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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 of 2 / 4 . A fraction that is reducible can be reduced by dividing both the numerator ...
6 1 2 1 1 −1 4 5 9. and would be written in modern notation as 6 1 / 4 , 1 1 / 5 , and 2 − 1 / 9 (i.e., 1 8 / 9 ). The horizontal fraction bar is first attested in the work of Al-Hassār (fl. 1200), [35] a Muslim mathematician from Fez, Morocco, who specialized in Islamic inheritance jurisprudence.
As for fractions, the simplest form is considered that in which the numbers in the ratio are the smallest possible integers. Thus, the ratio 40:60 is equivalent in meaning to the ratio 2:3, the latter being obtained from the former by dividing both quantities by 20. Mathematically, we write 40:60 = 2:3, or equivalently 40:60∷2:3.
This last non-simple continued fraction (sequence A110185 in the OEIS), equivalent to = [;,,,,,...], has a quicker convergence rate compared to Euler's continued fraction formula [clarification needed] and is a special case of a general formula for the exponential function:
1 ⁄ 9: 0.111... Vulgar Fraction One Ninth 2151 8529 ⅒ 1 ⁄ 10: 0.1 Vulgar Fraction One Tenth 2152 8530 ⅓ 1 ⁄ 3: 0.333... Vulgar Fraction One Third 2153 8531 ⅔ 2 ⁄ 3: 0.666... Vulgar Fraction Two Thirds 2154 8532 ⅕ 1 ⁄ 5: 0.2 Vulgar Fraction One Fifth 2155 8533 ⅖ 2 ⁄ 5: 0.4 Vulgar Fraction Two Fifths 2156 8534 ⅗ 3 ⁄ 5: 0 ...
The simplified equation is not entirely equivalent to the original. For when we substitute y = 0 and z = 0 in the last equation, both sides simplify to 0, so we get 0 = 0 , a mathematical truth. But the same substitution applied to the original equation results in x /6 + 0/0 = 1 , which is mathematically meaningless .
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 ).
Since P 2 (x) < 0 for x = 1 / 9 , and P 2 (x) > 0 for all x > 1 / 8 , the next term in the greedy expansion is 1 / 9 . If x 3 is the remaining fraction after this step of the greedy expansion, it satisfies the equation P 2 (x 3 + 1 / 9 ) = 0, which can again be expanded as a polynomial equation with integer ...