View Poll Results: So what happens indeed in the second case?

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Thread: abc conjecture

  1. #1
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    abc conjecture

    По предположению ABC , мы говорим, что:
    5 + 16 = 21
    Таким образом,
    2 x 3 x 5 x 7 = 210
    210> 21
    Однако, по-другому, мы можем выразить это:
    [2 ^ 0 + 2 ^ 1 + (2 ^ 1)] + [2 ^ 0 + 2 ^ 1 + 2 ^ 2 + 2 ^ 3 + (2 ^ 0)] = 2 ^ 0 + 2 ^ 1 + 2 ^ 2 + 2 ^ 3 + (2 ^ 1 + 2 ^ 2)
    В этом случае мы получить, что:
    2 ^ 1 + 2 ^ 0 <2 ^ 1 + 2 ^ 2

    Далее, по предположению abc , мы полагаем, что:
    5 + 27 = 32
    Таким образом,
    2 x 3 x 5 = 30
    30 <32
    Тем не менее, по-другому, мы можем выразить это:
    [2 ^ 0 + 2 ^ 1 + (2 ^ 1 )] + [2 ^ 0 + 2 ^ 1 + 2 ^ 2 + 2 ^ 3 + (2 ^ 2 + 2 ^ 3)] = 2 ^ 0 + 2 ^ 1 + 2 ^ 2 + 2 ^ 3 + 2 ^ 4 + (2 ^ 0)
    Итак, в этом случае мы получаем, что:
    2 ^ 1 + (2 ^ 2 + 2 ^ 3)> 2 ^ 0


    Так что же на самом деле происходит во втором случае?
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: abc conjecture

    Quote Originally Posted by velikh View Post
    По предположению ABC , мы говорим, что:
    5 + 16 = 21
    Таким образом,
    2 x 3 x 5 x 7 = 210
    210> 21
    Однако, по-другому, мы можем выразить это:
    [2 ^ 0 + 2 ^ 1 + (2 ^ 1)] + [2 ^ 0 + 2 ^ 1 + 2 ^ 2 + 2 ^ 3 + (2 ^ 0)] = 2 ^ 0 + 2 ^ 1 + 2 ^ 2 + 2 ^ 3 + (2 ^ 1 + 2 ^ 2)
    В этом случае мы получить, что:
    2 ^ 1 + 2 ^ 0 <2 ^ 1 + 2 ^ 2

    Далее, по предположению abc , мы полагаем, что:
    5 + 27 = 32
    Таким образом,
    2 x 3 x 5 = 30
    30 <32
    Тем не менее, по-другому, мы можем выразить это:
    [2 ^ 0 + 2 ^ 1 + (2 ^ 1 )] + [2 ^ 0 + 2 ^ 1 + 2 ^ 2 + 2 ^ 3 + (2 ^ 2 + 2 ^ 3)] = 2 ^ 0 + 2 ^ 1 + 2 ^ 2 + 2 ^ 3 + 2 ^ 4 + (2 ^ 0)
    Итак, в этом случае мы получаем, что:
    2 ^ 1 + (2 ^ 2 + 2 ^ 3)> 2 ^ 0


    Так что же на самом деле происходит во втором случае?
    According to Google translate:
    According to ABC, we say that:
    5 + 16 = 21
    In this way,
    2 x 3 x 5 x 7 = 210
    210> 21
    However, in a different way, we can express it:
    [2 ^ 0 + 2 ^ 1 + (2 ^ 1)] + [2 ^ 0 + 2 ^ 1 + 2 ^ 2 + 2 ^ 3 + (2 ^ 0)] = 2 ^ 0 + 2 ^ 1 + 2 ^ 2 + 2 ^ 3 + (2 ^ 1 + 2 ^ 2)
    In this case, we get that:
    2 ^ 1 + 2 ^ 0 <2 ^ 1 + 2 ^ 2
    Further, by the abc assumption, we assume that:
    5 + 27 = 32
    In this way,
    2 x 3 x 5 = 30
    30 <32
    However, in a different way, we can express it:
    [2 ^ 0 + 2 ^ 1 + (2 ^ 1)] + [2 ^ 0 + 2 ^ 1 + 2 ^ 2 + 2 ^ 3 + (2 ^ 2 + 2 ^ 3)] = 2 ^ 0 + 2 ^ 1 + 2 ^ 2 + 2 ^ 3 + 2 ^ 4 + (2 ^ 0)
    So, in this case we get that:
    2 ^ 1 + (2 ^ 2 + 2 ^ 3)> 2 ^ 0

    So what actually happens in the second case?
    Even though I can use Google translate, most that work on the site speak English, not Russian. I'm not trying to be elitist or anything but you would be better off if you did the translating with something (or someone) better equipped for it.

    -Dan
    Thanks from velikh
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  3. #3
    Newbie
    Joined
    Jun 2019
    From
    Kiev,Ukraine
    Posts
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    Re: abc conjecture

    By the abc conjecture, we put that:
    5 + 16 = 21
    Thus,
    2 x 3 x 5 x 7 = 210
    210 > 21
    However, in a different way, we can put that:
    [2^0 + 2^1 + (2^1)] + [2^0 + 2^1 + 2^2 + 2^3 + (2^0)] =
    = 2^0 + 2^1 + 2^2 + 2^3 + (2^1 + 2^2)
    In this case, we obtain that:
    2^1 + 2^0 < 2^1 + 2^2
    Further by the abc conjecture, we put that:
    5 + 27 = 32
    Thus,
    2 x 3 x 5 = 30
    30 < 32
    Nevertheless, in a different way, we can put that:
    [2^0 + 2^1 + (2^1)] + [2^0 + 2^1 + 2^2 + 2^3 + (2^2 + 2^3)] =
    = 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + (2^0)
    So in this case, we obtain that:
    2^1 + (2^2 + 2^3) > 2^0

    So what happens in the second case?
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