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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 = Which method is faster "by hand" depends on the ...
For example, 3 is a divisor of 21, since 21/7 = 3 (and therefore 7 is also a divisor of 21). If m is a divisor of n , then so is − m . The tables below only list positive divisors.
[14] [15] It is said to be an improper fraction, or sometimes top-heavy fraction, [16] if the absolute value of the fraction is greater than or equal to 1. Examples of proper fractions are 2/3, −3/4, and 4/9, whereas examples of improper fractions are 9/4, −4/3, and 3/3.
Sometimes this remainder is added to the quotient as a fractional part, so 10 / 3 is equal to 3 + 1 / 3 or 3.33..., but in the context of integer division, where numbers have no fractional part, the remainder is kept separately (or exceptionally, discarded or rounded). [5] When the remainder is kept as a fraction, it leads to a rational ...
For example, in duodecimal, 1 / 2 = 0.6, 1 / 3 = 0.4, 1 / 4 = 0.3 and 1 / 6 = 0.2 all terminate; 1 / 5 = 0. 2497 repeats with period length 4, in contrast with the equivalent decimal expansion of 0.2; 1 / 7 = 0. 186A35 has period 6 in duodecimal, just as it does in decimal.
Arthur Eddington argued that the fine-structure constant was a unit fraction. He initially thought it to be 1/136 and later changed his theory to 1/137. This contention has been falsified, given that current estimates of the fine structure constant are (to 6 significant digits) 1/137.036. [30]
where c 1 = 1 / a 1 , c 2 = a 1 / a 2 , c 3 = a 2 / a 1 a 3 , and in general c n+1 = 1 / a n+1 c n . Second, if none of the partial denominators b i are zero we can use a similar procedure to choose another sequence {d i} to make each partial denominator a 1:
To test for divisibility by D, where D ends in 1, 3, 7, or 9, the following method can be used. [12] Find any multiple of D ending in 9. (If D ends respectively in 1, 3, 7, or 9, then multiply by 9, 3, 7, or 1.) Then add 1 and divide by 10, denoting the result as m. Then a number N = 10t + q is divisible by D if and only if mq + t is divisible ...