Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dedekind cut

  1. Jul 13, 2008 #1
    How do I show that A={r in Q: r^3<2} is a Dedekind cut.


    Here is the definition I am working with.

    A subset A of Q is a Dedekind cut if and only if A satisfies the
    following 3 properties:

    (i) A is a proper nonempty subset of Q.
    (ii) If r is in A, s in Q, and s<r, then s is in A.
    (iii) A contains no greatest rational.

    I showed that A satisfies (i) and (ii). I noticed that 5/4 is in A
    and I tried to find a rational greater than 5/4 whose cube is less
    than 2. I looked at the sequence (5n+1)/4n, and I found that n=26
    works. That is, (4/5)<(131/104) and (131/104)^3 < 2.

    So I think that if a/b is any rational with b>0 and (a/b)^3 < 2, then
    there should be some positive integer n such that [(an+1)/bn]^3 < 2. But I don't know
    how to show that this n exists.

    I tried contradiction:
    Let a/b be in A with b>0 , and assume that [(an+1)/bn]^3 >= 2 for all
    positive n. Then (a/b)^3 < 2 <= [(an+1)/bn]^3 for all positive n. I
    think this may imply that 2^(1/3) is rational, which I know is not true.

    Am I on the right track?
     
  2. jcsd
  3. Jul 13, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I would start with some specific numbers. Since the cube root of 2 is approximately 1.26 (from a calculator) note that 1.253= 1.953125< 2 so 1.25 is in the set. On the other hand 1.263= 2.000376> 2 so 1.26 is not in the set.

    The point of that is that any r in the set that is less than 1.25 has 1.25 larger so if there were a largest member it would have to be between 1.25 and 1.26. Let [itex]\delta= 2- x^3[/itex]. Since 1.253= 1.953125, [itex]\delta< 2- 1.95325= 0.045875[/itex].

    Now, try to find some number, n (a positive integer just for simplicity) so that [itex](x+ \delta/n)^3= x^3+ 3(\delta/n)x^2+ 3(\delta/n)^2 x+ (\delta/n)^3= x^3+ \delta(3x^2/n+ 3\delta x/n^2+ \delta^2/n^3)< 2[/itex]
    Remembering the bounds on the sized for x and [itex]\delta[/itex] that should be easy.

    Proof by contradiction: Suppose there were some number, x, which were the largest member of {x| x3< 2}. That, as above, [itex]1.25< x< 1.27
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?