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In the case of two nested square roots, the following theorem completely solves the problem of denesting. [2]If a and c are rational numbers and c is not the square of a rational number, there are two rational numbers x and y such that + = if and only if is the square of a rational number d.
The simplified equation is not entirely equivalent to the original. For when we substitute y = 0 and z = 0 in the last equation, both sides simplify to 0, so we get 0 = 0, a mathematical truth. But the same substitution applied to the original equation results in x/6 + 0/0 = 1, which is mathematically meaningless.
A necessary (but not sufficient) condition for solvability is that n is not divisible by 4 or by a prime of form 4k + 3. [note 3] Thus, for example, x 2 − 3 y 2 = −1 is never solvable, but x 2 − 5 y 2 = −1 may be. [27] The first few numbers n for which x 2 − n y 2 = −1 is solvable are with only one trivial solution: 1
For example, 6 is highly composite because d(6)=4 and d(n)=1,2,2,3,2 for n=1,2,3,4,5 respectively. A related concept is that of a largely composite number , a positive integer that has at least as many divisors as all smaller positive integers.
Simplification is the process of replacing a mathematical expression by an equivalent one that is simpler (usually shorter), according to a well-founded ordering. Examples include:
The first 15 superior highly composite numbers, 2, 6, 12, 60, 120, 360, 2520, 5040, 55440, 720720, 1441440, 4324320, 21621600, 367567200, 6983776800 (sequence A002201 in the OEIS) are also the first 15 colossally abundant numbers, which meet a similar condition based on the sum-of-divisors function rather than the number of divisors. Neither ...
49 is the smallest discriminant of a totally real cubic field. [2] 49 and 94 are the only numbers below 100 whose all permutations are composites but they are not multiples of 3, repdigits or numbers which only have digits 0, 2, 4, 5, 6 and 8, even excluding the trivial one digit terms. 49 = 7^2 and 94 = 2 * 47
He concluded that the area between the secant line and the curve is 4 / 3 the area of triangle ACE. Archimedes encounters the series in his work Quadrature of the Parabola . He finds the area inside a parabola by the method of exhaustion, and he gets a series of triangles; each stage of the construction adds an area 1 / 4 times ...