View source for Euclid's theorem

{{About|the theorem on the infinitude of prime numbers|the theorem on perfect numbers and Mersenne primes|Euclid–Euler theorem}}
{{Infobox mathematical statement
| name = Euclid's theorem
| image = 
| caption = 
| field = [[Number theory]]
| conjectured by = 
| conjecture date =  
| first proof by = [[Euclid]]
| first proof date = {{circa|300 BCE}}
| generalizations = [[Dirichlet's theorem on arithmetic progressions]]<br>[[Prime number theorem]]
}}
'''Euclid's theorem''' is a fundamental statement in [[number theory]] that asserts that there are [[Infinite set|infinitely]] many [[prime number|prime]] numbers. It was first proved by [[Euclid]] in his work ''[[Euclid's Elements|Elements]]''. There are several proofs of the theorem.

==Euclid's proof==
Euclid offered a proof published in his work ''Elements'' (Book IX, Proposition 20),<ref>James Williamson (translator and commentator), ''The Elements of Euclid, With Dissertations'', [[Clarendon Press]], Oxford, 1782, page 63.</ref> which is paraphrased here.<ref>{{citation|first=O.|last=Ore|title=Number Theory and its History|year=1988|origyear=1948|publisher=Dover|page=65}}</ref>

Consider any finite list of prime numbers ''p''<sub>1</sub>,&nbsp;''p''<sub>2</sub>,&nbsp;...,&nbsp;''p''<sub>''n''</sub>. It will be shown that at least one additional prime number not in this list exists. Let ''P'' be the product of all the prime numbers in the list: ''P''&nbsp;=&nbsp;''p''<sub>1</sub>''p''<sub>2</sub>...''p''<sub>''n''</sub>. Let ''q''&nbsp;=&nbsp;''P''&nbsp;+&nbsp;1. Then ''q'' is either prime or not:

*If ''q'' is prime, then there is at least one more prime that is not in the list.
*If ''q'' is not prime, then some [[prime factor]] ''p'' divides&nbsp;''q''. If this factor ''p'' were in our list, then it would divide ''P'' (since ''P'' is the product of every number in the list); but ''p'' divides ''P''&nbsp;+&nbsp;1&nbsp;=&nbsp;''q''. If ''p'' divides ''P'' and ''q,'' then ''p'' would have to 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''&nbsp;+&nbsp;1)&nbsp;−&nbsp;''P'' or just&nbsp;1. Since no prime number divides 1, ''p'' cannot be on the list. This means that at least one more prime number exists beyond those in the list.

This proves that for every finite list of prime numbers there is a prime number not in the list.<ref>The exact formulation of Euclid's assertion is: "The prime numbers are more numerous than any proposed multitude of prime numbers".</ref> In the original work, as Euclid had no way of writing an arbitrary list of primes, he used a method that he frequently applied, that is, the method of generalizable example. Namely, he picks just three primes and using the general method outlined above, proves that he can always find an additional prime. Euclid presumably assumes that his readers are convinced that a similar proof will work, no matter how many primes are originally picked.<ref>{{citation|first=Victor J.|last=Katz|title=A History of Mathematics/ an Introduction|edition=2nd|publisher=Addison Wesley Longman|year=1998|page=87}}</ref>

Euclid is often erroneously reported to have [[proof by contradiction|proved this result by contradiction]] beginning with the assumption that the finite set initially considered contains all prime numbers,<ref>Michael Hardy and Catherine Woodgold, "Prime Simplicity", ''[[Mathematical Intelligencer]]'', volume 31, number 4, fall 2009, pages 44–52.</ref> though it is actually a [[proof by cases]], a [[direct proof]] method. Philosopher Franzén, in a book on logic, states, "Euclid's proof that there are infinitely many primes is not an indirect proof [...] The argument is sometimes formulated as an indirect proof by replacing it with the assumption 'Suppose {{math|''q''<sub>1</sub>, ... ''q''<sub>''n''</sub>}} are all the primes'. However, since this assumption isn't even used in the proof, the reformulation is pointless."<ref>{{citation|first=T.|last=Franzén|title=Inexhaustibility: A Non-exhaustive Treatment|year=2004|publisher=A K Peters, Ltd|page=101}}</ref>

===Variations===
Several variations on Euclid's proof exist, including the following:

The [[factorial]] ''n''<nowiki>!</nowiki> of a positive integer ''n'' is divisible by every integer from 2 to ''n'', as it is the product of all of them. Hence, {{nowrap|''n''! + 1}} is not divisible by any of the integers from 2 to ''n'', inclusive (it gives a remainder of 1 when divided by each). Hence {{nowrap|''n''! + 1}} is either prime or divisible by a prime larger than&nbsp;''n''. In either case, for every positive integer ''n'', there is at least one prime bigger than&nbsp;''n''. The conclusion is that the number of primes is infinite.<ref>{{Cite book|url=|title=Further Pure Mathematics|last=Bostock|first=Linda|last2=Chandler|first2=Suzanne|last3=Rourke|first3=C.|date=2014-11-01|publisher=Nelson Thornes|year=|isbn=9780859501033|location=|pages=168|language=en}}</ref>

==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:

: <math>\prod_{p\in P} \frac{1}{1-\frac{1}{p}}=\prod_{p\in P} \sum_{k\geq 0} \frac{1}{p^k}=\sum_n\frac{1}{n}.</math>

The first equality is given by the formula for a [[geometric series]] in each term of the product. The second equality is a special case of the [[Euler product]] formula [[Proof of the Euler product formula for the Riemann zeta function|for the Riemann zeta function]]. To show this, distribute the product over the sum:

: <math>
\begin{align}
\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 \\[8pt]
& =\sum_{k,\ell,m,n,\ldots \geq 0} \frac{1}{2^k 3^\ell 5^m 7^n \cdots}=\sum_n\frac{1}{n}
\end{align}
</math>

In the result, every product of primes appears exactly once and so by the fundamental theorem of arithmetic the sum is equal to the sum over all integers.

The sum on the right is the [[harmonic series (mathematics)|harmonic series]], which diverges. Thus the product on the left must also diverge. Since each term of the product is finite, the number of terms must be infinite; therefore, there is an infinite number of primes.

==Erdős's proof==

[[Paul Erdős]] gave a third proof that also relies on the fundamental theorem of arithmetic. First every integer ''n'' can be uniquely written as

: <math>rs^2</math>

where ''r'' is ''square-free'', or not divisible by any [[square number]]s (let ''s''<sup>2</sup> be the largest square number that divides ''n'' and then let ''r''&nbsp;=&nbsp;''n''/''s''<sup>2</sup>). Now suppose that there are only finitely many prime numbers and call the number of prime numbers&nbsp;''k''. As each of the prime numbers factorizes any squarefree number at most once, by the fundamental theorem of arithmetic, there are only ''2''<sup>''k''</sup> square-free numbers (see [[Combination#Number of k-combinations for all k|Combination#Number of ''k''-combinations for all ''k'']]).

Now fix a positive integer ''N'' and consider the integers between 1 and&nbsp;''N''.  Each of these numbers can be written as ''rs''<sup>2</sup> where ''r'' is square-free and ''s''<sup>2</sup> is a square, like this:

:  ( 1×1, 2×1, 3×1, 1×4, 5×1, 6×1, 7×1, 2×4, 1×9, 10×1, ...)

There are ''N'' different numbers in the list. Each of them is made by multiplying a squarefree number, by a square number that is ''N'' or less. There are {{floor|{{radic|''N''}}}} such square numbers. Then, we form all the possible products of all squares less than ''N'' multiplied by all squarefrees everywhere. There are exactly 2<sup>''k''</sup>{{floor|{{radic|''N''}}}} such numbers, all different, and they include all the numbers in our list and maybe more. Therefore, 2<sup>''k''</sup>{{floor|{{radic|''N''}}}}&nbsp;≥&nbsp;''N''.  Here, {{floor|''x''}} denotes the [[floor function]].

Since this inequality does not hold for ''N'' sufficiently large, there must be infinitely many primes.

==Furstenberg's proof==
{{main|Furstenberg's proof of the infinitude of primes}}
In the 1950s, [[Hillel Furstenberg]] introduced a proof by contradiction using [[point-set topology]].<ref>{{cite journal
| last = Furstenberg
| first = Harry
| authorlink = Hillel Furstenberg
| title = On the infinitude of primes
| journal = [[American Mathematical Monthly]]
| volume = 62
| year = 1955
| pages = 353
| doi = 10.2307/2307043
| jstor = 2307043
| issue = 5
|mr=0068566}}</ref>

Define a topology on the integers '''Z''', called the [[evenly spaced integer topology]], by declaring a subset ''U''&nbsp;⊆&nbsp;'''Z''' to be an [[open set]] [[if and only if]] it is either the [[empty set]], ∅, or it is a [[union (set theory)|union]] of arithmetic sequences ''S''(''a'',&nbsp;''b'') (for ''a''&nbsp;≠&nbsp;0), where

:<math>S(a, b) = \{ a n  + b \mid n \in \mathbb{Z} \} = a \mathbb{Z} + b. </math>

Then a contradiction follows from the property that a finite set of integers cannot be open and the property that the basis sets ''S''(''a'',&nbsp;''b'') are [[clopen set|both open and closed]], since
:<math>\mathbb{Z} \setminus \{ -1, + 1 \} = \bigcup_{p \mathrm{\, prime}} S(p, 0)</math>
cannot be closed because its complement is finite, but is closed since it is a finite union of closed sets.

==Some recent proofs==

===Proof using the inclusion-exclusion principle===

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>

Let ''p''<sub>1</sub>,&nbsp;...,&nbsp;''p''<sub>''N''</sub> be the smallest ''N'' primes. Then by the [[inclusion–exclusion principle]], the number of positive integers less than or equal to ''x'' that are divisible by one of those primes is

: <math>
\begin{align}
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 \\
& \cdots \pm (-1)^{N+1} \left\lfloor \frac{x}{p_1 \cdots p_N} \right\rfloor. \qquad (1)
\end{align}
</math>

Dividing by ''x'' and letting ''x''&nbsp;→&nbsp;∞ gives

: <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>

This can be written as

: <math> 1 - \prod_{i=1}^N \left( 1 - \frac{1}{p_i} \right). \qquad (3) </math>

If no other primes than ''p''<sub>1</sub>,&nbsp;...,&nbsp;''p''<sub>''N''</sub> exist, then the expression in (1) is equal to&nbsp;<math>\lfloor x \rfloor </math> and the expression in (2) is equal to&nbsp;1, but clearly the expression in (3) is not equal to 1. Therefore, there must be more primes than &nbsp;''p''<sub>1</sub>,&nbsp;...,&nbsp;''p''<sub>''N''</sub>.

===Proof using de Polignac's formula===

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&nbsp;181.</ref>  Let ''k'' be any positive integer.  Then according to [[de Polignac's formula]] (actually due to [[Adrien-Marie Legendre|Legendre]])

: <math> k! = \prod_{p\text{ prime}} p^{f(p,k)} </math>

where

: <math> f(p,k) = \left\lfloor \frac{k}{p} \right\rfloor + \left\lfloor \frac{k}{p^2} \right\rfloor + \cdots. </math>

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

But if only finitely many primes exist, then

: <math> \lim_{k\to\infty} \frac{\left(\prod_p p\right)^k}{k!} = 0, </math>

(the numerator of the fraction would grow [[Exponential growth|singly exponentially]] while by [[Stirling's approximation]] the denominator grows more quickly than singly exponentially),
contradicting the fact that for each ''k'' the numerator is greater than or equal to the denominator.

===Proof by construction===

Filip Saidak gave the following [[proof by construction]], which does not use [[reductio ad absurdum]]<ref>{{cite journal |title=A New Proof of Euclid's Theorem |url=http://fermatslibrary.com/s/a-new-proof-of-euclids-theorem |first=Filip |last=Saidak |date=December 2006 |work=[[American Mathematical Monthly]] |volume=113 |issue=10|doi=10.2307/27642094}}</ref> or Euclid's Lemma (that if a prime ''p'' divides ''ab'' then it must divide ''a'' or&nbsp;''b'').

Since each natural number (>&nbsp;1) has at least one prime factor, and two successive numbers ''n'' and (''n''&nbsp;+&nbsp;1) have no factor in common, the product ''n''(''n''&nbsp;+&nbsp;1) has more different prime factors than the number ''n'' itself. &nbsp;So the chain of [[Pronic number|pronic numbers]]:<br>1×2 = 2 {2}, &nbsp; &nbsp;2×3 = 6 {2, 3}, &nbsp; &nbsp;6×7 = 42 {2, 3, 7}, &nbsp; &nbsp;42×43 = 1806 {2, 3, 7, 43}, &nbsp; &nbsp;1806×1807 = 3263442 {2, 3, 7, 43, 13, 139}, · · ·<br>provides a sequence of unlimited growing sets of primes.

===Proof using the irrationality of {{pi}}===
{{OR section|date=October 2019}}
Representing the [[Leibniz formula for π|Leibniz formula for {{pi}}]] 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=https://books.google.com/books?id=K2liU-SHl6EC&pg=PA214}}.</ref>
:<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>
The numerators of this product are the odd prime numbers, and each denominator is the multiple of four nearest to the numerator.

If there were finitely many primes this formula would show that {{pi}} is a rational number whose denominator is the product of all multiples of 4 that are one more or less than a prime number, contradicting the fact that {{pi}} is [[Irrational number|irrational]].

===Proof using information theory===
Alexander Shen and others{{who|date=October 2019}} have presented a proof that uses [[Incompressibility method|incompressibility]]:<ref>{{citation|title=Kolmogorov complexity and algorithmic randomness|first=Alexander|last=Shen|publisher=AMS|year=2016|url=http://www.lirmm.fr/~ashen/kolmbook-eng.pdf|page=245}}</ref>

Suppose there were only ''k'' primes (''p''<sub>1</sub>...&nbsp;''p''<sub>k</sub>). By the [[fundamental theorem of arithmetic]], any positive integer ''n'' could then be represented as:

:<math>n = {p_1}^{e_1} {p_2}^{e_2} \cdots {p_k}^{e_k},</math>

where the non-negative integer exponents ''e''<sub>i</sub> together with the finite-sized list of primes are enough to reconstruct the number. Since <math>p_i \geq 2</math> for all ''i'', it follows that all <math>e_i \leq \lg n</math> (where <math>\lg</math> denotes the base-2 logarithm).

This yields an encoding for ''n'' of the following size (using [[big O notation]]):

:<math>O(\text{prime list size} + k \lg \lg n) = O(\lg \lg n)</math> bits.

This is a much more efficient encoding than representing ''n'' directly in binary, which takes <math>N = O(\lg n)</math> bits. An established result in [[Lossless compression#Limitations|lossless data compression]] states that one cannot generally compress ''N'' bits of information into less than ''N'' bits. The representation above violates this by far when ''n'' is large enough since <math>\lg \lg n = o(\lg n)</math>.

Therefore, the number of primes must not be finite.

==A generalization: Dirichlet's theorem on arithmetic progressions==
{{main article|Dirichlet's theorem on arithmetic progressions}}

Dirichlet's theorem states that for any two positive [[coprime]] [[integer]]s ''a'' and&nbsp;''d'', there are infinitely many [[prime number|primes]] of the form ''a''&nbsp;+&nbsp;''nd'', where ''n'' is also a positive integer. In other words, there are infinitely many primes that are [[congruence relation|congruent]] to ''a'' [[modular arithmetic|modulo]] ''d''. Euclid's theorem is a special case of Dirichlet's theorem for ''a = d = 1''. Every case of Dirichlet's theorem yields Euclid's theorem.

==A stronger result: the prime number theorem==
{{main article|Prime number theorem}}

Let {{math|''π''(''x'')}} be the [[prime-counting function]] that gives the number of primes less than or equal to {{mvar|x}}, for any real number&nbsp;{{mvar|x}}. The prime number theorem then states that {{math|''x'' / log ''x''}} is a good approximation to {{math|''π''(''x'')}}, in the sense that the [[limit of a function|limit]] of the ''quotient'' of the two functions {{math|''π''(''x'')}} and {{math|''x'' / log ''x''}} as {{mvar|x}} increases without bound is 1:

:<math>\lim_{x\rightarrow\infty} \frac{\pi(x)}{x/\ln(x)}=1. </math>

Using [[asymptotic notation]] this result can be restated as

:<math>\pi(x)\sim \frac{x}{\log x}.</math>

This yields Euclid's theorem, since <math>\lim_{x\rightarrow\infty} \frac{x}{\log x}=\infty. </math>

==Notes and references==
{{reflist}}

==External links==
* {{MathWorld|urlname=EuclidsTheorems|title=Euclid's Theorem}}
* [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)
* [https://arxiv.org/abs/1202.3670 125 proofs of Euclid's theorem]

[[Category:Articles containing proofs]]
[[Category:Theorems about prime numbers]]