks the separation between the integer part and the fractional part of the number. In the quaterimaginary system a radix point can also be used. For a digit string [math]\displaystyle{ \dots d_{5}d_{4}d_{3}d_{2}d_{1}d_{0} . d_{1}d_{2}d_{3} \dots }[/math] the radix point marks the separation between nonnegative and negative powers of b. Using the radix point the general formula becomes:
 [math]\displaystyle{ d_5 b^5 + d_4 b^4 + d_3 b^3 + d_2 b^2 + d_1 b + d_0 + d_{1} b^{1} + d_{2} b^{2} + d_{3} b^{3} }[/math]
or
 [math]\displaystyle{ \begin{align} 32id_{5}+16d_{4}8id_{3}4d_{2}+2id_{1}+d_{0}+\frac{1}{2i}d_{1}+\frac{1}{4}d_{2}+\frac{1}{8i}d_{3}\\ =32id_{5}+16d_{4}8id_{3}4d_{2}+2id_{1}+d_{0}\frac{i}{2}d_{1}\frac{1}{4}d_{2}+\frac{i}{8}d_{3} \end{align} }[/math]
Example
If the quaterimaginary representation of the complex unit i has to be found, the formula without radix point will not suffice. Therefore, the above formula should be used. Hence:
 [math]\displaystyle{ \begin{align} i & = 32id_{5}+16d_{4}8id_{3}4d_{2}+2id_{1}+d_{0}+\frac{1}{2i}d_{1}+\frac{1}{4}d_{2}+\frac{1}{8i}d_{3}\\ & = i(32d_{5}8d_{3}+2d_{1}\frac{1}{2}d_{1}+\frac{1}{8}d_{3})+16d_{4}4d_{2}+d_{0}\frac{1}{4}d_{2}\\ \end{align} }[/math]
for certain coefficients d_{k}. Then because the real part has to be zero: d_{4} = d_{2} = d_{0} = d_{−2} = 0. For the imaginary part, if d_{5} = d_{3} = d_{−3} = 0 and when d_{1} = 1 and d_{−1} = 2 the digit string can be found. Using the above coefficients in the digit string the result is:
 [math]\displaystyle{ i = 10.2_{2i}. }[/math]
Addition and subtraction
It is possible to add and subtract numbers in the quaterimaginary system. In doing this, there are two basic rules that have to be kept in mind:
 Whenever a number exceeds 3, subtract 4 and "carry" −1 two places to the left.
 Whenever a number drops below 0, add 4 and "carry" +1 two places to the left.
Or for short: "If you add four, carry +1. If you subtract four, carry −1". This is the opposite of normal long addition, in which a "carry" in the current column requires adding 1 to the next column to the left, and a "borrow" requires subtracting. In quaterimaginary arithmetic, a "carry" subtracts from the nextbutone column, and a "borrow" adds.
Example: Addition
Below are two examples of adding in the quaterimaginary system:
1 − 2i 1031 3 − 4i 1023 1 − 2i 1031 1 − 8i 1001  + <=>  +  + <=>  + 2 − 4i 1022 4 − 12i 12320
In the first example we start by adding the two 1s in the first column (the "ones' column"), giving 2. Then we add the two 3s in the second column (the "2is column"), giving 6; 6 is greater than 3, so we subtract 4 (giving 2 as the result in the second column) and carry −1 into the fourth column. Adding the 0s in the third column gives 0; and finally adding the two 1s and the carried −1 in the fourth column gives 1.
In the second example we first add 3+1, giving 4; 4 is greater than 3, so we subtract 4 (giving 0) and carry −1 into the third column (the "−4s column"). Then we add 2+0 in the second column, giving 2. In the third column, we have 0+0+(−1), because of the carry; −1 is less than 0, so we add 4 (giving 3 as the result in the third column) and "borrow" +1 into the fifth column. In the fourth column, 1+1 is 2; and the carry in the fifth column gives 1, for a result of [math]\displaystyle{ 12320_{2i} }[/math].
Example: Subtraction
Subtraction is analogous to addition in that it uses the same two rules described above. Below is an example:
− 2 − 8i 1102 1 − 6i 1011  <=>  − 3 − 2i 1131
In this example we have to subtract [math]\displaystyle{ 1011_{2i} }[/math] from [math]\displaystyle{ 1102_{2i} }[/math]. The rightmost digit is 2−1 = 1. The second digit from the right would become −1, so add 4 to give 3 and then carry +1 two places to the left. The third digit from the right is 1−0 = 1. Then the leftmost digit is 1−1 plus 1 from the carry, giving 1. This gives a final answer of [math]\displaystyle{ 1131_{2i} }[/math].
Multiplication
For long multiplication in the quaterimaginary system, the two rules stated above are used as well. When multiplying numbers, multiply the first string by each digit in the second string consecutively and add the resulting strings. With every multiplication, a digit in the second string is multiplied with the first string. The multiplication starts with the rightmost digit in the second string and then moves leftward by one digit, multiplying each digit with the first string. Then the resulting partial products are added where each is shifted to the left by one digit. An example:
11201 20121 ×  11201 ←––– 1 × 11201 12002 ←––– 2 × 11201 11201 ←––– 1 × 11201 00000 ←––– 0 × 11201 12002 + ←––– 2 × 11201  120231321
This corresponds to a multiplication of [math]\displaystyle{ (98i)\cdot(29+4i) = 293196i }[/math].
Tabulated conversions
Below is a table of some decimal and complex numbers and their quaterimaginary counterparts.




Examples
Below are some other examples of conversions from decimal numbers to quaterimaginary numbers.
 [math]\displaystyle{ 5 = 16 + (3\cdot4) + 1 = 10301_{2i} }[/math]
 [math]\displaystyle{ i = 2i + 2\left(\frac{1}{2}i\right) = 10.2_{2i} }[/math]
 [math]\displaystyle{ 7 \frac{3}{4}  7 \frac{1}{2}i = 1(16) + 1(8i) + 2(4) + 1(2i) + 3\left(\frac{1}{2}i\right) + 1\left(\frac{1}{4}\right) = 11210.31_{2i} }[/math]
Zorder curve
The representation
 [math]\displaystyle{ z = \sum_{k\ge n} z_k \cdot (2i)^{k} }[/math]
of an arbitrary complex number [math]\displaystyle{ z\in \Complex }[/math] with [math]\displaystyle{ z_k\in \{0,1,2,3\} }[/math] gives rise to an injective mapping
 [math]\displaystyle{ \textstyle \begin{array}{llcl} \varphi \colon & \Complex & \to & \R \\ & \sum_{k \ge n} z_k \cdot (2i)^{k} & \mapsto & \sum_{k \ge n} z_k \cdot r^{k}\\ \end{array} }[/math]
with some suitable [math]\displaystyle{ r \in \Z }[/math]. Here [math]\displaystyle{ r = 4 }[/math] cannot be taken as base because of
 [math]\displaystyle{ \textstyle \sum_{k\gt 0} 3\cdot (2i)^{k} = \tfrac{36i}5 \; \; \; \; \ne \; \; \; \; 1 = \sum_{k\gt 0} 3 \cdot 4^{k}. }[/math]
The image [math]\displaystyle{ \varphi(\Complex) \subset \R }[/math] is a Cantor set which allows to linearly order [math]\displaystyle{ \Complex }[/math] similar to a Zorder curve. Since the image is disconnected, [math]\displaystyle{ \varphi }[/math] is not continuous.
See also
References
 ↑ Donald Knuth (April 1960). "An imaginary number system". Communications of the ACM 3 (4): 245–247. doi:10.1145/367177.367233.
Further reading
 Knuth, Donald Ervin. "Positional Number Systems". The Art of Computer Programming. 2 (3 ed.). AddisonWesley. p. 205.
 Warren Jr., Henry S. (2013). Hacker's Delight (2 ed.). Addison Wesley  Pearson Education, Inc.. p. 309. 0321842685. ISBN 9780321842688.
Original source: https://en.wikipedia.org/wiki/Quaterimaginary base.
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