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Decimal representation of real numbers

  1. Feb 12, 2007 #1
    I'm doing self study out of Apostol's Calculus vol. I and I got stuck trying to prove what the author writes is easy to verify, but I can't get my head around it. Basically, this is the problem statement from page 31, last paragraph:
    Given a positive real number x, let a0 denote the largest integer <= x. Having chosen a0, we let a1 denote the largest integer such that
    a0 + (a1)/10 <= x.
    More generally, having chosen a0, a1,...,an-1, we let an denote the largest integer such that
    a0 + (a1)/10 + ... + (an)/10^n <= x.
    Then let S denote the set of all numbers obtained this way for n = 0, 1, 2, ...

    I understand the construction of set S because I have proven before that the greatest integer in a positive real number exists, hence the set is nonempty. I also see that
    a0 + (a1)/10 + ... + (an + 1)/10^n is an upper bound, so this property with the nonemptyness of S guarantees that S has a supremum, say b. The punchline is that I want to verify that x = b.

    I'm not trying to prove here that .999... = 1. I know that and can prove it using geometric series and algebraic arguments. I want to rigorously prove the above using just the bare essentials (ie. axioms) of the real number system.

    I was thinking of arguing by contradiction that x > b or x < b is impossible. My other thought for a rough sketch was that to prove that the infinum of the set for decimal expansion of x - b is 0, but I don't know where to start really. Thanks in advance.
     
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  3. Feb 13, 2007 #2

    morphism

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    Let s_n = a0 + (a1)/10 + ... + (an)/10^n. Then {s_n} is a bounded, nondecreasing sequence of reals. It must converge to supS = b, i.e. lim s_n = b. We also know that:
    s_n <= x <= s_n + 1/10^n

    Applying the sequeeze theorem, we see that:
    lim s_n <= x <= lim s_n
    => x = lim s_n = b.
     
  4. Feb 13, 2007 #3
    Thanks for the reply, but I was looking for something even more elementary - no limits and no squeeze theorem. Is this proof possible to do with even more basic concepts? I'm curious because Apostol is very thorough and I'm confident he wouldn't write this if it couldn't be done with the material so far covered (which is before differentiation and integration are even introduced).
     
  5. Feb 13, 2007 #4

    StatusX

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    It's clear that all the numbers in the sequence are <=x, so if you can show their limit is >=x, you're done. To do this, assume its <x, and derive a contradiction (ie, show you haven't picked the a_n correctly).
     
  6. Feb 14, 2007 #5
    I think I have worked out a correct proof. I would appreciate it if anyone could review this and criticize if necessary.

    I have already established that S is nonempty and bounded above so I will not dwell on this fact anymore. As in my first post, by the law of trichotomy, the least upper bound b is one of the following:
    b < x, x = b or b > x
    and I will show by contradiction that the first and last are impossible.
    Suppose that b < x. Choose a c such that 0 < c < x - b, or written equivalently as (1) b < b + c < x. By the Archimedean property of real numbers, there exists a positive integer n such that (*) 1/(10^n) < c (Proof on the bottom of this post). Since b is an upper bound, it follows that:
    a0 + (a1)/10 + ... + (an)/(10^n) <= b. Combining this with (*), we have
    a0 + (a1)/10 + ... + (an)/(10^n) + 1/(10^n) < b + c
    or
    a0 + (a1)/10 + ... + (an + 1)/(10^n) < b + c.
    Here we have a contradiction because an was supposed to be the greatest integer such that the decimal expansion is <= x, but we have found that an + 1 is an even greater positive number such that the expansion is <= x due to (1).
    If b > x, then choose a c such that 0 < c < b -x or x < c + x < b. It is clear that c + x is another upper bound for S that is less than the least upper bound b, which is a contradiction

    (*) To prove that that there exits a positive n such that 1/(10^n) < y, I use the fact that for every real number x, there exists a positive integer such that x < n and by induction I will show that n < 10^n for every n.
    Let A(n) = n < 10^n.
    A is clearly true for 1.
    Now assuming that A(k) = k < 10^k is true, I multiply both sides by 10 to obtain
    10k < 10^k * 10 or (1) 10k < 10^(k+1)
    Now, (2) k + 1 < 10k is true because 10k - (k + 1) = 9k + 1 > 0.
    By the transitive property of inequalities, (1) and (2) combine and we obtain
    k + 1 < 10^(k+1), as asserted.
    Therefore we have that x < n < 10^n or x < 10^n. Replacing x by 1/y (where y > 0), we have
    1/y < 10^n or 1/(10^n) < y, just as stated.

    QED?
    Thanks in advance.
     
  7. Feb 14, 2007 #6

    StatusX

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    Yea, you've proven it in painstaking detail. Normally you can leave out well known results you've already seen the proof of. Still, you've only shown the sup of the sequence is x, not that it converges to x. The proof that the first implies the second in this case is simple, but if you're going to include all the other details, you might as well throw that in.
     
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