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Leonhard Euler introduced the function in 1763. [7] [8] [9] However, he did not at that time choose any specific symbol to denote it.In a 1784 publication, Euler studied the function further, choosing the Greek letter π to denote it: he wrote πD for "the multitude of numbers less than D, and which have no common divisor with it". [10]
The three triangular numbers are not necessarily distinct, or nonzero; for example 20 = 10 + 10 + 0. This is a special case of the Fermat polygonal number theorem . The largest triangular number of the form 2 k − 1 is 4095 (see Ramanujan–Nagell equation ).
Van der Waerden's theorem is a theorem in the branch of mathematics called Ramsey theory.Van der Waerden's theorem states that for any given positive integers r and k, there is some number N such that if the integers {1, 2, ..., N} are colored, each with one of r different colors, then there are at least k integers in arithmetic progression whose elements are of the same color.
Of the cleanly formulated Hilbert problems, numbers 3, 7, 10, 14, 17, 18, 19, and 20 have resolutions that are accepted by consensus of the mathematical community. Problems 1, 2, 5, 6, [g] 9, 11, 12, 15, 21, and 22 have solutions that have partial acceptance, but there exists some controversy as to whether they resolve the problems.
Euclid's Elements was read by anyone who was considered educated in the West until the middle of the 20th century. [10] In addition to theorems of geometry, such as the Pythagorean theorem, the Elements also covers number theory, including a proof that the square root of two is irrational and a proof that there are infinitely many prime numbers.
Hilbert's tenth problem is the tenth on the list of mathematical problems that the German mathematician David Hilbert posed in 1900. It is the challenge to provide a general algorithm that, for any given Diophantine equation (a polynomial equation with integer coefficients and a finite number of unknowns), can decide whether the equation has a solution with all unknowns taking integer values.
This is perhaps the simplest known proof, requiring the least mathematical background. It is an attractive example of a combinatorial proof (a proof that involves counting a collection of objects in two different ways). The proof given here is an adaptation of Golomb's proof. [1] To keep things simple, let us assume that a is a positive integer.
Then mq + t = 10 × 3 + 91 = 121; this is divisible by 11 (with quotient 11), so 913 is also divisible by 11. As another example, to determine whether 689 = 10 × 68 + 9 is divisible by 53, find that m = (53 × 3 + 1) ÷ 10 = 16. Then mq + t = 16 × 9 + 68 = 212, which is divisible by 53 (with quotient 4); so 689 is also divisible by 53.