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Math Help - "Denseness" and addition of irrationals

  1. #1
    Member Jason Bourne's Avatar
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    "Denseness" and addition of irrationals

    I'm not sure on the following parts, I insert what I'm not sure on in () brackets.

    (a) Let a and b be ratonal numbers satisfying a < b. To which of the following three sets does the number \frac{2a +3b}{5} belong:

    A = \{x \in \mathbb{Q} : x<a\},<br />
B = \{x \in \mathbb{Q} : a<x<b\},<br />
C = \{x \in \mathbb{Q} : b<x\}

    (It seems fairly obvious that it's in set B but I'm not sure exactly why...)

    (b) Let X be a set and let \leq be a total order on X. We say that X is dense if for any two x,y \in X with x < y there exists z \in X such that x<z<y

    Consider the set

    A = \{\frac{a}{5^n} : a,n \in \mathbb{Z}, n \geq 0\}

    Is A dense?

    (it looks as though it is dense because for every two elements you can incerease n so that you find an element between them but I wasn't 100% sure on this.)

    (c) Prove that r=1 is the only positive rational number such that r + \frac{1}{r} is an integer.

    (I'm not sure what to do here, I tried to suppose that r + \frac{1}{r} is an integer and that r = \frac{a}{b} and then \frac{a^2 +b^2}{ab} is an integer but that this only happens when a=b=1, is any of that right?)

    (d) How many positive real numbers x are there such that x + \frac{1}{x} is an integer? Answer should be one of: 0,1,2,3,..., countable infinite, uncountably infinite.

    (x=1 is the only rational from the previous part so its certainly not 0. But are there irrationals x such that x + \frac{1}{x} is an integer? I don't know...)

    Thanks for any help!
    Last edited by Jason Bourne; January 13th 2009 at 11:06 PM.
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    Quote Originally Posted by Jason Bourne View Post
    I'm not sure on the following parts, I insert what I'm not sure on in () brackets.

    (a) Let a and b be ratonal numbers satisfying a < b. To which of the following three sets does the number \frac{2a +3b}{5} belong:

    A = \{x \in \mathbb{Q} : x<a\},<br />
B = \{x \in \mathbb{Q} : a<x<b\},<br />
C = \{x \in \mathbb{Q} : b<x\}

    (It seems fairly obvious that it's in set B but I'm not sure exactly why...)
    As a<b we have:

    \frac{2a +3b}{5}<\frac{2b +3b}{5}=b

    and:

    \frac{2a +3b}{5}>\frac{2a +3a}{5}=a

    hence \frac{2a +3b}{5} \in B

    (and anything in B cannot be in A or C)

    .
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  3. #3
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    Quote Originally Posted by Jason Bourne View Post
    (b) Consider the set
    A = \{\frac{a}{5^n} : a,n \in \mathbb{Z}, n \geq 0\}
    Is A dense?
    Here is an outline of the trick in this proof.
    First a lemma without proof: 1 < y - x\, \Rightarrow \,\left( {\exists n \in \mathbb{Z}} \right)\left[ {x < n < y} \right].

    Suppose that \left\{ {\frac{a}{{5^J }},\frac{b}{{5^K }}} \right\} \subset A\;\& \;\frac{a}{{5^J }} < \frac{b}{{5^K }}.
    So \left( {\exists N \in \mathbb{Z^+}} \right)\left[ {\frac{b}{{5^K }} - \frac{a}<br />
{{5^J }} > \frac{1}{{5^N }}} \right].
    But this means that \frac{{5^N b}}{{5^K }} - \frac{{5^N a}}{{5^J }} > 1 which by the lemma gives \left( {\exists M \in \mathbb{Z}^ +  } \right)\left[ {\frac{{5^N a}}<br />
{{5^J }} < M < \frac{{5^N b}}{{5^K }}} \right]\; \Rightarrow \;\left[ {\frac{a}<br />
{{5^J }} < \frac{M}{{5^N }} < \frac{b}{{5^K }}} \right].
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  4. #4
    Member Jason Bourne's Avatar
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    for (d) do you think that there is an irrational x such that x + \frac{1}{x} is equal to an integer?
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    Quote Originally Posted by Jason Bourne View Post
    for (d) do you think that there is an irrational x such that x + \frac{1}{x} is equal to an integer?
    Suppose for some integer n that:

    x+\frac{1}{x}=n

    then:

    x^2-nx+1=0

    and x is rational if and only if the discriminant of this quadraic is a perfect square, that is for some k \in \mathbb{N} :

    n^2-4=k^2

    As the gap between consecutive squares is greater than 4 when the lesser is the square of any number of absolute value greater than or equal 2, and the other cases are easily checked manualy the only solutions to this are k=0,\ n=\pm 1

    Now for irrational x 's. Any n \in \mathbb{Z} such that:

    n^2-4>0

    but not a perfect square then x=\frac{n+\sqrt{n^2-4}}{2} is an irrational such that x+1/x is an integer.

    In particular let n=5, then x=\frac{5 + \sqrt{21}}{2} is irrational and a solution of:

    x+\frac{1}{x}=5

    .
    Last edited by Constatine11; January 14th 2009 at 02:10 AM.
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  6. #6
    Member Jason Bourne's Avatar
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    Quote Originally Posted by Constatine11 View Post
    There are a countable infinitude of such triangles, and so there is a countable infinity of rational x 's such that x+1/x is an integer.
    But in part (c) I prove that:

    (c) Prove that r=1 is the only positive rational number such that r + \frac{1}{r} is an integer.

    so x = 1 is the only rational such that x +\frac{1}{x} is an integer. So there can't be an infinite number, there is just x=1.

    otherwise I guess there are countable infinite irrationals.
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    Quote Originally Posted by Jason Bourne View Post
    But in part (c) I prove that:

    (c) Prove that r=1 is the only positive rational number such that r + \frac{1}{r} is an integer.

    so x = 1 is the only rational such that x +\frac{1}{x} is an integer. So there can't be an infinite number, there is just x=1.

    otherwise I guess there are countable infinite irrationals.
    There is a mistake in that post that I will have to sort out.

    Should be fixed now.

    .
    Last edited by Constatine11; January 14th 2009 at 12:20 AM.
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