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The harmonic mean of a set of positive integers is the number of numbers times the reciprocal of the sum of their reciprocals. The optic equation requires the sum of the reciprocals of two positive integers a and b to equal the reciprocal of a third positive integer c. All solutions are given by a = mn + m 2, b = mn + n 2, c = mn.
Symbolab is an answer engine [1] that provides step-by-step solutions to mathematical problems in a range of subjects. [2] It was originally developed by Israeli start-up company EqsQuest Ltd., under whom it was released for public use in 2011. In 2020, the company was acquired by American educational technology website Course Hero. [3] [4]
Slices of approximately 1/8 of a pizza. A unit fraction is a positive fraction with one as its numerator, 1/ n.It is the multiplicative inverse (reciprocal) of the denominator of the fraction, which must be a positive natural number.
The product of a non-zero fraction and its reciprocal is 1, hence the reciprocal is the multiplicative inverse of a fraction. The reciprocal of a proper fraction is improper, and the reciprocal of an improper fraction not equal to 1 (that is, numerator and denominator are not equal) is a proper fraction.
Multiplying by a number is the same as dividing by its reciprocal and vice versa. For example, multiplication by 4/5 (or 0.8) will give the same result as division by 5/4 (or 1.25). Therefore, multiplication by a number followed by multiplication by its reciprocal yields the original number (since the product of the number and its reciprocal is 1).
If a does have an inverse modulo m, then there are an infinite number of solutions of this congruence, which form a congruence class with respect to this modulus. Furthermore, any integer that is congruent to a (i.e., in a 's congruence class) has any element of x 's congruence class as a modular multiplicative inverse.
To find a solution for , just divide all of the unit fractions in the solution for by : = + + = + +. If 4 n {\displaystyle {\tfrac {4}{n}}} were a counterexample to the conjecture, for a composite number n {\displaystyle n} , every prime factor p {\displaystyle p} of n {\displaystyle n} would also provide a counterexample 4 p {\displaystyle ...
This produces a sequence of approximations, all of which are rational numbers, and these converge to the starting number as a limit. This is the (infinite) continued fraction representation of the number. Examples of continued fraction representations of irrational numbers are: √ 19 = [4;2,1,3,1,2,8,2,1,3,1,2,8,...] (sequence A010124 in the ...