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anonymous

  • one year ago

Number theory help. How did they conclude gh = 1? (at the bottom of page 17)

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  1. anonymous
    • one year ago
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    https://books.google.com/books?id=eVwvvwZeBf4C&lpg=PA3&pg=PA17#v=onepage&q&f=false

  2. UsukiDoll
    • one year ago
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    Oh I love that book. I've used it last year.

  3. freckles
    • one year ago
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    how far did you get because right before they say that they have d=ghd and dividing d on both sides gives 1=gh (Assuming d isn't 0 )

  4. anonymous
    • one year ago
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    @UsukiDoll yeah I like it too. Certainly better than the one wrote by Stein

  5. freckles
    • one year ago
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    also we know d isn't zero since it is given in the theorem that a and b are not 0

  6. anonymous
    • one year ago
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    O: oh, I see it now.

  7. anonymous
    • one year ago
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    The whole time i was reading d2 = gh d1 .

  8. anonymous
    • one year ago
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    This question is a false alarm XD. I was having a brain fart :P sorry people lol

  9. UsukiDoll
    • one year ago
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    d = ghd ... when d was divided it left 1 = gh so g = 1 and h = 1 which are the only values that satisfy 1=gh

  10. UsukiDoll
    • one year ago
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    xD I'm still wondering how people master proofs x.x like I don't mind easy ones, but the hard ones are crazy

  11. anonymous
    • one year ago
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    @freckles thanks for clarifying :)

  12. anonymous
    • one year ago
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    You know what. While we're at it, can any one explain how they got 0 <= r_i <= b-i?

  13. mathstudent55
    • one year ago
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    \(gd_1 = d_2\) and \(hd_2 = d_1\) \(d_2 = \dfrac{d_1}{h} \) Equate the two expressions for \(d_2\) \(gd_1 = \dfrac{d_1}{h} \) \(ghd_1 = d_1\) \(gh = 1\)

  14. anonymous
    • one year ago
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    They said it's by induction though :/ @mathstudent55 thank you. I got that part :)

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