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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.
3 ⁄ 5: 0.6 Vulgar Fraction Three Fifths 2157 8535 ⅘ 4 ⁄ 5: 0.8 Vulgar Fraction Four Fifths 2158 8536 ⅙ 1 ⁄ 6: 0.166... Vulgar Fraction One Sixth 2159 8537 ⅚ 5 ⁄ 6: 0.833... Vulgar Fraction Five Sixths 215A 8538 ⅛ 1 ⁄ 8: 0.125 Vulgar Fraction One Eighth 215B 8539 ⅜ 3 ⁄ 8: 0.375 Vulgar Fraction Three Eighths 215C 8540 ⅝ 5 ...
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 ...
Figure 2 is used for the multiples of 2, 4, 6, and 8. These patterns can be used to memorize the multiples of any number from 0 to 10, except 5. As you would start on the number you are multiplying, when you multiply by 0, you stay on 0 (0 is external and so the arrows have no effect on 0, otherwise 0 is used as a link to create a perpetual cycle).
A non-nested radical expression is said to be in simplified form if no factor of the ... old p times 10 plus x. Subtract ... 1·1612 0 ·5 3 + 10 1 ·3·1612 1 ·5 2 ...
4/3: −5/2: 4/3: −1/12: 6 1/90: −3/20: 3/2: −49/18: 3/2: −3/20: 1/90: 8 −1/560: 8/315: −1/5: 8/5: ... The order of accuracy of the approximation takes ...
The 1/3–2/3 conjecture states that, at each step, one may choose a comparison to perform that reduces the remaining number of linear extensions by a factor of 2/3; therefore, if there are E linear extensions of the partial order given by the initial information, the sorting problem can be completed in at most log 3/2 E additional comparisons. [4]