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Thread: Prove:

  1. #1
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    Prove:

    For any integer $\displaystyle n>1$, prove that there is exactly one power of 2 having exactly n digits with leading digit 1
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  2. #2
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    Quote Originally Posted by newtoinequality View Post
    For any integer $\displaystyle n>1$, prove that there is exactly one power of 2 having exactly n digits with leading digit 1
    Hi

    It can be done by induction
    The property is true for n=2 (16 = 2^4)

    Suppose that it is true for n. This means that there exists p integer such that $\displaystyle 10^{n-1} \leq 2^p < 2 \cdot 10^{n-1}$

    You have to demonstrate that there exists q integer such that $\displaystyle 10^{n} \leq 2^q < 2 \cdot 10^{n}$
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  3. #3
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    Quote Originally Posted by running-gag View Post
    Hi

    It can be done by induction
    The property is true for n=2 (16 = 2^4)

    Suppose that it is true for n. This means that there exists p integer such that $\displaystyle 10^{n-1} \leq 2^p < 2 \cdot 10^{n-1}$

    You have to demonstrate that there exists q integer such that $\displaystyle 10^{n} \leq 2^q < 2 \cdot 10^{n}$
    You mean $\displaystyle 10^n\leq 2^{n+1}< 2\cdot 10^{n}$, don't you?
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  4. #4
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    Quote Originally Posted by newtoinequality View Post
    For any integer $\displaystyle n>1$, prove that there is exactly one power of 2 having exactly n digits with leading digit 1
    First, show existence. If the statement were false, then there would exist $\displaystyle m,\,n$ such that $\displaystyle 2^m<10^{n-1}$ and $\displaystyle 2^{m+1}\ge2\cdot10^{n-1}.$ But the former gives $\displaystyle 2^{m+1}<2\cdot10^{n-1}$ which is a contradiction of the latter.

    Hence for each $\displaystyle n$ there must be $\displaystyle m$ such that $\displaystyle 10^{n-1}\le2^m<2\cdot10^{n-1},$ i.e.

    $\displaystyle 2^m\ <\ 2\cdot10^{n-1}\quad\ldots\ \fbox1$

    $\displaystyle 2^m\ \ge\ 10^{n-1}\quad\ldots\ \fbox2$

    To prove uniqueness, observe that $\displaystyle 2^{m-1}<10^{n-1}$ (from $\displaystyle \fbox1)$ and $\displaystyle 2^{m+1}\ge2\cdot10^{n-1}$ (from $\displaystyle \fbox2).$ The result follows since if $\displaystyle m'\ne m$ then either $\displaystyle 2^{m'}\le2^{m-1}$ or $\displaystyle 2^{m'}\ge2^{m+1}$ and so $\displaystyle 2^{m'}$ cannot be between $\displaystyle 10^{n-1}$ and $\displaystyle 2\cdot10^{n-1}.$
    Last edited by TheAbstractionist; Jul 23rd 2009 at 12:08 PM.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    You mean $\displaystyle 10^n\leq 2^{n+1}< 2\cdot 10^{n}$, don't you?
    No

    For instance $\displaystyle 10^3 \leq 2^{10} < 2 \cdot 10^3$
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