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Every terminating decimal representation can be written as a decimal fraction, a fraction whose denominator is a power of 10 (e.g. 1.585 = 1585 / 1000 ); it may also be written as a ratio of the form k / 2 n ·5 m (e.g. 1.585 = 317 / 2 3 ·5 2 ).
Unit fractions can also be expressed using negative exponents, as in 2 −1, which represents 1/2, and 2 −2, which represents 1/(2 2) or 1/4. A dyadic fraction is a common fraction in which the denominator is a power of two , e.g. 1 / 8 = 1 / 2 3 .
Two fractions a / b and c / d are equal or equivalent if and only if ad = bc.) For example, 1 / 4 , 5 / 6 , and −101 / 100 are all irreducible fractions. On the other hand, 2 / 4 is reducible since it is equal in value to 1 / 2 , and the numerator of 1 / 2 is less than the numerator ...
A fixed-point representation of a fractional number is essentially an integer that is to be implicitly multiplied by a fixed scaling factor. For example, the value 1.23 can be stored in a variable as the integer value 1230 with implicit scaling factor of 1/1000 (meaning that the last 3 decimal digits are implicitly assumed to be a decimal fraction), and the value 1 230 000 can be represented ...
The prefix 0o also follows the model set by the prefix 0x used for hexadecimal literals in the C language; it is supported by Haskell, [19] OCaml, [20] Python as of version 3.0, [21] Raku, [22] Ruby, [23] Tcl as of version 9, [24] PHP as of version 8.1, [25] Rust [26] and ECMAScript as of ECMAScript 6 [27] (the prefix 0 originally stood for ...
For the folded general continued fractions of both expressions, the rate convergence μ = (3 − √ 8) 2 = 17 − √ 288 ≈ 0.02943725, hence 1 / μ = (3 + √ 8) 2 = 17 + √ 288 ≈ 33.97056, whose common logarithm is 1.531... ≈ 26 / 17 > 3 / 2 , thus adding at least three digits per two terms. This is because the ...
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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]