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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.
Starting from the rightmost digit, each pair of digits is added together. The rightmost digit of the sum is written below them. If the sum is a two-digit number then the leftmost digit, called the "carry", is added to the next pair of digits to the left. This process is repeated until all digits have been added. [65]
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.
Place the digit as the next digit of the root, i.e., above the two digits of the square you just brought down. Thus the next p will be the old p times 10 plus x. Subtract y from c to form a new remainder. If the remainder is zero and there are no more digits to bring down, then the algorithm has terminated.
Here, 7 − 9 = −2, so try (10 − 9) + 7 = 8, and the 10 is got by taking ("borrowing") 1 from the next digit to the left. There are two ways in which this is commonly taught: The ten is moved from the next digit left, leaving in this example 3 − 1 in the tens column.
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