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 p_{n}. Now, the number N, N = p_{1}p_{2}...p_{n} + 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 p_{1,}p_{2,}...p_{n}. And in this case, N = p_{1}p_{2}...p_{n} + 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^{-s}
where 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 - (p_{n})^{-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 + a_{n}), from 1 to ∞, is the convergence of the serie Σ¦a_{n}¦, where n runs from 1 to ∞.) Therefore, according to the previous theorem the product converge uniformly for σ ≥ σ_{0} > 1, is the same is valid for the infinite serie Σ¦(p_{n})^{-s}¦=Σ(p_{n})^{-σ}. Since the latter term is obtained by omitting terms of Σ n^{-σ}, its uniform convergence for σ ≥ σ_{0 }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^{-s} where 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^{-s} where 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-p_{N}^{-s}) = Σ m^{-s} where the sum of the right beign over all integers that contain none of the prime factors 2,3, ..., p_{N}. The first term in the sum is 1, and the next is p_{N+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 - (p_{n})^{-s}) = 1. Now, we will use the last reasoning to prove that there are infinitely many primes. Suppose that p_{n }is the latest prime. Then, ζ(s) (1-2^{-s})(1-3^{-s}) ... (1-p_{N}^{-s}) = Σ m^{-s} would become, ζ(s) (1-2^{-s})(1-3^{-s}) ... (1-p_{N}^{-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. Олександр Фільчаков прокуроркупить хороший игровой ноутбукотдых в карпатах летом с детьми отзывыбуковель жилье частный сектордверь касапорте ливорно 07 купитьсобираем игровой компьютер 2016отбеливание зубов ценамарки детского питанияноутбук асус цена украинакупить литые диски r16приготовление блюд с помощью блендеракомпромат Данильченко Юрий Брониславович |