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Apostol's Analysis Problem 1.22

  1. Jan 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Given x>0, suppose a_0 = [x], the largest integer less than or equal to x. Assuming a_0, a_1, . . . , a_n-1 are defined, define a_n as the largest integer such that a_0.a_1...a_n less than or equal to x. Now define r_n = a_0.a_1...a_n, r_0 = a_0. Prove that the sup of the set of all the r_n's is x.


    2. Relevant equations



    3. The attempt at a solution
    I did proof by contradiction but does not work. I've got stuck. Just give me a useful hint, don need full solution. I've analyzed all the meanings carefully, but seems I need a trick to solve this, or maybe not. I even tried to use the notion of limit: I thought about |r_n - x|. It goes 0. But this fact it is even harder to prove.

    Okay, while I'm writing this thread, I've got an idea. Please have a look:
    Suppose there exists an upper bound p of the set such that p is less than x. Then, since |p-x|>0, for some n, 1/(10^n) < |p-x|. Thus p + 1/10^n < x. Note that a_0.a_1...a_n <= p. Thus a_0.a_1...(a_n + 1) < x; but this is a contradiction.

    I've used a huge assumption that "for some n, 1/(10^n) < |p-x|." But I didn't learn this yet in this book. Is there any other way not using it?
     
  2. jcsd
  3. Jan 4, 2012 #2
    I don't think you haven't copied the problem correctly.
    The book i'm looking at also gives a fixed integer [itex]k\geq2,[/itex] and [itex]a_n[/itex] is defined to be the largest integer s.t. [itex]\displaystyle \sum_{i=0}^n{\frac{a_i}{k^i}} \leq x.[/itex] You should be able to prove that [itex]\sup\{r_n:n\geq 0\}=x[/itex] by first proving that [itex]0\leq a_i \leq k-1, \forall i\geq 1[/itex] and that [itex]r_n\leq r_{n+1},\ \forall n.[/itex] After these I think you can use the idea you got to finish the proof, but don't give [itex]k[/itex] any particular value.
     
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