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Homework Help: Prove that every real number x in [0,1] has a decimal expansion.

  1. Mar 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that every real number x in [0,1] has a decimal expansion.

    2. Relevant equations

    Let [itex]x\in{[0,1]}[/itex], then the decimal expansion for x is an infinite sequence [itex](k_{i})^{\infty}_{i=1}[/itex] such that for all i, [itex]k_i[/itex] is an integer between 0 and 9 and such that [itex]x\in\left[\frac{k_1}{10}+\frac{k_2}{10^2}+\cdots +\frac{k_n}{10^n},\frac{k_1}{10} +\frac{k_2}{10^2}+ \cdots +\frac{k_{n}+1}{10^n}\right][/itex].

    We call that interval above [itex]I_{k_1,k_2,\ldots,k_n}[/itex]

    3. The attempt at a solution

    Assume BWOC that there exists a real number [itex]t\in[0,1][/itex] with no decimal expansion. That means there exists an [itex]N\in{\mathbb{Z}}[/itex] such that for all sequences [itex](k_{i})^{N}_{i=1}[/itex], [itex]t{\notin}\left[\frac{k_1}{10}+\frac{k_2}{10^2}+\cdots +\frac{k_N}{10^N},\frac{k_1}{10}+ \frac{k_2}{10^2}+ \cdots +\frac{k_{N}+1}{10^N}\right][/itex].

    But [itex]I_{0,0,\ldots,0}\cup I_{0,0,\ldots,0,1} \cup I_{0,0,\ldots,0,2} \cup \cdots \cup I_{9,9,9,\ldots,9} = [0,1][/itex]
    (That big string of unions is supposed to denote breaking up [0,1] into the union of intervals of size 10-N, but I didn't know how exactly to write it... you get the idea though.)

    So [itex]t{\notin}I_{0,0,\ldots,0}\wedge t{\notin}I_{0,0,\ldots,0,1} \wedge t{\notin}I_{0,0,\ldots,0,2} \wedge \cdots \wedge t{\notin}I_{9,9,9,\ldots,9}[/itex] implies [itex]t{\notin}[0,1][/itex], which is a contradiction.

    How does this look? Advice?
    Last edited by a moderator: Mar 22, 2013
  2. jcsd
  3. Mar 22, 2013 #2


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    Staff: Mentor

    You have to be careful with the negation. "For all x exists a sequence such that for all i [...]" negated is "There is a t without a sequence such that for all i [...]", or "There is a t such that for all sequences, there is an i where [...] is wrong."
    The problem here: Those intervals for different i are not independent. If you consider some specific i, it is possible to find an appropriate interval, sure. But that restricts the intervals for other i. The proof is still possible, but you have to take care of this. You could consider the first i where t is not in the interval, for example - if there are i with that property, there is a first one as well.
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