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Thread: Good Day! just need Guidance on my process.

  1. #1
    rcs
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    Good Day! just need Guidance on my process.

    If a|b+c then a|b and b|c

    Suppose a|(b+c) = t for some t element of Z such that t = (k+p) for some k and p elements of Z

    b+c = a(t)
    b+c = a (k+p) by substitution
    b+c = ak + ap by distributive property

    then b = ak and c = ap
    this implies a|b = k and a|c =p

    thus if a|b and b|c , then a|(b+c)

    please correct me if im mistaken... thnaks
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  2. #2
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    Re: Good Day! just need Guidance on my process.

    If a|b+c then a|b and b|c
    did you mean ...

    If a|(b+c) then a|b and b|c ?

    are you proving this statement or its converse?


    "If a|(b+c) then a|b and b|c." is false ... consider a = 2, b = 3, and c = 5


    The converse "If a|b and b|c, then a|(b+c)" is true.
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    Re: Good Day! just need Guidance on my process.

    Quote Originally Posted by rcs View Post
    If a|b+c then a|b and b|c

    Suppose a|(b+c) = t for some t element of Z such that t = (k+p) for some k and p elements of Z

    b+c = a(t)
    b+c = a (k+p) by substitution
    b+c = ak + ap by distributive property

    then b = ak and c = ap THIS DOES NOT FOLLOW
    this implies a|b = k and a|c =p

    thus if a|b and b|c , then a|(b+c)

    please correct me if im mistaken... thnaks
    $a= 2,\ b = 13,\ c = 19,\ k = 11,\ and\ p = 5.$

    $b + c = 13 + 19 = 32 \implies \dfrac{b + c}{a} = \dfrac{32}{2} = 16 = 11 + 5 = k + p.$

    $ak = 2 * 11 = 22.$

    $ap = 2 * 5 = 10.$

    $\therefore ak + ap = 22 + 10 = 32 = b + c.$

    $But\ 13 \ne 2 * 11 = 22\ and\ 19 \ne 2 * 5 = 10.$
    Last edited by JeffM; Feb 21st 2017 at 08:29 AM.
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  4. #4
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    Re: Good Day! just need Guidance on my process.

    Your basic problem is that you are trying to prove something that isn't true! 7 divides 4+ 3 but does not divide either 4 or 3.
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