Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By johng

Math Help - Need help for a proof, related to decimal representation of a rational number

  1. #1
    Member
    Joined
    Apr 2011
    From
    Somwhere in cyberspace.
    Posts
    85
    Thanks
    4

    Need help for a proof, related to decimal representation of a rational number

    The problem appears in Tom Apostol's Calculus, Volume 1, pages 31,32. This is introductory material, related to foundations of real number system.
    Quote:
    Since we shall do very little with decimals in this book, we shall not develop their properties in any further detail except to mention how decimal expansions may be defined analytically with the help of the Least-Upper-Bound axiom.
    If x is a given positive real number, let a_{0} denote the largest integer \le x Having chosen a_{0}, we let a_{1} denote the largest integer such that
     a_{0}+\frac{a_{1}}{10}\le x
    More generally, having chosen a_{0},a_{1},\ldots,a_{n-1} we let a_{n} denote the largest integer such that
    a_{0}+\frac{a_{1}}{10}+\frac{a_{2}}{10^{2}}+\ldots  +\frac{a_{n}}{10^{n}}\le x
    Let S denote the set of all numbers
    a_{0}+\frac{a_{1}}{10}+\frac{a_{2}}{10^{2}}+\ldots  +\frac{a_{n}}{10^{n}}
    obtained this way for n=0,1,2... Then S is nonempty and bounded above, and it is easy to verify that x is actually the least upper bound of S.
    No proof is given and as often happens to me, contrary to the claim, I could not do the thing easily.

    So I need a proof that sup(S)=x.
    For this purpose I should use Least-Upper-Bound axiom which roughly states that any set of real numbers that has upper bound has supremum too (which is real number).

    I did try to prove it, resulting somewhat ugly and cumbersome proof, which most surely has weaknesses. I'm giving it here in hope someone could
    comment on it. My attempt at the proof uses two intermediate facts.

    a/ If a,x,y are real numbers which satisfy equalities a\le x \le a + \frac{y}{n} for any n\ge 1 then a=x. It is proven in the textbook as consequence of the Least-Upper-Bound axiom
    b/ \frac{1}{n}>\frac{1}{10^{n}} for any n \ge 1.

    The proof is by induction on n. Case n=1 is obviously true, so assuming
    \frac{1}{n}>\frac{1}{10^{n}} is true we need to prove \frac{1}{n+1}>\frac{1}{10^{n+1}}
    \frac{1}{n+1}>\frac{1}{10^{n+1}} \iff \frac{1}{n+1}> \frac{1}{10^{n}} \frac{1}{10}
    Since \frac{1}{n}>\frac{1}{10^{n}}
    if we manage to prove that
    \frac{1}{n+1}> \frac{1}{n} \frac{1}{10} we shall have
    \frac{1}{n+1}> \frac{1}{n} \frac{1}{10} > \frac{1}{10^{n}} \frac{1}{10}.
    This is easy
    \frac{1}{n+1} > \frac{1}{n} \frac{1}{10} \iff 10n > n+1 \iff 9n>1 which is true for all n\ge 1.

    Now to the proof of the main statement.

    By construction of x,

    a_{0}+\frac{a_{1}}{10}+\frac{a_{2}}{10^{2}}+\ldots  +\frac{a_{n}}{10^{n}} \le x < a_{0}+\frac{a_{1}}{10}+\frac{a_{2}}{10^{2}}+\ldots  +\frac{a_{n}}{10^{n}} + \frac{1}{10^{n}} for any n \ge 0 and x is an upper bound of S.
    By Least-Upper-Bound axiom, S has supremum and it's a real number. Denote y=sup(S). Then
    a_{0}+\frac{a_{1}}{10}+\frac{a_{2}}{10^{2}}+\ldots  +\frac{a_{n}}{10^{n}} \le y for any n \ge 0.
    Combining the two last inequalities with b/ we can conclude that
     x < a_{0}+\frac{a_{1}}{10}+\frac{a_{2}}{10^{2}}+\ldots  +\frac{a_{n}}{10^{n}} + \frac{1}{10^{n}} \le y + \frac{1}{10^{n}} \le y + \frac{1}{n} for any n \ge 0.
    And because y is least upper bound we have
     y \le x < y + \frac{1}{n} for any positive integer n. Applying a/ gives us the equality x=y.

    I'd appreciate any comments as well as one-liner proof by contradiction for example.
    Live long and prosper ;-).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2012
    From
    Athens, OH, USA
    Posts
    639
    Thanks
    257

    Re: Need help for a proof, related to decimal representation of a rational number

    Hi,
    Your proof looks fine to me. I think maybe Tom was thinking along these lines:
    Let s_n=a_0+{a_1\over10}+\cdots+{a_n\over10^n} and S=\{s_n:n\geq0\}. Then sup S=x:

    As you point out, for any n\geq0,s_n\leq x<s_n+{1\over10^n and so x is an upper bound of S. Let \epsilon>0. Choose n with {1\over10^n}<\epsilon -- n can be such that \text{log}_{10}(\epsilon^{-1})<n. Then s_n\in S and x-\epsilon<s_n\leq x. Done.
    Thanks from mrproper
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2011
    From
    Somwhere in cyberspace.
    Posts
    85
    Thanks
    4

    Re: Need help for a proof, related to decimal representation of a rational number

    Thank You very much for the proof and sorry for the delayed feedback. This seems quite much like one-liner I asked for!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: April 1st 2013, 05:33 PM
  2. Proof with divisibility and decimal representation
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: November 12th 2010, 12:25 AM
  3. Rational number if n is even proof
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: January 15th 2009, 11:19 AM
  4. rational number representation
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 4th 2008, 07:20 PM
  5. Rational Number Proof
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: September 14th 2006, 05:05 PM

Search Tags


/mathhelpforum @mathhelpforum