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Math Help - Base-Ten Representaion of Integers proof

  1. #1
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    Base-Ten Representaion of Integers proof

    To Prove:

    For all n in the Natural Numbers, v(n) = k if and only if 10^(k-1) <= n < 10^k

    v(n) is the number of digits of n with respect to base 10....so for example, v(d) = 1 for all digits d, and v(10) = 2.

    I can't seem to figure out how to work this proof. Any help would be appreciated!
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  2. #2
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    It's kind of obvious... For example, three-digit numbers are between 100 and 1000.

    If v(n) is defined as \lfloor\log_{10}n\rfloor+1, then

    \lfloor\log n\rfloor+1=k\Leftrightarrow
    \lfloor\log n\rfloor=k-1\Leftrightarrow
    k-1\le\log n<k\Leftrightarrow
    10^{k-1}\le n<10^k.
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  3. #3
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    Hint - What is the smallest and larget 'k' digit number (in decimal representation)?
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  4. #4
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    Can we prove this using the fact that n < 10^n ? Instead of the log10n ? The procedure seems to be correct, but I need to use the face that if n is in the Naturals, then n < 10^n

    Thanks for all the help!
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  5. #5
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    Why in the world would you want to use that?

    Perhaps the simplest way to prove "v(n)= k if and only if 10^{k-1}\le n< 10^k" is by induction on k.

    If k= 1, the statement becomes "v(n)= 1 if and only if 1\le n< 10.

    Given that the statement for k= K is true, can you prove it for k= K+1?
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