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Math Help - Divisibility 13

  1. #1
    Sea
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    Divisibility 13

    Show that:

    k\in\mathbf{Z^{+}} and a|b \Rightarrow a^{k}|b^{k}
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  2. #2
    Moo
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    Quote Originally Posted by Sea View Post
    Show that:

    k\in\mathbf{Z^{+}} and a|b \Rightarrow a^{k}|b^{k}
    a|b means that there exists an integer n such that b=an.
    Hence b^k=a^kn^k

    Therefore...
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  3. #3
    Senior Member JaneBennet's Avatar
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    Are you trying to prove the converse, namely a^k\mid b^k\ \Rightarrow\ a\mid b\,?
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  4. #4
    Sea
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    Quote Originally Posted by JaneBennet View Post
    Are you trying to prove the converse, namely a^k\mid b^k\ \Rightarrow\ a\mid b\,?


    I don't know......

    But good question......

    a^3|b^3 \Rightarrow a|b (Why?) ...write...

    http://www.mathhelpforum.com/math-help/number-theory/65693-divisibility-14-a.html



    I think...Maybe use binomial theorem...
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  5. #5
    Senior Member JaneBennet's Avatar
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    I would suggest maybe using prime factorization.
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  6. #6
    Sea
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    I would suggest maybe using prime factorization.

    <br />
a^k|b^k (\Rightarrow):



    I think;


    (p_1 \times p_2 \times p_3 \times ....\times p_m,N)=1


    and


     (p_1 , p_2 , p_3 , ... , p_m)=1




    a^k=(p_1^{y_1}\times p_2^{y_2}\times p_3^{y_3} \times .......p_m^{y_S})^k

           =(p_1^{y_1})^k\times (p_2^{y_2})^k\times (p_3^{y_3})^k\times .......(p_m^{y_S})^k

    <br />
          =(p_1^{ky_1}\times p_2^{ky_2}\times p_3^{ky_3} \times .......p_m^{ky_S})<br />




    <br />
b^k=(p_1^{z_1}\times p_2^{z_2}\times p_3^{z_3} \times .......p_m^{z_S})^k \times N


    <br />
=(p_1^{z_1})^k\times (p_2^{z_2})^k\times (p_3^{z_3})^k \times .......(p_m^{z_S})^k \times N<br />


    =p_1^{kz_1}\times p_2^{kz_2}\times p_3^{kz_3} \times .......p_m^{kz_S} \times N


    a^k|b^k \Rightarrow p_i^{ky_i} |  p_i^{kz_i} and     i=1,2,3,....,s   \Rightarrow ky_i \leq kz_i \Rightarrow y_i \leq z_i , k\in Z^+ \Rightarrow  p_i^{y_i} |  p_i^{z_i}


    I think...?
    Last edited by Sea; December 26th 2008 at 12:34 AM.
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  7. #7
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    If your'e trying to prove that a^k|b^k => a|b then
    a^k=nb^k
    (a/b)^k=n
    a/b=n^1/k
    Now a|b iff n=m^k for some integer m, the problem is to show that it is so.

    You can assume by contradiction that a!|b which means there exists p and r such that:
    a=pb+r r<=b
    Now write a^k=(pb+r)^k and see that it contradicts our assumption.
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