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This series was used as a representation of two of Zeno's paradoxes. [2] For example, in the paradox of Achilles and the Tortoise, the warrior Achilles was to race against a tortoise. The track is 100 meters long. Achilles could run at 10 m/s, while the tortoise only 5. The tortoise, with a 10-meter advantage, Zeno argued, would win.
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.
This is denoted as 20 / 5 = 4, or 20 / 5 = 4. [2] In the example, 20 is the dividend, 5 is the divisor, and 4 is the quotient. Unlike the other basic operations, when dividing natural numbers there is sometimes a remainder that will not go evenly into the dividend; for example, 10 / 3 leaves a remainder of 1, as 10 is not a multiple of 3.
1 + 2 = 3 + 3 = 6 + 4 = 10 + 5 = 15. Structurally, this is shorthand for ([(1 + 2 = 3) + 3 = 6] + 4 = 10) + 5 = 15, but the notation is incorrect, because each part of the equality has a different value. If interpreted strictly as it says, it would imply that 3 = 6 = 10 = 15 = 15. A correct version of the argument would be 1 + 2 = 3, 3 + 3 = 6 ...
One half is the rational number that lies midway between 0 and 1 on the number line. Multiplication by one half is equivalent to division by two, or "halving"; conversely, division by one half is equivalent to multiplication by two, or "doubling". A square of side length one, here dissected into rectangles whose areas are successive powers of ...
The Amateur Athletics Union even went as far as prohibiting women from running more than 1.5 miles (2.4 km) and the organizers of the Boston Marathon did not want to “take the liability” of ...
The idea becomes clearer by considering the general series 1 − 2x + 3x 2 − 4x 3 + 5x 4 − 6x 5 + &c. that arises while expanding the expression 1 ⁄ (1+x) 2, which this series is indeed equal to after we set x = 1.
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.