Stroke (engine)

Euclid's theorem is a fundamental statement in number theory that asserts that infinitely many numbers are prime. Several well-known proofs of the theorem exist.

Euclid's proof

Euclid offered the following proof published in his work Elements (Book IX, Proposition 20); the proof is here paraphrased.

For any finite list of prime numbers p₁, p₂, ..., p_n. at least one additional prime number exists. Let P be the product of all the prime numbers in the list: P = p₁p₂...p_n. Let q = P + 1. Therefore q is either prime or not:

• If q is prime, then at least one prime is not listed.

• 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 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.

Therefore for any finite list of prime numbers at least one additional prime number exists, and therefore infinitely many numbers are prime.

Euclid is often erroneously reported to have proved this result by 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. 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.

Euler's proof

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:

$\prod _{p\in P}\frac{1}{1-1/p}=\prod _{p\in P}\sum _{k\ge 0}\frac{1}{p^k}=\sum _n\frac{1}{n}.$

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:

$\prod _{p\in P}\sum _{k\ge 0}\frac{1}{p^k}=\sum _{k\ge 0}\frac{1}{2^k}\times \sum _{k\ge 0}\frac{1}{3^k}\times \sum _{k\ge 0}\frac{1}{5^k}\times \sum _{k\ge 0}\frac{1}{7^k}\times \cdots =\sum _n\frac{1}{n}$

in the result, every product of primes once appears; therefore by the fundamental theorem of arithmetic the sum equals the sum over all integers.

The sum on the right is the 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.

Erdős's proof

Paul Erdős gave a third proof that relies on the fundamental theorem of arithmetic. Every integer n can be uniquely written as

$r{s}^2$

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:

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 s² are both less than N. By the fundamental theorem of arithmetic, there are only 2^k 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^k√N; i.e.:

$2^k\sqrt{N}\ge N$

This inequality does not hold for N sufficiently large; therefore primes are infinitely many.

Furstenberg's proof

In the 1950s, Hillel Furstenberg introduced a proof using point-set topology. See Furstenberg's proof of the infinitude of primes.

Some recent proofs

Pinasco

Juan Pablo Pinasco has written the following proof.

Let p₁, ..., p_N 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

$\begin{array}{rl}1+\sum _i\lfloor \frac{x}{p_i}\rfloor -\sum _{i<j}\lfloor \frac{x}{p_i{p}_j}\rfloor & +\sum _{i<j<k}\lfloor \frac{x}{p_i{p}_j{p}_k}\rfloor -\cdots \\ & \cdots \pm (-1{)}^{N+1}\lfloor \frac{x}{p_1\cdots p_N}\rfloor .(1)\end{array}$

Divide by x and let x → ∞

$\sum _i\frac{1}{p_i}-\sum _{i<j}\frac{1}{p_i{p}_j}+\sum _{i<j<k}\frac{1}{p_i{p}_j{p}_k}-\cdots \pm (-1{)}^{N+1}\frac{1}{p_1\cdots p_N}.(2)$

Rewrite

$1-\prod _{i=1}^N(1-\frac{1}{p_i}).(3)$

If no other primes than p₁, ..., p_N exist, then the expression in (1) equals $\lfloor x\rfloor$ and the expression in (2) equals 1, but 1 exceeds the expression in (3). Therefore more primes than p₁, ..., p_N exist.

Whang

In 2010, Junho Peter Whang published the following proof by contradiction. Let k be any positive integer. Then according to de Polignac's formula (actually due to Legendre)

$k!=\prod _{p\text{ prime}}{p}^{f(p,k)}$

where

$f(p,k)=\lfloor \frac{k}{p}\rfloor +\lfloor \frac{k}{p^2}\rfloor +\cdots .$

$f(p,k)<\frac{k}{p}+\frac{k}{p^2}+\cdots =\frac{k}{p-1}\le k.$

If only finitely many primes exist, then

$\underset{k\to \mathrm{\infty }}{\lim }\frac{{\left(\prod _{p}p\right)}^k}{k!}=0,$

(the numerator of the fraction would grow singly exponentially whereas by Stirling's approximation the denominator grows more quickly than singly exponentially), contradicting the fact that for each k the numerator exceeds or equals the denominator.

Proof using the irrationality of π

Representing the Leibniz formula for π as an Euler product gives

$\frac{\pi }{4}=\frac{3}{4}\times \frac{5}{4}\times \frac{7}{8}\times \frac{11}{12}\times \frac{13}{12}\times \frac{17}{16}\times \frac{19}{20}\times \frac{23}{24}\times \frac{29}{28}\times \frac{31}{32}\times \cdots$

The numerators of this product are the prime numbers, and each denominator is the multiple of four nearest to the numerator.

If primes were finitely many, then this formula would show that Potter or Ceramic Artist Harry Rave from Cobden, spends time with hobbies for instance magic, property developers house in singapore singapore and fitness. Finds inspiration through travel and just spent 7 months at Keoladeo National Park. is rational, but Potter or Ceramic Artist Harry Rave from Cobden, spends time with hobbies for instance magic, property developers house in singapore singapore and fitness. Finds inspiration through travel and just spent 7 months at Keoladeo National Park. is not rational.

See also

• Dirichlet's theorem on arithmetic progressions
• Prime number theorem

Notes and references

43 year old Petroleum Engineer Harry from Deep River, usually spends time with hobbies and interests like renting movies, property developers in singapore new condominium and vehicle racing. Constantly enjoys going to destinations like Camino Real de Tierra Adentro.

External links

• 
  
  I had like 17 domains hosted on single account, and never had any special troubles. If you are not happy with the service you will get your money back with in 45 days, that's guaranteed. But the Search Engine utility inside the Hostgator account furnished an instant score for my launched website. Fantastico is unable to install WordPress in a directory which already have any file i.e to install WordPress using Fantastico the destination directory must be empty and it should not have any previous installation files. When you share great information, others will take note. Once your hosting is purchased, you will need to setup your domain name to point to your hosting. Money Back: All accounts of Hostgator come with a 45 day money back guarantee. If you have any queries relating to where by and how to use Hostgator Discount Coupon, you can make contact with us at our site. If you are starting up a website or don't have too much website traffic coming your way, a shared plan is more than enough. Condition you want to take advantage of the worldwide web you prerequisite a HostGator web page, -1 of the most trusted and unfailing web suppliers on the world wide web today. Since, single server is shared by 700 to 800 websites, you cannot expect much speed.
  
  
  
  Hostgator tutorials on how to install Wordpress need not be complicated, especially when you will be dealing with a web hosting service that is friendly for novice webmasters and a blogging platform that is as intuitive as riding a bike. After that you can get Hostgator to host your domain and use the wordpress to do the blogging. Once you start site flipping, trust me you will not be able to stop. I cut my webmaster teeth on Control Panel many years ago, but since had left for other hosting companies with more commercial (cough, cough) interfaces. If you don't like it, you can chalk it up to experience and go on. First, find a good starter template design. When I signed up, I did a search for current "HostGator codes" on the web, which enabled me to receive a one-word entry for a discount. Your posts, comments, and pictures will all be imported into your new WordPress blog.
• Euclid's Elements, Book IX, Prop. 20 (Euclid's proof, on David Joyce's website at Clark University)