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In Latin and Greek, the ordinal forms are also used for fractions for amounts higher than 2; only the fraction 1 / 2 has special forms. The same suffix may be used with more than one category of number, as for example the orginary numbers second ary and terti ary and the distributive numbers bi nary and ter nary .
1 ⁄ 5: 0.2 Vulgar Fraction One Fifth 2155 8533 ⅖ 2 ⁄ 5: 0.4 Vulgar Fraction Two Fifths 2156 8534 ⅗ 3 ⁄ 5: 0.6 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 ...
A simple fraction (also known as a common fraction or vulgar fraction) [n 1] is a rational number written as a/b or , where a and b are both integers. [9] As with other fractions, the denominator (b) cannot be zero. Examples include 1 / 2 , − 8 / 5 , −8 / 5 , and 8 / −5 .
In terms of partition, 20 / 5 means the size of each of 5 parts into which a set of size 20 is divided. For example, 20 apples divide into five groups of four apples, meaning that "twenty divided by five is equal to four". 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 ...
A handy chart of decimal-fraction equivalents, 0 to 1 by 64ths. Prints nicely as 11x17 in landscape orientation. Useful for machinists who work with inch-based measurements. Date: 24 October 2007: Source: Own work: Author: Three-quarter-ten
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 ...
The largest number that the divisor 4 can be multiplied by without exceeding 5 is 1, so the digit 1 is put above the 5 to start constructing the quotient. Next, the 1 is multiplied by the divisor 4, to obtain the largest whole number that is a multiple of the divisor 4 without exceeding the 5 (4 in this case).
k=1 yielding 21 is the smallest positive number that can be successively divided by 5 twice with remainder 1. If there are 5 divisions, then multiples of 5 5 =3125 are required; the smallest such number is 3125 – 4 = 3121. After 5 divisions, there are 1020 coconuts left over, a number divisible by 5 as required by the problem.