1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question about Dedekind Cuts

  1. Apr 22, 2014 #1
    1. The problem statement, all variables and given/known data
    I don't take issue with the 'cuts' but am having difficulty understanding an example given in my text for √2. After going over the right and left classes the text jumps into,

    "Then this section is such that the R-class has no least member and the L-class has no greatest member; for, if 'x' be any positive rational fraction and,

    $$y=\frac{x(x^2+6)}{3x^2+2}$$ then $$y-x=\frac{2x(2-x^2)}{3x^2+2}$$ and $$y^2-2=\frac{(x^2-2)^3}{(3x^2+2)^2}$$, so x^2,y^2 and 2 are in order of magnitude"

    2. Relevant equations
    Given above

    3. The attempt at a solution
    A look on wikipedia shows that the last part (where it states, '...are in order of magnitude') seems to represent x^2<y^2<2, other than that I don't understand what the author is getting at or where these equations come from.

    ***EDITED*** To correct a mistake pointed out by Petek.
     
    Last edited: Apr 22, 2014
  2. jcsd
  3. Apr 22, 2014 #2
    There appears to be a typo in your above equation for [itex]y^2[/itex]. Both [itex]x[/itex] and [itex]y[/itex] are supposed to be rational numbers. However, if [itex]x=1[/itex], then $$y^2= \frac{-1}{25}$$ so [itex]y[/itex] can't even be real.
     
  4. Apr 22, 2014 #3
    You are correct, I missed a minus 2. Thanks for the correction!
     
  5. Apr 22, 2014 #4

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I think the author is asserting in somewhat obscure (Hardyish?) language that either ##x^2 < y^2 < 2## or ##2 < y^2 < x^2##, depending on whether ##x \in L## or ##x \in R##.

    To see that this is true, consider the two possibilities:

    Case 1: ##x \in L##

    Then ##x^2 < 2##, so the formulas for ##y-x## and ##y^2 - 2## imply that ##y-x > 0## and ##y^2 - 2 < 0##.

    Case 2: ##x \in R##

    Then ##x^2 > 2##, so the formulas imply that ##y-x < 0## and ##y^2 - 2 > 0##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Question about Dedekind Cuts
  1. Dedekind Cuts (Replies: 9)

  2. Dedekind cut (Replies: 0)

Loading...