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

    $\displaystyle A = \{x \in \mathbb{Q} : x<a\},
    B = \{x \in \mathbb{Q} : a<x<b\},
    C = \{x \in \mathbb{Q} : b<x\}$

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

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

    Consider the set

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

    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 $\displaystyle r=1$ is the only positive rational number such that $\displaystyle r + \frac{1}{r}$ is an integer.

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

    (d) How many positive real numbers x are there such that $\displaystyle x + \frac{1}{x}$ 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 $\displaystyle x + \frac{1}{x}$ is an integer? I don't know...)

    Thanks for any help!
    Last edited by Jason Bourne; Jan 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 $\displaystyle a$ and $\displaystyle b$ be ratonal numbers satisfying $\displaystyle a < b$. To which of the following three sets does the number $\displaystyle \frac{2a +3b}{5}$ belong:

    $\displaystyle A = \{x \in \mathbb{Q} : x<a\},
    B = \{x \in \mathbb{Q} : a<x<b\},
    C = \{x \in \mathbb{Q} : b<x\}$

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

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

    and:

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

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

    (and anything in $\displaystyle B$ cannot be in $\displaystyle A$ or $\displaystyle C$)

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

    Suppose that $\displaystyle \left\{ {\frac{a}{{5^J }},\frac{b}{{5^K }}} \right\} \subset A\;\& \;\frac{a}{{5^J }} < \frac{b}{{5^K }}$.
    So $\displaystyle \left( {\exists N \in \mathbb{Z^+}} \right)\left[ {\frac{b}{{5^K }} - \frac{a}
    {{5^J }} > \frac{1}{{5^N }}} \right]$.
    But this means that $\displaystyle \frac{{5^N b}}{{5^K }} - \frac{{5^N a}}{{5^J }} > 1$ which by the lemma gives $\displaystyle \left( {\exists M \in \mathbb{Z}^ + } \right)\left[ {\frac{{5^N a}}
    {{5^J }} < M < \frac{{5^N b}}{{5^K }}} \right]\; \Rightarrow \;\left[ {\frac{a}
    {{5^J }} < \frac{M}{{5^N }} < \frac{b}{{5^K }}} \right]$.
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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 $\displaystyle x + \frac{1}{x}$ 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 $\displaystyle x + \frac{1}{x}$ is equal to an integer?
    Suppose for some integer $\displaystyle n$ that:

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

    then:

    $\displaystyle x^2-nx+1=0$

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

    $\displaystyle n^2-4=k^2$

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

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

    $\displaystyle n^2-4>0$

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

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

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

    .
    Last edited by Constatine11; Jan 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 $\displaystyle x$ 's such that $\displaystyle x+1/x$ is an integer.
    But in part (c) I prove that:

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

    so x = 1 is the only rational such that $\displaystyle x +\frac{1}{x}$ 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 $\displaystyle r=1$ is the only positive rational number such that $\displaystyle r + \frac{1}{r}$ is an integer.

    so x = 1 is the only rational such that $\displaystyle x +\frac{1}{x}$ 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; Jan 14th 2009 at 12:20 AM.
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