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Quater-imaginary base
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==Addition and subtraction== It is possible to [[addition|add]] and [[subtraction|subtract]] numbers in the quater-imaginary 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" β1 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 '''β1'''". 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 quater-imaginary arithmetic, a "carry" ''subtracts'' from the next-but-one column, and a "borrow" ''adds''. ===Example: Addition=== Below are two examples of adding in the quater-imaginary system: <pre> 1 β 2i 1031 1 β 2i 1031 ------ + <=> ---- + 2 β 4i 1022 3 β 4i 1023 1 β 8i 1001 ------ + <=> ----- + 4 β12i 12320 </pre> 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 "2''i''s column"), giving 6; 6 is greater than 3, so we subtract 4 (giving 2 as the result in the second column) and carry β1 into the fourth column. Adding the 0s in the third column gives 0; and finally adding the two 1s and the carried β1 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 β1 into the third column (the "β4s column"). Then we add 2+0 in the second column, giving 2. In the third column, we have 0+0+(β1), because of the carry; β1 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>12320_{2i}</math>. ===Example: Subtraction=== Subtraction is analogous to addition in that it uses the same two rules described above. Below is an example: <pre> β 2 β 8i 1102 1 β 6i 1011 ------- <=> ----- β 3 β 2i 1131 </pre> In this example we have to subtract <math>1011_{2i}</math> from <math>1102_{2i}</math>. The rightmost digit is 2β1 = 1. The second digit from the right would become β1, so add 4 to give 3 and then carry +1 two places to the left. The third digit from the right is 1β0 = 1. Then the leftmost digit is 1β1 plus 1 from the carry, giving 1. This gives a final answer of <math>1131_{2i}</math>.
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