Are there infinitely many primes? ... (Three Proofs)   Written by Administrator Monday, 22 September 2008 07:27 A prime number is a number whose divisors are only 1 and the number itself. The numbers, 2, 3 and 5 and 101 are prime. The number 6 is not a prime because it admit divisors like 2 and 3, different than 6.Are there infinitely many primes? Yes! I will give three proof for this well-know fact. First proof (Euclid)This is the classical proof of infinite primes and use the absurd logical assertion, negating the thesis. So, we suppose there are a last prime number, called pn. Now, the number N,N = p1p2...pn + 1, is not in the list of previous primes and therefore the hypothesis is incorrect. So, there are infinitely many primes.Second proof (Original)I simply note that instead of "1", in the previous proof we can choose M, such that M belong to the sequence p1,p2,...pn. And in this case, N = p1p2...pn + M,we will be sure that N is a new number and prime. We can prove that there are blocks of number of any length without any prime inside. So, this proof is a little more precise, than Euclid proof.Third proof (Using Riemann Zeta Function)This proof includes also the proof of the Product as a representation of the Riemann Zeta Function.The Riemann Zeta Function, is defined as follows:Let be s = σ + it, where Re s > 1. So, ζ(s) = Σ n-swhere s runs from 1 to ∞Now, we will prove that there are another formula for the zeta function. In other words, we will prove,Theorem. For σ=Re s > 1,1/ ζ(s) = Π (1 - (pn)-s)where n runs from 1 to ∞.Proof. To prove this theorem we will use (without to prove it, the following theorem).(Theorem: A necessary and sufficient condition for the absolute convergence of the product Π (1 + an), from 1 to ∞, is the convergence of the serie Σ¦an¦, where n runs from 1 to ∞.)Therefore, according to the previous theorem the product converge uniformly for σ ≥ σ0 > 1, is the same is valid for the infinite serie Σ¦(pn)-s¦=Σ(pn)-σ.Since the latter term is obtained by omitting terms of Σ n-σ, its uniform convergence for σ ≥ σ0 is obvious.The uniform convergence assure us, that we can multiply the serie by numbers. Now, under the assumption σ > 1 it is seen at once that, ζ(s) (1-2-s)=Σ n-s - Σ (2n)-s =Σ m-swhere m runs thought the odd integers, that is numbers that does not accept 2 has a divisor.By the same reasoning, ζ(s) (1-2-s)(1-3-s) = Σ m-swhere this time m runs thought all integers that are neither divisible by 2 or 3. More generally, ζ(s) (1-2-s)(1-3-s) ... (1-pN-s) = Σ m-swhere the sum of the right beign over all integers that contain none of the prime factors 2,3, ..., pN. The first term in the sum is 1, and the next is pN+1-s. Therefore we conclude that all the terms except the first tends to zero as N ->∞, and we conclude that,lim ζ(s) Π (1 - (pn)-s) = 1.Now, we will use the last reasoning to prove that there are infinitely many primes. Suppose that pn is the latest prime. Then,  ζ(s) (1-2-s)(1-3-s) ... (1-pN-s) = Σ m-swould become,   ζ(s) (1-2-s)(1-3-s) ... (1-pN-s) = 1.and it would follows that ζ(s) has a finite limit where σ -> 1. This contradict that the harmonic serie, Σ n-1, diverge. Last Updated on Monday, 22 September 2008 09:21