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If necessary, simplify the long division problem by moving the decimals of the divisor and dividend by the same number of decimal places, to the right (or to the left), so that the decimal of the divisor is to the right of the last digit. When doing long division, keep the numbers lined up straight from top to bottom under the tableau.
In abstract algebra, given a magma with binary operation ∗ (which could nominally be termed multiplication), left division of b by a (written a \ b) is typically defined as the solution x to the equation a ∗ x = b, if this exists and is unique. Similarly, right division of b by a (written b / a) is the solution y to the equation y ∗ a = b ...
The Dewey Decimal Classification (DDC) is structured around ten main classes covering the entire world of knowledge; each main class is further structured into ten hierarchical divisions, each having ten divisions of increasing specificity. [1]
To change a common fraction to decimal notation, do a long division of the numerator by the denominator (this is idiomatically also phrased as "divide the denominator into the numerator"), and round the result to the desired precision. For example, to change 1 / 4 to a decimal expression, divide 1 by 4 (" 4 into 1 "), to obtain exactly ...
In the division of 43 by 5, we have: 43 = 8 × 5 + 3, so 3 is the least positive remainder. We also have that: 43 = 9 × 5 − 2, and −2 is the least absolute remainder. These definitions are also valid if d is negative, for example, in the division of 43 by −5, 43 = (−8) × (−5) + 3, and 3 is the least positive remainder, while,
In arithmetic, Euclidean division – or division with remainder – is the process of dividing one integer (the dividend) by another (the divisor), in a way that produces an integer quotient and a natural number remainder strictly smaller than the absolute value of the divisor. A fundamental property is that the quotient and the remainder ...
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.
The word problem for an algebra is then to determine, given two expressions (words) involving the generators and operations, whether they represent the same element of the algebra modulo the identities. The word problems for groups and semigroups can be phrased as word problems for algebras. [1]
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