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The Goldbuch's conjecture is right,the prove as follow

  1. Jan 6, 2004 #1
    The Goldbach's conjecture is right,the prove as follow

    The Goldbach's conjecture right as follow don't worse the time so as it is hard so
    Axiom 1: every natural number factorization is sole.
    No same factors theorem: between every both odd number from 2n to 2n- square root(2n), it is not same factorization.
    The certification summary as follow:
    To consider both odd numbers A1, A2. Its all primary factors are same.
    As this reason: A1+p1=2n; A2+p2=2n; p1 The one surplus number of p1 and p2 corresponding the A1, A2 common factor, should be large than the A1, A2 common factor.
    It is imposible.
    As the next condition axiom1,
    the no same factors theorem is right.

    The reasoning result 1: every even number, it is equal to both primary number add, one primary of the both less than the even number square root.
    The right key as follow:
    2n, the even number,
    From 2n to 2n - square root 2n, every both odd number factor group is different.
    As 2 is beside the add, according to the primary number serious from 1 to 2n square root, there is a odd number according to the factor 1 also, that is right.
    There is a both add primary numbers to equal to the 2n, it is also right.
    The Goldbach's conjecture ir right.

    [The ¸çµÂ°ÍºÕ suspect is right.

    ×öΪÎÞÏàͬÒò×Ó¶¨ÀíµÄÒ»¸öÍÆÂÛ£¬Ò»¸ö´óżÊý 2n ¿ÉÖÁÉÙ·Ö½âΪ int£¨(2nµÄƽ·½¸ù£©/2£©¶ÔËØÊýÖ®ºÍ¡£ ]

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    Last edited: Jan 6, 2004
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