Euclid's 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 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 p1p2, ..., pn. 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...pn. Let q = P + 1. Then q is either prime or not:

This proves that for every finite list of prime numbers there is a prime number not in the list.[4] 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.[5]

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,[6] though it is actually a proof by cases, a direct proof method. The philosopher Torkel 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 q1, ... qn are all the primes'. However, since this assumption isn't even used in the proof, the reformulation is pointless."[7]

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.[8]

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:

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:

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.

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

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 2k 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:

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 2kN⌋ such numbers, all different, and they include all the numbers in our list and maybe more. Therefore, 2kN⌋ ≥ N. Here, ⌊x⌋ denotes the floor function.

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

In the 1950s, Hillel Furstenberg introduced a proof by contradiction using point-set topology.[9]

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(ab) (for a ≠ 0), where

Then a contradiction follows from the property that a finite set of integers cannot be open and the property that the basis sets S(ab) are both open and closed, since

cannot be closed because its complement is finite, but is closed since it is a finite union of closed sets.

Let p1, ..., pN 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

In 2010, Junho Peter Whang published the following proof by contradiction.[11] Let k be any positive integer. Then according to de Polignac's formula (actually due to Legendre)

(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.

Filip Saidak gave the following proof by construction, which does not use reductio ad absurdum[12] or Euclid's Lemma (that if a prime p divides ab then it must divide a or b).

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.

Alexander Shen and others[who?] have presented a proof that uses incompressibility:[14]

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

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

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.

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: