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Proving a sequence is a cauchy sequence in for the 7 -adic metric

  1. May 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that the sequence (xn)n[itex]\in[/itex]N [itex]\in[/itex]Z given by xn = Ʃ from k=0 to n (7n) for all n [itex]\in[/itex] N is a cauchy sequence for the 7 adic metric.


    2. Relevant equations
    In a metric space (X,dx) a sequence (xn)n[itex]\in[/itex]N in X is a cauchy sequence if for all ε> 0 there exists some M[itex]\in[/itex]N such that dx(xn,xm)<ε for all m,n ≥ M.

    the 7-adic metric is defined as follows:
    p(m,n)= 1/(the largest power of 7 dividing m-n) if m[itex]\neq[/itex]n or 0 if m=n

    3. The attempt at a solution

    I am struggling with proving sequences are cauchy because I am not sure how to go about finding the 'M'? I am not even sure how to start the question apart from assuming m<n. Just a hint at how to start it or how to approach the question would be appreciated.

    Assuming m<n am I able to write d7(xn,xm)=Ʃ from k=m+1 to n (7n)?

    Thank you.
     
  2. jcsd
  3. May 8, 2013 #2

    Fredrik

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    So to calculate ##p(x_n,x_m)##, you need to find the largest power of 7 that divides ##x_n-x_m##?

    That sum at the end is ##x_n-x_m##, right? Can you take ##7^{m+1}## outside of the sum, and then do something fun with the sum you have left.
    $$x_n-x_m=\sum_{k=m+1}^n 7^k=7^{m+1}(1+7+\cdots+7^{n-m-1}).$$
     
    Last edited: May 8, 2013
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