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To calculate the whole number quotient of dividing a large number by a small number, the student repeatedly takes away "chunks" of the large number, where each "chunk" is an easy multiple (for example 100×, 10×, 5× 2×, etc.) of the small number, until the large number has been reduced to zero – or the remainder is less than the small ...
Thought of quotitively, a division problem can be solved by repeatedly subtracting groups of the size of the divisor. [1] For instance, suppose each egg carton fits 12 eggs, and the problem is to find how many cartons are needed to fit 36 eggs in total. Groups of 12 eggs at a time can be separated from the main pile until none are left, 3 groups:
Instead, the division is reduced to small steps. Starting from the left, enough digits are selected to form a number (called the partial dividend) that is at least 4×1 but smaller than 4×10 (4 being the divisor in this problem). Here, the partial dividend is 9. The first number to be divided by the divisor (4) is the partial dividend (9).
In the example, 20 is the dividend, 5 is the divisor, and 4 is the quotient. Unlike the other basic operations, when dividing natural numbers there is sometimes a remainder that will not go evenly into the dividend; for example, 10 / 3 leaves a remainder of 1, as 10 is not a multiple of 3.
Long division is the standard algorithm used for pen-and-paper division of multi-digit numbers expressed in decimal notation. It shifts gradually from the left to the right end of the dividend, subtracting the largest possible multiple of the divisor (at the digit level) at each stage; the multiples then become the digits of the quotient, and the final difference is then the remainder.
For example, can be expanded to the periodic continued fraction [;,,,...]. This article considers only the case of periodic regular continued fractions . In other words, the remainder of this article assumes that all the partial denominators a i ( i ≥ 1) are positive integers.
In cases where one or more of the b terms has more than two digits, the final quotient value b cannot be constructed simply by concatenating the digit pairs. Instead, each term, starting with b 1 , {\displaystyle b_{1},} should be multiplied by 100, and the next term added (or, if negative, subtracted).
When a partial fraction term has a single (i.e. unrepeated) binomial in the denominator, the numerator is a residue of the function defined by the input fraction. We calculate each respective numerator by (1) taking the root of the denominator (i.e. the value of x that makes the denominator zero) and (2) then substituting this root into the ...