A Full and Detailed Proof for the Riemann Hypothesis & the Simple Inductive proof of Goldbach’s Conjecture

Lam Kai Shun
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Abstract

As in my previous two papers [2] & [3] about the boundary of the prime gap still cause some misunderstanding, I here in this paper tries to clarify those detailed steps in proving such boundary of the prime gap for a contradiction. Indeed, the general idea of my designed proof is to make all of the feasible case of the Riemann Zeta function with exponents ranged from 1 to s = u + v*I becomes nonsense (where u, v are real numbers with I is imaginary equals to (-1)1/2 except that u = 0.5 with some real numbers v as the expected zeta roots. Once if we can exclude all other possibilies unless u = 0.5 with some real numbers v in the Riemann Zeta function’s exponent “s”, then the Riemann Hypothesis will be proved immediately. The truth of the hypothesis further implies that there is a need for the shift from the line x = 0 to the line x = 0.5 as all of the zeta roots lie on it. However, NOT all of the points on x = 0.5 are zeros as we may find from the model equation that has been well established in [2]. One of my application is in the quantum filtering for an elimination of noise in a quantum system but NOT used to filter human beings like the political counter-parts.In general, this author suggests that for all of the proof or disproof to any cases of hypothesis, one may need to point out those logical contradictions [14] among them. Actually, my proposition works very well for the cases in my disproof of Continuum Hypothesis [15] together with the proof in Riemann Hypothesis
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黎曼假设的一个完整而详细的证明哥德巴赫猜想的简单归纳证明
正如我之前的两篇论文[2]&[3]关于素数间隙的边界仍然引起一些误解,我在这里试图澄清证明一个矛盾的素数间隙的边界的那些详细步骤。实际上,我设计的证明的一般思想是使所有可行的黎曼ζ函数的情况,指数范围从1到s = u + v*I变得毫无意义(其中u, v是实数,I是虚数等于(-1)1/2,除了u = 0.5,一些实数v作为期望的ζ根。一旦我们能够排除除u = 0.5和一些实数v在Riemann Zeta函数的指数“s”之外的所有其他可能性,那么Riemann假设将立即得到证明。假设的真实性进一步表明,由于所有的根都在x = 0上,因此需要从x = 0平移到x = 0.5。然而,并非x = 0.5上的所有点都是零,我们可以从[2]中建立的模型方程中发现。我的一个应用是在量子滤波中消除量子系统中的噪声,但不用于像政治对手那样过滤人类。总的来说,笔者认为对于任何假设的证明或反证,都可能需要指出其中的逻辑矛盾[14]。实际上,我的命题对于我对连续统假设的反证[15]以及对黎曼假设的证明都是非常有效的
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