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mathfan文库  否定一个数学证明

必被实筛.现观察闭区间B: 内的奇数.
(1).  故知:闭区间B内的奇数均小于
(2)∵  ∴

故知闭区间B内必含的奇倍数
又因实筛,故知,必被实筛而被虚筛.
(3)与相邻的的奇倍数为:.且∵
故知:
 .
 .
故知,闭区间B内有且仅有一个的奇倍数,且为虚筛.
且因

故知,在闭区间  内至少有个同时又属于闭区间B内的受虚筛的连续奇数.故由推论四知,在这个大于,小于的连续奇数中,至少有二个,至少有一个
故知: 当时,
由1、2、知:
由一、二、知:当时,
故由Ⅰ.Ⅱ.知:当时,.(证毕)

参考文献:
①张 忠     在间至少有二个素数.连云港师范高等专科学校学报.2001.4.
②闵嗣鹤、严士健     初等数论[M]北京.人民教育出版社出版  1982.
③张 忠  论同余式方程组正整数解的构造式[ J ],南通职业大学学报,2000.14(4).
④张 忠  谈谈素数分布的两个规律[ J ] 《中华学术论坛》2003. 9总第六期

Elementary discussion on the two laws of the distribution of prime numbers
ZHANG  Zhong
(Nantong South-City Zhong-Yian Bookshop    226001 NT Jiangsu,China)

Abstract: In this paper, we use the theory of congruence, group the integer segment () which consist of  ordered continuous integers intoresidue segment classes of module , and also use the n-dimensional sieve method, analyse the segment  which in the minimal nonnegative complete system of residue segment classes of module  real-sieved or false-sieved by . And then,we obtained the conclusions: there are at least two numbers prime to among  continuous integers, and from this,obtain the two laws of the distribution of prime numbers: 1. there are at least two prime numbers between  and ; 2.the difference of a prime number and its successor  .

 



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