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通过构造满足毕达哥拉斯定理的GB【哥德巴赫猜想】,
===得出
===满足毕达哥拉斯定理的GB的解是唯一的,这个特称命题
== ,对此【特称命题】真假进行判断!
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易一个偶数为两素数之和[A=I+B]
YIG一个奇数为三素数之和[C=I+Q+B]
S设满足勾股定理的联立方程
A=I+B
C=I+Q+B
A*A+B*B=C*C
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以B为未知数求解
B*B-2Q*B-(2QI+Q*Q)=0
用求根公式求的B1, B2
B1=Q+2Q=3Q=3I
B2=-Q-I
======================
因为假设B是素数,而解出B是合数!
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1)SAP=任何一个偶数可以由两个素数之和构成,任何一个奇数可以由三个素数之和构成
2)SEP=任何一个偶数不可以由两个素数之和构成
3)SIP=任何一个奇数可以由三个素数之和构成(有些数的哥德巴赫猜想是成立)
4)SOP==满足毕达哥拉斯定理的GB的解是唯一的【GB(SOP)】
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逻辑表达特称命题为:
【GB(SOP)】==P(A)&q(YIG)
===在特称命题中解出满足q(YIG)中的解 B=3I
因为规定了B是素数,一个奇数可以表达为三个素数之和是已经证明是成立
=====在毕达哥拉斯的联立方程中,q(YIG)为假
====有维特根斯坦的p&q规则辨别
====p(A)&q(YIG)为假
==========既特称命题是假
=========================
因为【GB(SOP)】为假
===则SAP为真,既哥德巴赫猜想成立
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====================
不在规律不在爻,
而在载之莲素王;
不在载之连数王;
而在规律而在爻;
载之规律爻数王
@
不在规律不在爻,
而在载之莲数王;
不在载之连数王;
而在规律而在爻;
载之规律爻数王。
谷歌翻译
Satisfy the Pythagorean theorem by constructing the GB [Goldbach Conjecture],
=== Come
=== Satisfy the Pythagorean theorem of GB is the only solution, said this particular proposition
==, [This] special called true and false propositions to judge!
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Easy one even for the two prime numbers, and [A = I + B]
YIG an odd number of three prime numbers, and [C = I + Q + B]
S set of simultaneous equations satisfy the Pythagorean theorem
A = I + B
C = I + Q + B
A * A + B * B = C * C
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To B for the unknown solution
B * B-2Q * B-(2QI Q * Q) = 0
Roots of the formula with the demand of the B1, B2
B1 = Q 2Q = 3Q = 3I
B2 =- Q-I
======================
Reason is that B is prime, and the solution of B is a composite number!
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1) SAP = even number can be any one of two prime numbers, and constitute, by any odd number and composition of three prime numbers
2) SEP = any even number can not be made up of two prime numbers and composition
3) SIP = odd number can be any one of three prime numbers and composition of (some number of Goldbach';s conjecture is established)
4) SOP == satisfy the Pythagorean theorem is the only solution of GB [GB (SOP)]
================================================== ====
Dart said the proposition is a logical table:
[GB (SOP)] == P (A) & q (YIG)
=== Proposition in the special, said the solution to meet the q (YIG) in the solution B = 3I
It is stipulated that the B is prime, an odd prime number can be expressed as the sum of the three has proven to be established
===== In the Pythagorean simultaneous equation, q (YIG) is false
==== With Wittgenstein';s p & q rules to identify
==== p (A) & q (YIG) is false
========== Special, said both propositions are false
=========================
Because [GB (SOP)] is false
=== Then SAP is true, both Goldbach conjecture
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