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This is denoted as 20 / 5 = 4, or 20 / 5 = 4. [2] 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
An orange that has been sliced into two halves. In mathematics, division by two or halving has also been called mediation or dimidiation. [1] The treatment of this as a different operation from multiplication and division by other numbers goes back to the ancient Egyptians, whose multiplication algorithm used division by two as one of its fundamental steps. [2]
Euclidean division can also be extended to negative dividend (or negative divisor) using the same formula; for example −9 = 4 × (−3) + 3, which means that −9 divided by 4 is −3 with remainder 3.
[1] For example, the expression "5 mod 2" evaluates to 1, because 5 divided by 2 has a quotient of 2 and a remainder of 1, while "9 mod 3" would evaluate to 0, because 9 divided by 3 has a quotient of 3 and a remainder of 0. Although typically performed with a and n both being integers, many computing systems now allow other types of numeric ...
Integer arithmetic is not closed under division. This means that when dividing one integer by another integer, the result is not always an integer. For instance, 7 divided by 2 is not a whole number but 3.5. [73] One way to ensure that the result is an integer is to round the result to a whole number.
[1] [2] [3] Spreadsheets were developed as computerized analogs of paper accounting worksheets. [4] The program operates on data entered in cells of a table. Each cell may contain either numeric or text data, or the results of formulas that automatically calculate and display a value based on the contents of other cells.
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In more generality: For all p ≥ 1 and odd h, f p − 1 (2 p h − 1) = 2 × 3 p − 1 h − 1. (Here f p − 1 is function iteration notation.) For all odd h, f(2h − 1) ≤ 3h − 1 / 2 The Collatz conjecture is equivalent to the statement that, for all k in I, there exists an integer n ≥ 1 such that f n (k) = 1.