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Metric Space and Topology HW help

  1. Feb 17, 2013 #1
    Metric Space and Topology HW help!!!

    Let X be a metric space and let (sn
    )n be a sequence whose terms are in X. We say that (sn
    )n converges to s [itex]\ni[/itex] X if
    [itex]\forall[/itex] [itex]\epsilon[/itex] > 0 [itex]\exists[/itex] N [itex]\forall[/itex] n ≥ N : d(sn,s) < [itex]\epsilon[/itex]

    For n ≥ 1, let jn = 2[(5^(n) - 5^(n-1))/4].

    (Convince yourself that 5^(n) - 5^(n-1) is always divisible by
    4, so the exponent in the definition is always a positive integer.) The first few terms of this
    sequence are
    2; 32; 33554432; 42535295865117307932921825928971026432
    so you would reasonably expect this sequence to diverge with respect to the usual metric on
    Q (the one given by the usual absolute value).

    However, show that |j2n - (-1)|5 ≤ 5^(-n) where ||5 is the 5-adic absolute value.


    My Attempt:
    I started by writing the claim in terms of v5(j2n + 1). Then i tried to find a recurrence that looks like this:
    (j2n+ 1)^5 = (2n+1+1) + (some other stuff).

    I was thinking I can show that the sequence (jn)n is also Cauchy with respect to ||5, so in Q5,the completion of Q with respect to ||5, the sequence (jn)n converges to a number j [itex]\ni[/itex]Q5 such that j2 = -1. It follows that Q5 is not an ordered field, unlike the completion of Q with respect to the usual ||5, which is our old friend R.
     
  2. jcsd
  3. Feb 19, 2013 #2

    Fredrik

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    Re: Metric Space and Topology HW help!!!

    More people will be able to attempt to solve the problem if you define "5-adic absolute value".
     
  4. Feb 19, 2013 #3

    micromass

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    Re: Metric Space and Topology HW help!!!

    http://en.wikipedia.org/wiki/P-adic_number

    Basically, you take the rational numbers [itex]\mathbb{Q}[/itex] and you define some very weird norm on it. For a rational number [itex]q=a/b[/itex], you factor 5 out of it, so you get

    [tex]q=5^n \frac{a^\prime}{b^\prime}[/tex]

    Then you define

    [tex]|q|_5 = 5^{-n}[/tex]

    For example

    [tex]\frac{23}{10} = 5^{-1} \frac{23}{2}[/tex]

    So [itex]|23/10|_5=5[/itex].

    Also, [itex]|5^n|_5 = 5^{-n}[/itex]. So [itex]5^n[/itex] converges to 0 in that norm.

    Anyway. The OP should try to look at the expression

    [tex]j_n^2+1[/tex]

    and he should try to factor out 5 as many times as he can. Maybe try it first for small n.
     
  5. Feb 19, 2013 #4

    Fredrik

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    Re: Metric Space and Topology HW help!!!

    That is one crazy norm. 5n converges to 0. :smile:
     
  6. Feb 19, 2013 #5

    micromass

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    Re: Metric Space and Topology HW help!!!

    If p=10 (that's not a prime, I know), then you can show crazy things like

    [tex]...9999999999 = -1[/tex]

    In the sense that

    [tex]\sum_{k=0}^{+\infty} 9\cdot 10^k = -1[/tex]

    where convergence of the series is n the 10-adic norm.

    It actually isn't so crazy. A naive student (who doesn't know that natural numbers have to have finitely many digitis), might do something like


    Let [itex]x=...99999[/itex]. Then [itex]10x+9=x[/itex]. So [itex]x=-1[/itex].

    In some sense, the 10-adic numbers are a formalization of that (wrong) argument. In the 10-adic numbers, the argument does work :tongue2:
     
  7. Feb 19, 2013 #6
    Re: Metric Space and Topology HW help!!!


    starting with this
    [tex]j_n^2+1[/tex]
    where n=1 which is 2 from the first term from the given sequence
    gives |22 +1| = |5|5 factored out 5 once.

    repeated for the next term n=2, which gives |1025|5
    where 5 could be factored out twice.

    repeated for the next term |335544322+1| which could be factored out 9 times.

    the last term given could have been factored at least 4 times.

    how do i show that
    5-V5[j2n+1] ≤ 5-n
     
  8. Feb 19, 2013 #7

    micromass

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    Re: Metric Space and Topology HW help!!!

    You'll need to show that you can factor 5 from [itex]j_n^2 +1[/itex] at least n times.
     
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