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The nines' complement of a decimal digit is the number that must be added to it to produce 9; the nines' complement of 3 is 6, the nines' complement of 7 is 2, and so on, see table. To form the nines' complement of a larger number, each digit is replaced by its nines' complement. Consider the following subtraction problem:
Five additions problems for each heat, each problem consisting of 50 three- to six-digit numbers. The soroban won in two successive heats. Five subtraction problems for each heat, each problem having six- to eight-digit minuends and subtrahends. The soroban won in the first and third heats; the second heat was a no contest.
Both these methods break up the subtraction as a process of one digit subtractions by place value. Starting with a least significant digit, a subtraction of the subtrahend: s j s j−1... s 1. from the minuend m k m k−1... m 1, where each s i and m i is a digit, proceeds by writing down m 1 − s 1, m 2 − s 2, and so forth, as long as s i ...
A subtraction problem such as is solved by borrowing a 10 from the tens place to add to the ones place in order to facilitate the subtraction. Subtracting 9 from 6 involves borrowing a 10 from the tens place, making the problem into +. This is indicated by crossing out the 8, writing a 7 above it, and writing a 1 above the 6.
Catastrophic cancellation may happen even if the difference is computed exactly, as in the example above—it is not a property of any particular kind of arithmetic like floating-point arithmetic; rather, it is inherent to subtraction, when the inputs are approximations themselves.
For instance, if the number π is rounded to 4 decimal places, the result is 3.142 because the following digit is a 5, so 3.142 is closer to π than 3.141. [107] These methods allow computers to efficiently perform approximate calculations on real numbers. [108]
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