用逻辑真值表扩展证明黎曼假设

Kai Shun Lam
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摘要

有许多数学家试图证明或反驳黎曼假设。然而,他们中没有一个人成功地获得了克莱数学研究所的认可。此外,据笔者所知,这些数学家在证明过程中都没有使用逻辑真值表技术。参照笔者之前在 [1] 中的证明,笔者采用了乘法望远镜法和质数边界间隙法。在笔者对黎曼假设的这个扩展证明中,笔者试图通过真值表中条件语句的四种情况来证明黎曼假设为真。其中三种情况(I、II、IV)被认定为黎曼假设中条件语句的真,而只有一种情况(情况 III)被认定为假(并作为反例反证)。此外,在这四种情况中还有三种子情况(i,ii,iii)[1]。其主要思想是,我们可以通过先找到一个反例来反证与 RH 假设相似的假设语句,而这个反例显然是对(黎曼)假设的反证(情形 III)。但这并不符合哥德尔不完备性定理。否则,要么是对陈述的反证,要么是哥德尔不完备性定理不正确,而这是不可能的。因此,反证与哥德尔不完备性定理是不相容的。另一方面,该陈述的所有其他真值情况(I、II、IV)确实是黎曼假说陈述的正向结果的例子,并且与哥德尔定理相容。一般来说,对于任何像黎曼假说这样的条件语句结构的假说,我们也可以用类似的技术和真值表的论证来证明它们的条件语句,并用哥德尔不完备性定理来强制证明假说语句的正结果。事实上,真值表有很多应用,特别是在语言(结构和建模)或工程(逻辑门和编程)等领域。
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An Extension Proof of Riemann Hypothesis by a Logical Entails Truth Table
There were many mathematicians who tried to prove or disprove the statement of Riemann Hypothe-sis. However, none of them have been successfully approved by the Clay Mathematical Institute. In addition, to the best of this author’s knowledge, these mathematicians haven’t employed the technique of logical truth table during their proofs. With reference to this author’s previous proof in [1], this author have employed the method of multiplicative telescope together with the prime boundary gaps. In this extended version of my proof to the Riemann Hypothesis, this author tries to show that RH statement is true through the four cases of the conditional statements in the truth table. Three of the cases (I, II, IV) are found to be true for the conditional statement in the Riemann Hypothesis while only one (case III) is found to be false (and acts as the disproof by a counter-example). Moreover, there are also three sub-cases (i, ii, iii) [1] among these four tabled cases. The main idea is that the we may disproof the hypothesis statement that is similar to the RH one by first find a counter-example which is obviously a disproof (case III) to the (Riemann) hypothesis. But it is NOT compatible with the GÖdel’s Incompleteness Theorem. Otherwise either the disproof to the statement or the Gödel is incorrect which is impossible. Hence, the disproof is said to be incompatible with the Gödel. On the other hand, all of the other truth cases (I, II, IV) for the statement are indeed the examples for the pos-itive results to the Riemann Hypothesis statement and are compatible with the Gödel. Therefore, the only way to make a conclusion is to say or force the Riemann Hypothesis statement to be correct.In general, for any hypothesis with the conditional statements structure like the Riemann one, we may also prove them by the similar techniqe and the arguments of the truth table for their conditional statements together with the Gödel’s Incompleteness theorem to force the positive result for the hy-pothesis statement. Actually, there are many applications for the truth tables especially in the fields like language (structure & modeling) or in engineering (logic gates & programming) etc during our everyday usage.
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