3x+1数学猜想的证明

duanbaoyuan

lusishun

看哥德巴赫猜想的证明见，可免费下载的《倍数含量筛法与恒等式的妙用》，
论文顺便把孪生素数猜想也证明了。

duanbaoyuan

我这是很严密的证明，不是胡搅蛮缠哈，更不是玄学！

duanbaoyuan

duanbaoyuan 发表于 2022-1-25 15:33
我这是很严密的证明，不是胡搅蛮缠哈，更不是玄学！

3x+1不是数学猜想？醒醒吧，只不过我说得通俗而异，我的重点在证明，不在语言描述上，
文章写得仓促，有的甚至没打草稿，但证明我认为是很严密的！

duanbaoyuan

摘要补充:
Build a special odd tree model according with (*3+1)/2^k algorithm, to depart odd numbers in different groups.Carry out research into the tree,found that counts of elements in the tree reduce and converge downward one by one layer,values of elements converge downward to 1,and all odd numbers can appear in the tree only one time.Then prove the “3x+1”guess indirectly.At last,find a logic error of deduction of the tree,and prove 3x+1 guess in another ways.In the last section of last version of this paper,build a special identical equation,use its calculation characters prove and search for solution of any odd converge to 1 equation through (*3+1)/2^k operation,and give a solution for this equation,which is exactly same with calculating directly.And give a specific example to verify it,indicate that we can estimate the value of convergence steps n during some middle procedures.Thus prove 3x+1 guess--Collatz Conjecture strictly. Some supplements:The last section is not very detailed due to time,i think omitting part is not hard and critical.we can fully use characters of odd multiplying 3,and no more +1 interference again.To build model for estimating steps n,we can use characters of formula (3) is bigger than corresponding part in formula (2),but should avoid trap:when converge to 1,steps can still increase forever.Add convergence condition:if highest bit of t(i) is 2^k,k should be odd,then all steps n>=1 can satisfy with the equation,i say n>=4 in paper is just want enough parts in formula (2) can appear. Some more supplements:watch t(i),its odd part add corresponding odd produced by odd x calculate directly in each step through (*3+1)/2^k should exactly be 2^k;watch 0 bits in odd part in t(i),because of characters of odd multiplying 3,it should shift right or bit-count reduce in each step,and its weight in total t(i) should reduce step by step till to 0,when odd part converge to 1...1.Build weight model:value of all 0 bits in odd part/2^(2k),which 2^(2k) is corresponding part in each step,then model value should reduce step by step,could not exist loop!and model value can and must converge to 0,because obviously there is not possible to exist a convergence value,which its corresponding odd part in t(i) is not 1...1,and its model value can remain unchanged in next steps through multiply 3 operation.Then odd part must converge to 1...1,could not diverge or converge to other odds.

duanbaoyuan

Build a special odd tree model according with (*3+1)/2^k algorithm, to depart odd numbers in different groups.Carry out research into the tree,found that counts of elements in the tree reduce and converge downward one by one layer,values of elements converge downward to 1,and all odd numbers can appear in the tree only one time.Then prove the “3x+1”guess indirectly.At last,find a logic error of deduction of the tree,and prove 3x+1 guess in another ways.In the last section of last version of this paper,build a special identical equation,use its calculation characters prove and search for solution of any odd converge to 1 equation through (*3+1)/2^k operation,and give a solution for this equation,which is exactly same with calculating directly.And give a specific example to verify it,indicate that we can estimate the value of convergence steps n during some middle procedures.Thus prove 3x+1 guess--Collatz Conjecture strictly. Some supplements:The last section is not very detailed due to time,i think omitting part is not hard and critical.we can fully use characters of odd multiplying 3,and no more +1 interference again.To build model for estimating steps n,we can use characters of formula (3) is bigger than corresponding part in formula (2),but should avoid trap:when converge to 1,steps can still increase forever.Add convergence condition:if MSB bit of t(i) is 2^k,k should be odd,then all steps n>=1 can satisfy with the equation,i say n>=4 in paper is just want enough parts in formula (2) can appear. Some more supplements:watch t(i),its odd part add corresponding odd produced by odd x calculating directly in each step through (*3+1)/2^k should exactly be 2^k;watch 0 bits in odd part in t(i),because of characters of odd multiplying 3,it should shift right or bit-count reduce in each step,and its weight in total t(i) should reduce step by step till to 0,when odd part converge to 1...1.Build simple weight model:value of all 0 bits in odd part/2^(2k),which 2^(2k) is corresponding part in each step,then model value should reduce step by step,could not exist loop!and model value can and must converge to 0,because obviously there is no possible to exist a convergence value,which its corresponding odd part in t(i) is not 1...1,and its model value can remain unchanged in next steps through multiplying 3 operation.Then odd part must converge to 1...1,could not diverge or converge to other odds.t(i) has many other characters,for example,its odd part should be with form 3*y after first step(can be proved easily),then within each few steps(normally<=3),should be back to form 3*y(suppose it has not converged yet before this step),and after a few steps,y should be with binary form 101...1,since this time,with each some steps,head part of y increase a 01 pair,tail part is different from previous,until finally y become form of 10101...01,which is just convergence form we need.This states again that this kind of iteration calculation has determined direction,and converge regularly.

duanbaoyuan

duanbaoyuan

duanbaoyuan

任在深

楼主您辛苦啦！
       您费劲容易吗？简单的问题复杂化！
看《中华单位论》的论证！！
        证：
            
            令：
                     (1)  3X+1=4Y
         所以：
                     (2) (3X+1)/Y=4→2→1

            当    X              Y
                    1              1
                    2             7/4
                    3             10/4
                    4             13/4
                    5             16/4
                    *                 *
                    *                 *
                    *                 *
                    n              n/4
  注意！ 分数在纯粹数学中也是单位数，
         如图：
                     
                                              証毕。