Euclid's theorem

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Euclid's theorem

Field	Number theory
First proof by	Euclid
First proof in	c. 300 BCE
Generalizations	Dirichlet's theorem on arithmetic progressions Prime number theorem

Euclid's theorem is a fundamental statement in number theory that asserts that there are infinitely many prime numbers. It was first proved by Euclid in his work Elements. There are several proofs of the theorem.

Euclid's proof

Euclid offered a proof published in his work Elements (Book IX, Proposition 20), which is paraphrased here.

Consider any finite list of prime numbers p₁, p₂, ..., p_n. 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 = p₁p₂...p_n. Let q = P + 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 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 + 1 = q. If p divides P and q, then p would have to divide the difference of the two numbers, which is (P + 1) − P or just 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, and therefore there must be infinitely many prime numbers. Euclid is often erroneously reported to have proved this result by contradiction beginning with the assumption that the finite set initially considered contains all prime numbers. While such a proof does follow from Euclid's method, Euclid's proof deduces the infinitude directly.

Variations

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

The factorial n! of a positive integer n is divisible by every integer from 2 to n, as it is the product of all of them. Hence, 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 n! + 1 is either prime or divisible by a prime larger than n. In either case, for every positive integer n, there is at least one prime bigger than n. The conclusion is that the number of primes is infinite.

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:

${\displaystyle \prod _{p\in P}{\frac {1}{1-{\frac {1}{p}}}}=\prod _{p\in P}\sum _{k\geq 0}{\frac {1}{p^{k}}}=\sum _{n}{\frac {1}{n}}.}$

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 for the Riemann zeta function. To show this, distribute the product over the sum:

${\displaystyle {\begin{aligned}\prod _{p\in P}\sum _{k\geq 0}{\frac {1}{p^{k}}}&=\sum _{k\geq 0}{\frac {1}{2^{k}}}\times \sum _{k\geq 0}{\frac {1}{3^{k}}}\times \sum _{k\geq 0}{\frac {1}{5^{k}}}\times \sum _{k\geq 0}{\frac {1}{7^{k}}}\times \cdots \\[8pt]&=\sum _{k,\ell ,m,n,\ldots \geq 0}{\frac {1}{2^{k}3^{\ell }5^{m}7^{n}\cdots }}=\sum _{n}{\frac {1}{n}}\end{aligned}}}$

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, 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

${\displaystyle rs^{2}}$

where r is square-free, or not divisible by any square numbers (let s² be the largest square number that divides n and then let r = n/s²). Now suppose that there are only finitely many prime numbers and call the number of prime numbers k. As each of the prime numbers factorizes any squarefree number at most once, by the fundamental theorem of arithmetic, there are only 2^k square-free numbers. 

Now fix a positive integer N and consider the integers between 1 and N. Each of these numbers can be written as rs² where r is square-free and s² 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 ⌊√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^k⌊√N⌋ such numbers, all different, and they include all the numbers in our list and maybe more. Therefore, 2^k⌊√N⌋ ≥ N. Here, ⌊x⌋ denotes the floor function.

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

Furstenberg's proof

In the 1950s, Hillel Furstenberg introduced a proof by contradiction using point-set topology.

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

${\displaystyle S(a,b)=\{an+b\mid n\in \mathbb {Z} \}=a\mathbb {Z} +b.}$

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, b) are both open and closed, since

${\displaystyle \mathbb {Z} \setminus \{-1,+1\}=\bigcup _{p\mathrm {\,prime} }S(p,0)}$

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.

Let p₁, ..., p_N 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

${\displaystyle {\begin{aligned}1+\sum _{i}\left\lfloor {\frac {x}{p_{i}}}\right\rfloor -\sum _{i$

Dividing by x and letting x → ∞ gives 

${\displaystyle \sum _{i}{\frac {1}{p_{i}}}-\sum _{i$

 This can be written as

${\displaystyle 1-\prod _{i=1}^{N}\left(1-{\frac {1}{p_{i}}}\right).\qquad (3)}$

If no other primes than p₁, ..., p_N exist, then the expression in (1) is equal to ${\displaystyle \lfloor x\rfloor }$ and the expression in (2) is equal to 1, but clearly the expression in (3) is not equal to 1. Therefore, there must be more primes than  p₁, ..., p_N.

Proof using de Polignac's formula

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) 

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

where

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

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

But if only finitely many primes exist, then

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

(the numerator of the fraction would grow 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 or Euclid's Lemma (that if a prime p divides ab then it must divide a or b). 

Since each natural number (> 1) has at least one prime factor, and two successive numbers n and (n + 1) have no factor in common, the product n(n + 1) has more different prime factors than the number n itself.  So the chain of pronic numbers:

1×2 = 2 {2},    2×3 = 6 {2, 3},    6×7 = 42 {2,3, 7},    42×43 = 1806 {2,3,7, 43},    1806×1807 = 3263443 {2,3,7,43,13,139}, · · ·
 

provides a sequence of unlimited growing sets of primes.

Proof using the irrationality of π

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Representing the Leibniz formula for π as an Euler product gives

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

The numerators of this product are the 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 π 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 π is irrational.

Proof using information theory

Alexander Shen and others have presented a proof that uses incompressibility:

Suppose there were only k primes (p₁... p_k). By the fundamental theorem of arithmetic, any positive integer n could then be represented as:

${\displaystyle n={p_{1}}^{e_{1}}{p_{2}}^{e_{2}}\cdots {p_{k}}^{e_{k}},}$

where the non-negative integer exponents e_i together with the finite-sized list of primes are enough to reconstruct the number. Since ${\displaystyle p_{i}\geq 2}$ for all i, it follows that all ${\displaystyle e_{i}\leq \lg n}$ (where ${\displaystyle \lg }$ denotes the base-2 logarithm). 

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

${\displaystyle O({\text{prime list size}}+k\lg \lg n)=O(\lg \lg n)}$ bits.

This is a much more efficient encoding than representing n directly in binary, which takes ${\displaystyle N=O(\lg n)}$ bits. An established result in 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 ${\displaystyle \lg \lg n=o(\lg n)}$.

Therefore, the number of primes must not be finite.

A generalization: Dirichlet's theorem on arithmetic progressions

Dirichlet's theorem states that for any two positive coprime integers a and d, there are infinitely many primes of the form a + nd, where n is also a positive integer. In other words, there are infinitely many primes that are congruent to a 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

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

${\displaystyle \lim _{x\rightarrow \infty }{\frac {\pi (x)}{x/\ln(x)}}=1.}$

Using asymptotic notation this result can be restated as

${\displaystyle \pi (x)\sim {\frac {x}{\log x}}.}$

This yields Euclid's theorem, since ${\displaystyle \lim _{x\rightarrow \infty }{\frac {x}{\log x}}=\infty .}$