Euclid's theorem

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

Euclid's proof

Euclid offered a proof published in his work Elements (Book IX, Proposition 20),^[1] which is paraphrased here.^[2]

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 than is in the list.
• 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 (since 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 have to divide the difference^[3] of the two numbers, which is (P + 1) − P or just 1. Since no prime number divides 1, this would be a contradiction and so 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 on 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 set initially considered contains all prime numbers, or that it contains precisely the n smallest primes, rather than any arbitrary finite set of primes.^[4] Although the proof as a whole is not by contradiction (it does not assume that only finitely many primes exist), a proof by contradiction is within it, which is that none of the initially considered primes can divide the number q above.

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:

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

The first equality is given by the formula for a geometric series in each term of the product. To show the second equality, distribute the product over the sum:

$\prod_{p\in P} \sum_{k\geq 0} \frac{1}{p^k}=\sum_{k\geq 0} \frac{1}{2^k}\times\sum_{k\geq 0} \frac{1}{3^k}\times\sum_{k\geq 0} \frac{1}{5^k}\times\sum_{k\geq 0} \frac{1}{7^k}\times\cdots=\sum_{k,l,m,n,\cdots \geq 0} \frac{1}{2^k 3^l 5^m 7^n \cdots}=\sum_n\frac{1}{n}$

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 note that every integer n can be uniquely written as

$rs^2$

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 (see Combination#Number of k-combinations for all k).

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 floor(√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^kfloor(√N) such numbers, all different, and they include all the numbers in our list and maybe more. Therefore 2^kfloor(√N)$\ge$N.

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 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.^[5]

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

$\begin{align} 1 + \sum_{i} \left\lfloor \frac{x}{p_i} \right\rfloor - \sum_{i < j} \left\lfloor \frac{x}{p_i p_j} \right\rfloor & + \sum_{i < j < k} \left\lfloor \frac{x}{p_i p_j p_k} \right\rfloor - \cdots \\ & \cdots \pm (-1)^{N+1} \left\lfloor \frac{x}{p_1 \cdots p_N} \right\rfloor. \qquad (1) \end{align}$

Dividing by x and letting x → ∞ gives

$\sum_{i} \frac{1}{p_i} - \sum_{i < j} \frac{1}{p_i p_j} + \sum_{i < j < k} \frac{1}{p_i p_j p_k} - \cdots \pm (-1)^{N+1} \frac{1}{p_1 \cdots p_N}. \qquad (2)$

This can be written as

$1 - \prod_{i=1}^N \left( 1 - \frac{1}{p_i} \right). \qquad (3)$

If no other primes than p₁, ..., p_N exist, then the expression in (1) is equal to $\lfloor x \rfloor$ and the expression in (2) is equal to 1, but clearly the expression in (3) exceeds 1. Therefore there must be more primes than  p₁, ..., p_N.

Whang

In 2010, Junho Peter Whang published the following proof by contradiction.^[6] 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

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

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

But if only finitely many primes exist, then

$\lim_{k\to\infty} \frac{\left(\prod_p p\right)^k}{k!} = 0,$

(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.

Saidak

Filip Saidak gave the following proof which does not use reductio ad absurdum:.^[7] It also does not use Euclid's Lemma that if a prime p divides ab then it must divide a or b.

For any n > 1, n and n+1 have no common factors, they are coprime.

Therefore N₂=n(n+1) must have at least two different prime factors.

N₂ and N₂+1 are coprime and N₂ has at least two prime factors

Therefore N₃=N₂(N₂+1) has at least three prime factors.

By induction for any k, N_k=N_k-1(N_k-1+1) has at least k prime factors.

Therefore there is an infinity of primes.

Thus for example 2×(2+1)×(2(2+1)+1)×((2×(2+1)×(2(2+1)+1))+1) = 1806 has at least four prime factors.

Proof using the irrationality of π

Representing the Leibniz formula for π as an Euler product gives^[8]

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

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 factorials

Assume that the number of prime numbers is finite. There is thus an integer, p which is the largest prime.

p! (p-factorial) is divisible by every integer from 2 to p, as it is the product of all of them. Hence, p! + 1 is not divisible by every integer from 2 to p (it gives a remainder of 1 when divided by each). p! + 1 is therefore either prime or is divisible by a prime larger than p.

This contradicts the assumption that p is the largest prime. The conclusion is that the number of primes is infinite.^[9]

See also

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

Notes and references

1. James Williamson (translator and commentator), The Elements of Euclid, With Dissertations, Clarendon Press, Oxford, 1782, page 63.
2. The exact formulation of Euclid's assertion is: "The prime numbers are more numerous than any proposed multitude of prime numbers". In the proof, from the assumption that there are at least three prime numbers, Euclid deduces the existence of a fourth one.
3. In general, for any integers a, b, c if $a \mid b$ and $a \mid c$, then $a \mid (b - c)$. For more information, see Divisibility.
4. Michael Hardy and Catherine Woodgold, "Prime Simplicity", Mathematical Intelligencer, volume 31, number 4, fall 2009, pages 44–52.
5. Juan Pablo Pinasco, "New Proofs of Euclid's and Euler's theorems", American Mathematical Monthly, volume 116, number 2, February, 2009, pages 172–173.
6. Junho Peter Whang, "Another Proof of the Infinitude of the Prime Numbers", American Mathematical Monthly, volume 117, number 2, February 2010, page 181.
7. Saidak, Filip (December 2006). "A New Proof of Euclid's Theorem". American Mathematical Monthly 113 (10).
8. Debnath, Lokenath (2010), The Legacy of Leonhard Euler: A Tricentennial Tribute, World Scientific, p. 214, ISBN 9781848165267.
9. Further Pure Mathematics, L Bostock, F S Chandler and C P Rourke

External links

• Weisstein, Eric W., "Euclid's Theorem", MathWorld.
• Euclid's Elements, Book IX, Prop. 20 (Euclid's proof, on David Joyce's website at Clark University)