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| <!-- [[File:Pacioli.jpg|thumb|300px|[[Jacopo de' Barbari]] (?). ''A Portrait of [[Luca Pacioli]]'' (1495) with scheme of Euclid's theorem (left)]] -->
| | The writer is called Irwin. For years I've been working as a payroll clerk. For years he's been living in North Dakota and his family members loves it. One of the extremely best things in the world for me is to do aerobics and I've been doing it for quite a while.<br><br>Feel free to surf to my page :: [http://www.animecontent.com/user/QKPA over the counter std test] |
| '''Euclid's theorem''' is a fundamental statement in [[number theory]] that asserts that [[Infinite set|infinitely]] many numbers are [[prime number|prime]]. Several well-known proofs of the [[theorem]] exist.
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| ==Euclid's proof==
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| [[Euclid]] offered the following proof published in his work ''[[Euclid's Elements|Elements]]'' (Book IX, Proposition 20)<ref>James Williamson (translator and commentator), ''The Elements of Euclid, With Dissertations'', [[Clarendon Press]], Oxford, 1782, page 63.</ref>; the proof is here paraphrased.
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| For any finite list of prime numbers ''p''<sub>1</sub>, ''p''<sub>2</sub>, ..., ''p''<sub>''n''</sub>. at least one additional prime number exists. Let ''P'' be the product of all the prime numbers in the list: ''P'' = ''p''<sub>1</sub>''p''<sub>2</sub>...''p''<sub>''n''</sub>. Let ''q'' = ''P'' + 1. Therefore ''q'' is either prime or not: | |
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| *If ''q'' is prime, then at least one prime is not listed.
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| *If ''q'' is not prime, then some [[prime factor]] ''p'' divides ''q''. If this factor ''p'' were on our list, then it would divide ''P'' (because ''P'' is the product of every number on the list); but ''p'' divides ''P'' + 1 = ''q''. If ''p'' divides ''P'' and ''q,'' then ''p'' would divide the difference<ref>In general, for any integers a, b, c if <math>a \mid b</math> and <math>a \mid c</math>, then <math> a \mid (b - c)</math>. For more information, see [[Divisibility]].</ref> of the two numbers, which is (''P'' + 1) − ''P'' or just 1, but no prime number divides 1, and therefore ''p'' cannot be on the list and at least one more prime number is not listed.
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| Therefore for any finite list of prime numbers at least one additional prime number exists, and therefore infinitely many numbers are prime.
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| Euclid is often erroneously reported to have proved this result by [[proof by contradiction|contradiction]], beginning by assuming that the set initially considered contains all prime numbers, or that it contains precisely the ''n'' smallest primes, rather than any arbitrary finite set of primes.<ref>Michael Hardy and Catherine Woodgold, "Prime Simplicity", ''[[Mathematical Intelligencer]]'', volume 31, number 4, fall 2009, pages 44–52.</ref> Although the proof as a whole is not by contradiction, insofar as it begins without assuming that only finitely many primes exist, a proof by contradiction is within it: that none of the initially considered primes can divide ''q''.
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| ==Euler's proof==
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| Another proof, by the Swiss mathematician [[Leonhard Euler]], relies on the [[fundamental theorem of arithmetic]]: that every integer has a unique prime factorization. If ''P'' is the set of all prime numbers, Euler wrote that:
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| : <math>\prod_{p\in P} \frac{1}{1-1/p}=\prod_{p\in P} \sum_{k\geq 0} \frac{1}{p^k}=\sum_n\frac{1}{n}.</math>
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| The first equality is given by the formula for a [[geometric series]] in each term of the product. To show the second equality, distribute the product over the sum:
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| : <math>\prod_{p\in P} \sum_{k\geq 0} \frac{1}{p^k}=\sum_{k\geq 0} \frac{1}{2^k}\times\sum_{k\geq 0} \frac{1}{3^k}\times\sum_{k\geq 0} \frac{1}{5^k}\times\sum_{k\geq 0} \frac{1}{7^k}\times\cdots=\sum_n\frac{1}{n}</math>
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| in the result, every product of primes once appears; therefore by the fundamental theorem of arithmetic the sum equals the sum over all integers.
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| The sum on the right is the [[harmonic series (mathematics)|harmonic series]], which diverges; therefore the product on the left must also diverge. Each term of the product is finite; therefore the number of terms is infinite; therefore primes are infinitely many.
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| ==Erdős's proof==
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| [[Paul Erdős]] gave a third proof that relies on the fundamental theorem of arithmetic. Every integer ''n'' can be uniquely written as
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| : <math>rs^2</math>
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| where ''r'' is ''square-free'' - not divisible by any square numbers (let ''s²'' be the largest square number that divides ''n'' and then let ''r=n/s²''). Suppose that a finite number ''k'' of prime numbers exists:
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| Fix a positive integer ''N'' and try to count the number of integers between 1 and ''N''. Each of these numbers can be written as ''rs²'' where ''r'' is square-free and ''r'' and ''s<sup>2</sup>'' are both less than ''N''. By the fundamental theorem of arithmetic, there are only ''2<sup>k</sup>'' square-free numbers ''r'' (see [[Combination#Number of k-combinations for all k]]) as each of the prime numbers factorizes ''r'' at most once, and we must have ''s<√N''. So the total number of integers less than ''N'' is at most ''2<sup>k</sup>√N''; i.e.:
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| : <math>2^k\sqrt N\ge N</math>
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| This inequality does not hold for ''N'' sufficiently large; therefore primes are infinitely many.
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| ==Furstenberg's proof==
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| In the 1950s, [[Hillel Furstenberg]] introduced a proof using [[point-set topology]]. See [[Furstenberg's proof of the infinitude of primes]].
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| ==Some recent proofs==
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| ===Pinasco===
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| Juan Pablo Pinasco has written the following proof.<ref>Juan Pablo Pinasco, "New Proofs of Euclid's and Euler's theorems", ''[[American Mathematical Monthly]]'', volume 116, number 2, February, 2009, pages 172–173.</ref>
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| Let ''p''<sub>1</sub>, ..., ''p''<sub>''N''</sub> be the smallest ''N'' primes. By the [[inclusion–exclusion principle]], the number of positive integers that are divisible by one of those primes and that ''x'' equals or exceeds is
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| : <math>
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| \begin{align}
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| 1 + \sum_{i} \left\lfloor \frac{x}{p_i} \right\rfloor - \sum_{i < j} \left\lfloor \frac{x}{p_i p_j} \right\rfloor & + \sum_{i < j < k} \left\lfloor \frac{x}{p_i p_j p_k} \right\rfloor - \cdots \\
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| & \cdots \pm (-1)^{N+1} \left\lfloor \frac{x}{p_1 \cdots p_N} \right\rfloor. \qquad (1)
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| \end{align}
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| </math>
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| Divide by ''x'' and let ''x'' → ∞
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| : <math> \sum_{i} \frac{1}{p_i} - \sum_{i < j} \frac{1}{p_i p_j} + \sum_{i < j < k} \frac{1}{p_i p_j p_k} - \cdots \pm (-1)^{N+1} \frac{1}{p_1 \cdots p_N}. \qquad (2) </math>
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| Rewrite
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| : <math> 1 - \prod_{i=1}^N \left( 1 - \frac{1}{p_i} \right). \qquad (3) </math>
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| If no other primes than ''p''<sub>1</sub>, ..., ''p''<sub>''N''</sub> exist, then the expression in (1) equals <math>\lfloor x \rfloor </math> and the expression in (2) equals 1, but 1 exceeds the expression in (3). Therefore more primes than ''p''<sub>1</sub>, ..., ''p''<sub>''N''</sub> exist.
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| ===Whang===
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| In 2010, Junho Peter Whang published the following proof by contradiction.<ref>Junho Peter Whang, "Another Proof of the Infinitude of the Prime Numbers", ''[[American Mathematical Monthly]]'', volume 117, number 2, February 2010, page 181.</ref> Let ''k'' be any positive integer. Then according to [[de Polignac's formula]] (actually due to [[Adrien-Marie Legendre|Legendre]])
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| : <math> k! = \prod_{p\text{ prime}} p^{f(p,k)} \, </math>
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| where
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| : <math> f(p,k) = \left\lfloor \frac{k}{p} \right\rfloor + \left\lfloor \frac{k}{p^2} \right\rfloor + \cdots. </math>
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| : <math> f(p,k) < \frac{k}{p} + \frac{k}{p^2} + \cdots = \frac{k}{p-1} \le k. </math> | |
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| If only finitely many primes exist, then
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| : <math> \lim_{k\to\infty} \frac{\left(\prod_p p\right)^k}{k!} = 0, </math> | |
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| (the numerator of the fraction would grow [[Exponential growth|singly exponentially]] whereas by [[Stirling's approximation]] the denominator grows more quickly than singly exponentially),
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| contradicting the fact that for each ''k'' the numerator exceeds or equals the denominator.
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| ==Proof using the irrationality of π==
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| Representing the [[Leibniz formula for π]] as an [[Euler product]] gives<ref>{{citation|title=The Legacy of Leonhard Euler: A Tricentennial Tribute|first=Lokenath|last=Debnath|publisher=World Scientific|year=2010|isbn=9781848165267|page=214|url=http://books.google.com/books?id=K2liU-SHl6EC&pg=PA214}}.</ref>
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| :<math> \frac{\pi}{4} = \frac{3}{4} \times \frac{5}{4} \times \frac{7}{8} \times \frac{11}{12} \times \frac{13}{12} \times \frac{17}{16} \times \frac{19}{20} \times \frac{23}{24} \times \frac{29}{28} \times \frac{31}{32} \times \cdots \; </math>
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| The numerators of this product are the prime numbers, and each denominator is the multiple of four nearest to the numerator.
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| If primes were finitely many, then this formula would show that {{pi}} is rational, but {{pi}} is not rational.
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| ==See also==
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| * [[Dirichlet's theorem on arithmetic progressions]]
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| * [[Prime number theorem]]
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| ==Notes and references==
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| {{reflist}}
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| ==External links==
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| * {{MathWorld|urlname=EuclidsTheorems|title=Euclid's Theorem}}
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| * [http://aleph0.clarku.edu/~djoyce/java/elements/bookIX/propIX20.html Euclid's Elements, Book IX, Prop. 20] (Euclid's proof, on David Joyce's website at Clark University)
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| [[Category:Articles containing proofs]]
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| [[Category:Theorems about prime numbers]]
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