Results 1 to 7 of 7

Math Help - Dedekind Cut

  1. #1
    Banned
    Joined
    Sep 2009
    Posts
    502

    Dedekind Cut

    Let \alpha = 0^* \cup \{p \in \mathbf{Q}: p \geq 0 and p^2 <2 \}. Prove that \alpha is a Dedekind cut and also that it has the property \alpha \cdot \alpha = 2^*; that is, the square of \alpha is 2^*. Note: This seems to be surprisingly tricky, as pointed out by Linda Hill and Robert J. Fisher at Idaho State University. Their solution is available from them or from the author.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by novice View Post
    Let \alpha = 0^* \cup \{p \in \mathbf{Q}: p \geq 0 and p^2 <2 \}. Prove that \alpha is a Dedekind cut and also that it has the property \alpha \cdot \alpha = 2^*; that is, the square of \alpha is 2^*. Note: This seems to be surprisingly tricky, as pointed out by Linda Hill and Robert J. Fisher at Idaho State University. Their solution is available from them or from the author.
    Since you have a solution in what way is this a question?

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,306
    Thanks
    1283
    Obviously a homework problem.

    The only real difficulty is in showing that the set does NOT have a largest member.

    Suppose x is the largest member of the set of all rational numbers whose square is less than 2. Since 1.4^2= 1.96< 2 you can assume 1.4< x. Since 1.5^2= 2.25> 2 you can assume x< 1.5. Let d= 2- x^2 (d is, of course, rational). Then d< 2- 1.4^2= 2- 1.96= 0.04. Now find an integer, n such that (x+ d/n)^2= x^2+ 2 dx/n+ d^2/n^2< 2 showing that x is NOT the largest number in that set.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Sep 2009
    Posts
    502
    Thank you, sir. You instruction is more than adequate. I knew the set has no maximum, but didn't know how to commucate my idea in mathematical term. The last line in particular is of great help.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Sep 2009
    Posts
    502
    Quote Originally Posted by CaptainBlack View Post
    Since you have a solution in what way is this a question?

    CB
    Respectfully sir, I do not have the solution. It will take at least three weeks to get the book by the inter-library loan. I could eventually get my answer one way or the other, but I NOT am too proud to ask.
    Last edited by novice; September 22nd 2010 at 06:37 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Sep 2010
    Posts
    185
    Thanks
    13
    Quote Originally Posted by novice View Post
    Respectfully sir, I do not have the solution. It will take at least three weeks to get the book by the inter-library loan. I could eventually get my answer one way or the other, but I am too proud to ask.
    take a peek at wikipedia, eh?!
    Last edited by CaptainBlack; September 22nd 2010 at 01:23 PM. Reason: remove irony as justification was a typo
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Banned
    Joined
    Sep 2009
    Posts
    502
    Quote Originally Posted by HallsofIvy View Post
    Obviously a homework problem.

    The only real difficulty is in showing that the set does NOT have a largest member.

    Suppose x is the largest member of the set of all rational numbers whose square is less than 2. Since 1.4^2= 1.96< 2 you can assume 1.4< x. Since 1.5^2= 2.25> 2 you can assume x< 1.5. Let d= 2- x^2 (d is, of course, rational). Then d< 2- 1.4^2= 2- 1.96= 0.04. Now find an integer, n such that (x+ d/n)^2= x^2+ 2 dx/n+ d^2/n^2< 2 showing that x is NOT the largest number in that set.
    Since the domain of \sqrt{x} is \mathbb{Z}^+ \cup {0}, does it mean that the set \{p\in \mathbb{Q}: p \geq 0 and p^2 <2 \} is not a Dedekin cut in itself, unless it is unioned with 0^*?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: October 2nd 2011, 02:26 AM
  2. e as a Dedekind cut
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: February 14th 2011, 08:38 PM
  3. dedekind sets
    Posted in the Advanced Math Topics Forum
    Replies: 6
    Last Post: November 9th 2009, 12:07 PM
  4. sum of two Dedekind cuts
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: August 15th 2009, 02:13 AM
  5. Dedekind Cuts
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: January 7th 2009, 08:32 AM

Search Tags


/mathhelpforum @mathhelpforum