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A proof needed

  1. Jun 30, 2005 #1
    prove that between any two real numbers there is a number of the form
    where k is an integer and n is a natural number.
  2. jcsd
  3. Jun 30, 2005 #2


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    This should be very easy.

    Hint: Write the real numbers in binary.
  4. Jun 30, 2005 #3


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    Sounds like homework. What have you tried?
  5. Jul 3, 2005 #4
    Sounds interesting. I haven't been able to prove it, even with binary. :confused:
  6. Jul 3, 2005 #5


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    I did something similar in a proof showing that monotone convergence implies least upper bound property.

    Let the two distinct real numbers be a and b with a> b. Then a- b> 0 so [tex]\frac{1}{a-b}[/tex] exists and is positive. By the Archimedian property of integers (given any real number x, there exist an integer m with m>x) there exist an integer n with [tex]n> \frac{1}{a-b}[/tex].

    But it is easy to prove by induction that, for all positive integers n, 2n> n (Hint: first prove by induction that 2n>= n+1) so we have [tex]2^n> \frac{1}{a-b}[/tex]. That means that 2n(a- b)= 2na- 2nb> 1.
    Since the distance between those two numbers is greater than 1, there exist an integer, k, such that 2nb< k< 2na. Dividing through by 2n gives [tex]b< \frac{k}{2^n}< a[/tex].
    Last edited: Jul 3, 2005
  7. Jul 4, 2005 #6


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    That's very elegant.

    I was thinking more along the lines of:
    WOLOG [itex]a>b[/itex].
    Since [itex]a[/itex] and [itex]b[/itex] are real numbers, they can be written in binary notation in such a way that each of them has an infinite number of zeroes to the right of the binary (or is it still decimal?) point.

    Now, since [itex]a \neq b[/itex] there is a first digit where they differ. At that position, [itex]a[/itex] will have a 1, and [itex]b[/itex] will have a zero. Now, since there is an infinite number of zeros, there is a next location where b has a zero. Let's say that that's the [itex]2^{-k}[/itex]s place.

    Then the integer part of
    divided by
    Is between [itex]a[/itex] and [itex]b[/itex]
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