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Rational numbers, supremum (Is my proof correct?)

  1. Apr 13, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that the set ##\{x \in \mathbf Q; x^2< 2 \}## has no least upper bound in ##\mathbf Q##; using that if ##r## were one then ##r^2=2##. Do this assuming that the real field haven't been constructed.

    2. Relevant equations
    N/A
    3. The attempt at a solution
    Attempt at proof:
    ##r\in Q## is an upper bound of the set if and only if ##r^2 \ge 2##.
    Choose any ##r_1 \in \mathbf Q## that satisfies ##r_1^2 >2##. Then ##r_1## is an upper bound of the set.
    Set ##r_2 = r_1-\frac{r_1^2-2}{r_1+2}<r_1 \Longrightarrow r_2^2=2+\frac{2(r_1^2-2)}{(r_1+2)^2}>2##.
    Note that ##r_2 \in \mathbf Q## and ##r_2 < r_1##. ##r_2## is also an upper bound since ##r_2^2>2##.

    This shows that for every upper bound ##r_1## with ##r_1^2 > 2## there's always possible to find an upper bound ##r_2## where ##r_2<r_1##. Hence the only possible least upper bound satisfy ##r^2=2## but there is no ##r\in \mathbf Q## that satisfy that. Hence the set have no least upper bound in ##\mathbf Q##.

    Is the above correct? Anything I can do to improve it? I'm also wondering about the choice of ##r_2##. I remembered that from an earlier example in our book but I'm not sure I could've come up with a choice like that by myself without a lot of trial and error. How would I go about finding a choice like that in the first place?
     
  2. jcsd
  3. Apr 13, 2016 #2

    andrewkirk

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    How did you get the following?
    When I square ##r_2## I get 2 plus a fraction with the same denominator as yours but a messy quartic in ##r_1## in the numerator.
     
  4. Apr 13, 2016 #3

    andrewkirk

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    Given an upper bound ##r## such that ##r{}^2=2+h## for ##h>0##, you can write an equation for a lower upper bound ##s=r-k## such that ##s^2## is 2 plus some positive fraction of ##h##. Then, using the relationships you've established, solve for ##k##.
     
  5. Apr 14, 2016 #4
    Sorry, I should have clarified that step
    ##r_2 = r_1-\frac{r_1^2-2}{r_1+2} = \frac{r_1^2+2r_1-r_1^2+2}{r_1+2}=\frac{2r_1+2}{r_1+2}##
    square
    ##r_2^2 = \frac{4(r_1^2+2r_1+1)}{(r_1+2)^2} = \frac{2(r_1^2+4r_1+4)+2(r_1^2-2)}{(r_1+2)^2}= 2+\frac{2(r_1^2-2)}{(r_1+2)^2}##.
     
  6. Apr 14, 2016 #5
    Thanks for responding! Is there something missing in my proof where I need this to complete it? When you say the relationship do you mean my value of ##r_2## or something else? I'm not entirely sure I understand.
     
  7. Apr 14, 2016 #6

    andrewkirk

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    No. I think with the extra steps you've put in, your proof now looks good.
     
  8. Apr 14, 2016 #7
    Thanks for taking the time looking it over!
     
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