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The entire fraction may be expressed as a single composition, in which case it is hyphenated, or as a number of fractions with a numerator of one, in which case they are not. (For example, two-fifths is the fraction 2 / 5 and two fifths is the same fraction understood as 2 instances of 1 / 5 .) Fractions should always be ...
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 ...
By consequence, we may get, for example, three different values for the fractional part of just one x: let it be −1.3, its fractional part will be 0.7 according to the first definition, 0.3 according to the second definition, and −0.3 according to the third definition, whose result can also be obtained in a straightforward way by
The topic of Egyptian fractions has also seen interest in modern number theory; for instance, the Erdős–Graham problem [9] and the Erdős–Straus conjecture [10] concern sums of unit fractions, as does the definition of Ore's harmonic numbers. [11] A pattern of spherical triangles with reflection symmetry across each triangle edge.
A rational fraction is an algebraic fraction whose numerator and denominator are both polynomials. Thus 3 x x 2 + 2 x − 3 {\displaystyle {\frac {3x}{x^{2}+2x-3}}} is a rational fraction, but not x + 2 x 2 − 3 , {\displaystyle {\frac {\sqrt {x+2}}{x^{2}-3}},} because the numerator contains a square root function.
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 ...
Enjoy a classic game of Hearts and watch out for the Queen of Spades!
Simplifying this further gives us the solution x = −3. It is easily checked that none of the zeros of x (x + 1)(x + 2) – namely x = 0, x = −1, and x = −2 – is a solution of the final equation, so no spurious solutions were introduced.