## About the formalization of some results by Chebyshev in number theory (2009)

Venue: | Proceedings of TYPES’08, Vol. 5497 of LNCS |

Citations: | 8 - 4 self |

### BibTeX

@INPROCEEDINGS{Asperti09aboutthe,

author = {Andrea Asperti and Wilmer Ricciotti and Mura Anteo Zamboni},

title = {About the formalization of some results by Chebyshev in number theory},

booktitle = {Proceedings of TYPES’08, Vol. 5497 of LNCS},

year = {2009},

publisher = {Springer-Verlag}

}

### OpenURL

### Abstract

Abstract. We discuss the formalization, in the Matita Interactive Theorem Prover, of a famous result by Chebyshev concerning the distribution of prime numbers, essentially subsuming, as a corollary, Bertrand’s postulate. Even if Chebyshev’s result has been later superseded by the stronger prime number theorem, his machinery, and in particular the two functions ψ and θ still play a central role in the modern development of number theory. Differently from other recent formalizations of other results in number theory, our proof is entirely arithmetical. It makes use of most part of the machinery of elementary arithmetics, and in particular of properties of prime numbers, factorization, products and summations, providing a natural benchmark for assessing the actual development of the arithmetical knowledge base. 1

### Citations

928 |
An Introduction to the Theory of Numbers
- Hardy, Wright
- 1980
(Show Context)
Citation Context ... is sensibly simpler than the prime number theorem, already formalized by Avigad et al. in Isabelle [3] and by Harrison in HOL Light [5], it is far form trivial (in Hardy and Wright’s famous textbook =-=[7]-=-, it takes pages 340-344 of chapter 22). In particular, our point was to give a fully arithmetical (and constructive) proof of this theorem. Even if Selberg’s proof of the prime number theorem is “ele... |

476 |
Introduction to Analytic Number Theory
- Apostol
- 1979
(Show Context)
Citation Context ...prime numbers. The idea is that the numbers 1, 2,...,n include just n p multiples of p, n p2 multiples of p2 , an so on. Hence (the variable bound by the product is written in bold) n! = ∏ ∏ p n/pi+1 =-=(1)-=- p≤n i<logp n The previous one is a good example of a typical mathematical argumentation (see e.g. [7], p. 342). Looking more carefully, you see that it provides you (almost) no information, since it ... |

24 | Paul(2007) ‘A formally verified proof of the prime number theorem
- Avigad, Donnelly, et al.
(Show Context)
Citation Context ...constants c1 and c2 such that, for any n n n c1 ≤ π(n) ≤ c2 log(n) log n Even if Chebyshev’s theorem is sensibly simpler than the prime number theorem, already formalized by Avigad et al. in Isabelle =-=[3]-=- and by Harrison in HOL Light [5], it is far form trivial (in Hardy and Wright’s famous textbook [7], it takes pages 340-344 of chapter 22). In particular, our point was to give a fully arithmetical (... |

15 |
Beweis eines Satzes von Tschebyschef
- Erdős
- 1930
(Show Context)
Citation Context ...estimates, we could only prove the existence of a prime number between n and 5n, fornsufficiently large. There exists however an alternative approach to the proof of Bertrand’s postulate due to Erdös =-=[4]-=- (see also [7], p. 344) that is well suited to a formal encoding in arithmetics2 . Let k(n, p) = ∑ (n/p i+1 mod 2) Then, B can also be written as i<log p n B(n) = ∏ p k(n,p) p≤n We now split this prod... |

13 |
The Prime Numbers and Their Distribution
- France, Tenenbaum
(Show Context)
Citation Context ... far as possible. The relation between Ψ and π should be clear: Ψ(n) = ∏ p logp n ≤ ∏ n = n π(n) (9) p≤n p≤n Since moreover, n<a log a n+1 we also have n<a 2log a n , so that, easily, n π(n) ≤ Ψ(n) 2 =-=(10)-=- Let us now rewrite Ψ(n) in the following equivalent form: Ψ ′ (n) = ∏ ∏ p p≤n i<logp n It is then clear that, for any n, B(n) ≤ Ψ ′ (n) =Ψ(n) Hence, the lower bound for B immediately gives a lower bo... |

12 |
The prime number theorem
- Jameson
- 2003
(Show Context)
Citation Context ...(2n + 1)(2n)! ≤ (2n + 2)(2n +1)2 2n−2 n! 2 ≤ (2n + 2)(2n +2)2 2n−2 n! 2 =2 2n (n +1)! 2 So, by equation (4), we conclude that, for any n B(2n) ≤ 2 2n−1 (5) and when n is larger than 4, B(2n) ≤ 2 2n−2 =-=(6)-=- Similarly, we prove that, for any n>0, 2 2n n! 2 ≤ 2n(2n)!About the Formalization of Some Results by Chebyshev in Number Theory 23 The proof is by induction on n. Forn = 1 both sides reduce to 4. Fo... |

9 |
Formalizing an Analytic Proof of the Prime Number Theorem
- Harrison
- 2008
(Show Context)
Citation Context ...r any n n n c1 ≤ π(n) ≤ c2 log(n) log n Even if Chebyshev’s theorem is sensibly simpler than the prime number theorem, already formalized by Avigad et al. in Isabelle [3] and by Harrison in HOL Light =-=[5]-=-, it is far form trivial (in Hardy and Wright’s famous textbook [7], it takes pages 340-344 of chapter 22). In particular, our point was to give a fully arithmetical (and constructive) proof of this t... |

7 | A constructive and formal proof of Lebesgue’s dominated convergence theorem in the interactive theorem prover Matita
- Coen, C, et al.
- 2008
(Show Context)
Citation Context ...erian logarithm of Ψ, but as we mentioned in the introduction, we try to avoid the use of logarithms as far as possible. The relation between Ψ and π should be clear: Ψ(n) = ∏ p logp n ≤ ∏ n = n π(n) =-=(9)-=- p≤n p≤n Since moreover, n<a log a n+1 we also have n<a 2log a n , so that, easily, n π(n) ≤ Ψ(n) 2 (10) Let us now rewrite Ψ(n) in the following equivalent form: Ψ ′ (n) = ∏ ∏ p p≤n i<logp n It is th... |

6 |
A page in number theory
- Asperti, Armentano
(Show Context)
Citation Context ...alization of proofs of not trivial complexity (see . for another recent formalization effort). Although the Matita arithmetical library was already well developed at the time we started the work (see =-=[2]-=-), several integrations were required, concerning the following subjects: – logarithms, square root (632 lines) – inequalities involving integer division (339 lines) – magnitude of functions (255 line... |

5 |
Proving pearl: Knuth’s algorithm for prime numbers
- Théry
- 2003
(Show Context)
Citation Context ...rime between n and 2n. Moreover,if 2n 3 <p≤ n, then2n/p = 2 and for i>1and n ≥ 62n/pi =0since ( ) 2 2n 2n ≤ ≤ p 3 i 2 Erdös’ argument was already exploited by Théry in his proof of Bertrand postulate =-=[11]-=-; however he failed to provide a fully arithmetical proof, being forced to make use of the (classical, axiomatic) library of Coq reals to solve the remaining inequalities. Similarly, Riccardi’s formal... |

2 | Pocklington’s theorem and bertrand’s postulate
- Riccardi
(Show Context)
Citation Context ...ithmetical proof, being forced to make use of the (classical, axiomatic) library of Coq reals to solve the remaining inequalities. Similarly, Riccardi’s formalization of Bertrand’s postulate in Mizar =-=[8]-=- makes an essential use of real numbers.26 A. Asperti and W. Ricciotti so k(2n, p) = 0. Summing up, under the assumption that Bertrand postulate is false, B1(2n) = ∏ p p ≤ 2n k(2n, p) =1 = ∏ p≤2n/3 p... |