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時(shi)間︰2020-04-01 00:49:41  作者︰科學革(ge)命(ming)家(jia)  來(lai)源(yuan)︰歆竹(zhu)苑文學網  閱(yue)讀︰2525  評論︰0
內容摘要︰---當今(jin)中國數學超(chao)級難題佘赤求發表了(liao)《哥德巴赫猜想證(zheng)明(ming)及其成敗(bai)原因》,其論證(zheng)原理方法(fa)如下。論證(zheng)戰(zhan)略方案篩除2n=a+b的集(ji)合xian)興、b同(tong)時(shi)和分別為(wei)合數和1的式(shi)子,?有余式(shi),哥德巴赫猜想成立。試問,有可(ke)能錯嗎(ma)?論證(zheng)戰(zhan)術(shu)辦法(fa)假定2n的小(xiao)于2n的平方根(gen)的質因數只(zhi)有2能夠同(tong)時(shi)整除a和......

---當今(jin)中國數學超(chao)級難題

       佘赤求發表了(liao)《哥德巴赫猜想證(zheng)明(ming)及其成敗(bai)原因》,其論證(zheng)原理方法(fa)如下。

論證(zheng)戰(zhan)略方案 篩除2n=a+b的集(ji)合xian)興、b同(tong)時(shi)和分別為(wei)合數和1的式(shi)子,?有余式(shi),哥德巴赫猜想成立。

       試問,有可(ke)能錯嗎(ma)?

論證(zheng)戰(zhan)術(shu)辦法(fa) 假定2n的小(xiao)于2n的平方根(gen)的質因數只(zhi)有2能夠同(tong)時(shi)整除a和b,其余式(shi)子再無兩合數和?

總式(shi)數-合數和式(shi)數下限-非合數和式(shi)數上限?素數和式(shi)數下限

      試問,有可(ke)能錯嗎(ma)?

計算方法(fa) 根(gen)據合數性質?a、b同(tong)時(shi)和分別為(wei)合數的式(shi)子,應用乘法(fa)分配律運算分別減去它們?

G(1+1)=[???[「n/2」1/3〕3/5〕???(pr-1-2)/pr-1〕(pr-2)/pr〕-s+b’-1 (或(huo)0)

              [...[「n/2」1/3〕3/5〕...(pr-1-2)/pr-1〕(pr-2)/pr〕-s-1

(Pr表2n方根(gen)內最大(da)質數。b’表不該減去的式(shi)子數目(mu)。s表取整運算誤差。每次(ci)舍成整數? [r/2 ]s, 0≦ b’≦ r-1,0表示1所在式(shi)另(ling)一(yi)數是合數)

       加大(da)保險(xian)求下限,pr不小(xiao)時(shi)可(ke)不管(guan)s大(da)小(xiao)減去其上限,再視b’為(wei)0 ?G(1+1)的下限。

       試問,有可(ke)能錯嗎(ma)?

決定這個公式(shi)生死的“細節” 該式(shi)存在以下質疑猜想不成立的問題。因此,數學界不認可(ke)。

1、按公式(shi)計算,某些大(da)偶數的“答案數”大(da)于實際,或(huo)大(da)于小(xiao)偶數的“答案數”,而實際shi)刃xiao)偶數少(shao),即(ji)“波(bo)動”反例。此前,解決這個問題就是做無米之炊(chui),有待于數學基礎理論知(zhi)識進步發展(zhan)。

2、不管(guan)多麼小(xiao),公式(shi)存在(前面(mian)已解決的)取整計算誤差。

      試問,有可(ke)能錯嗎(ma)?

化解波(bo)動策(ce)略 先(xian)找(zhao)米下鍋,創(chuang)新(xin)發現(xian)基礎理論知(zhi)識。推證(zheng)“自然數N值區(qu)間定理”“連續(xu)合數定理”?數列2n=r個由素數統轄(xia)的“2n值區(qu)間”。

       再“特別限定”取每個“2n值區(qu)間”的下限prpr+1代入公式(shi)計算?G(1+1)的“區(qu)間下限”。

       試問,有可(ke)能錯嗎(ma)?

      公式(shi)中,相鄰兩因數後(hou)一(yi)個數的分子或(huo)=或(huo)大(da)于遠dui)洞da)于前一(yi)個數的分母,pr小(xiao)于n?結論 

        每個“2n值區(qu)間”的“1+1”bi)shi)數的“區(qu)間下限”不僅不小(xiao)于1,而且(qie)r稍yuan)缶筒簧shao)于該偶數平方根(gen)內的奇素數個數,r越大(da)還 pr的一(yi)半(ban)(甚至于大(da)于pr?)。有合數和1的式(shi)子已經減完?同(tong)一(yi)區(qu)間的偶數的“1+1”bi)shi)數比其區(qu)間下限只(zhi)多不少(shao)?作者不僅證(zheng)明(ming)了(liao)“1+1”,而且(qie)大(da)大(da)改進了(liao)該猜想、將(jiang)其逼(bi)近于實際。

       試問,有可(ke)能錯嗎(ma)? 誰能坐實論證(zheng)錯誤,否認佘赤求大(da)功告成?!

       然而,某些判官判斬論文的理由令人啼笑皆非。

       其一(yi) 作者學歷太低,文章短小(xiao),不可(ke)能解答個世界難題。作者認為(wei)這類(lei)意見(jian)與評判xin)壑zheng)錯誤風馬(ma)牛不相及。

       其二 “兩個定理”bi)譴da)家(jia)都(du)知(zhi)道的常識,沒有學術(shu)價值意義。“區(qu)間下限”設(she)定莫名其妙(有判官據此篡改公式(shi)後(hou)計算舉’反例’)。作者認為(wei)這類(lei)意見(jian)赤裸裸否定創(chuang)新(xin)發現(xian)。

       其三(san) 作者說prpr至其相鄰後(hou)一(yi)個素數平方之間的偶數,有一(yi)個共同(tong)的各(ge)自平方根(gen)內的最大(da)素數pr錯了(liao),它們的公共素因子是2。作者認為(wei)這類(lei)意見(jian)純粹是指驢為(wei)馬(ma)栽贓陷(xian)害。

       除此而外,目(mu)前還沒有任何其它否定意見(jian)。

       其四 作者當面(mian)或(huo)書(shu)面(mian)請教了(liao)幾乎所有的當今(jin)數學院士。僅僅有一(yi)位院士說,你的論文要獲得認可(ke)很(hen)難。另(ling)一(yi)位院士說,別說人家(jia)錯誤自樹敵人。其他院士不開金口(kou)。

       其五 作者上書(shu)科技主管(guan)部門請求鑒定,泥牛入海(hai)無消息。

       其六 也有判官認定作者大(da)功告成。可(ke)惜(xi)他們沒有終審裁判權。“1+2”權威們的“共識”早已頒令,惟有他們擁有生殺予奪權。

       尊敬的mu)垂  餃wei)哥猜是世界難題嗎(ma),鑒定哥猜答案對(dui)yuan)硎遣皇潛裙?爍綺祿鼓眩空饈塹苯jin)中國數學超(chao)級難題嗎(ma),有解嗎(ma)?

--- Today's Chinese mathematics super puzzle

       He has asked for the proof of Goldbach's conjecture and its reasons for success or failure. The principle of argumentation is as follows.

Demonstrating the strategic plan Screening out all the a and b in the set of 2n=a+b at the same time and the formulas of the sum and the respectively, and there is more than one, and the Goldbach conjecture is established.

       I ask, is it possible to be wrong?

Demonstrate tactical approach Assuming that 2n is less than 2n square root, the prime factor is only 2, which can divide a and b at the same time, and the rest of the equations have no more than two sums.

The total number of formulas - the number of sums and the lower limit of the number of formulas - the upper limit of the number of non-combination numbers and the lower limit of the number of formulas

      I ask, is it possible to be wrong?

The calculation method is based on the combination of the properties ?a, b and the equations that are combined, respectively, and the multiplication law is applied to subtract them.

G(1+1)=[???["n/2". 1/3]. 3/5]???(pr-1-2)/pr-1](pr-2)/pr]-s+b'-1 (or 0)

             [...["n/2". 1/3]. 3/5]...(pr-1-2)/pr-1](pr-2)/pr]-s-1

(The maximum prime number in the 2n square root of the Pr table. b' indicates the number of equations to be subtracted. The s table takes the integer operation error. Each time rounds to an integer ? [r/2 ]s, 0 ≦ b' ≦ r-1 , 0 means that 1 is another number is a composite number)

       Increase the lower limit of insurance, pr is not small, regardless of the size of s minus its upper limit, and then consider b' as the lower limit of 0 ? G (1 + 1).

       I ask, is it possible to be wrong?

The "details" that determine the life and death of this formula have the following questions that doubt the conjecture. Therefore, the mathematics community does not recognize it.

1. According to the formula, the "number of answers" of some large even numbers is larger than the actual, or larger than the "number of answers" of the small even number, and the actual number is less than the small even number, that is, the "fluctuation" counterexample. Previously, to solve this problem is to do without the shackles of rice, and to advance the development of basic knowledge of mathematics.

2. No matter how small, the formula has (previously solved) rounding calculation error.

      I ask, is it possible to be wrong?

Resolve the volatility strategy First find the rice pot, and discover the basic theoretical knowledge. It is proved that the "N-value interval theorem of natural number" and the "continuous joint theorem" are 2n = r "2n-value interval" governed by prime numbers.

       Further, "specially limited" takes the lower limit prpr+1 of each "2n value interval" into the formula to calculate the "period lower limit" of ?G(1+1).

       I ask, is it possible to be wrong?

      In the formula, the numerator of the next two factors or = or greater than the denominator of the previous number, pr is less than n? conclusion

        The "lower limit of the interval" of the "1+1" formula of each "2n value interval" is not only not less than 1, but r is slightly larger than the number of odd prime numbers in the square root of the even number, and the larger r is half of pr (even greater than pr?). The "1+1" formula with the number of conjunctions and the formula of 1 has been reduced by the even interval of the same interval. The author only proves "1+1" and greatly improves the conjecture. Approach it to reality.

       I ask, is it possible to be wrong? Who can sit down and argue with the mistakes and deny that the red is doing a good job? !

       However, the reasons for some judges to judge the paper are ridiculous.

       One of the authors' academic qualifications is too low, and the article is short and it is impossible to answer a world problem. The author believes that such opinions are inconsistent with the judgment arguments.

       The second two theorems are common sense that everyone knows and have no academic value. The “lower limit of the interval” is inexplicable (there is a counter-example after the judge has falsified the formula accordingly). The author believes that such opinions are naked to deny innovation discoveries.

       The authors say that the prpr to an even number between the squares of a prime after its neighbors has a common maximum prime pr in the respective square roots, and their common prime factor is 2. The author believes that such opinions are purely meant to be framed by horses.

       Other than that, there are no other negative opinions.

       The four authors consulted almost all of today's academicians in person or in writing. Only one academician said that it is difficult for your paper to be recognized. Another academician said, don't say that people are wrong with their own enemies. Other academicians do not open the gold mouth.

       The five authors of the science and technology administration department requested the identification, and there was no news of the mud cows entering the sea.

       Six of them also judged that the author was done. Unfortunately, they do not have the final jurisdiction. The "consensus" of the "1+2" authorities has already issued orders, but they have the right to kill and seize power.

       Dear clerk, do you think that Gu guess is a world problem? Is it difficult to identify whether the answer is wrong or not? Is this a super problem in Chinese mathematics today, is there a solution?



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