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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 ).
In the second step, they were divided by 3. The final result, 4 / 3 , is an irreducible fraction because 4 and 3 have no common factors other than 1. The original fraction could have also been reduced in a single step by using the greatest common divisor of 90 and 120, which is 30. As 120 ÷ 30 = 4, and 90 ÷ 30 = 3, one gets
Also the converse is true: The decimal expansion of a rational number is either finite, or endlessly repeating. Finite decimal representations can also be seen as a special case of infinite repeating decimal representations. For example, 36 ⁄ 25 = 1.44 = 1.4400000...; the endlessly repeated sequence is the one-digit sequence "0".
Stylistic impression of the repeating decimal 0.9999..., representing the digit 9 repeating infinitely In mathematics , 0.999... (also written as 0. 9 , 0. . 9 , or 0.(9) ) is a repeating decimal that is an alternative way of writing the number 1 .
Since e is an irrational number (see proof that e is irrational), it cannot be represented as the quotient of two integers, but it can be represented as a continued fraction. Using calculus , e may also be represented as an infinite series , infinite product , or other types of limit of a sequence .
A mathematical constant is a key number whose value is fixed by an unambiguous definition, often referred to by a symbol (e.g., an alphabet letter), or by mathematicians' names to facilitate using it across multiple mathematical problems. [1]
Because it is a divergent series, it should be interpreted as a formal sum, an abstract mathematical expression combining the unit fractions, rather than as something that can be evaluated to a numeric value. There are many different proofs of the divergence of the harmonic series, surveyed in a 2006 paper by S. J. Kifowit and T. A. Stamps. [13]
Squaring both sides of c = 2y yields c 2 = (2y) 2, or c 2 = 4y 2. Substituting 4y 2 for c 2 in the first equation (c 2 = 2b 2) gives us 4y 2 = 2b 2. Dividing by 2 yields 2y 2 = b 2. Since y is an integer, and 2y 2 = b 2, b 2 is divisible by 2, and therefore even. Since b 2 is even, b must be even. We have just shown that both b and c must be ...