Combinatorics: Prove that $x, y, z$ exist such that $frac {1}{2} leq frac {x^2}{yz} leq 2$












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Suppose we have a subset of $2n-1$ numbers from the set ${1, 2, 3, ..., 2^n-2}$. Prove that there exits $x, y, z$ such that $frac {1}{2} leq frac {x^2}{yz} leq 2$



I'm fairly new to this topic, and i honestly have no idea how to prove this. Any hint on how can i start the proof?



Thanks in advance.






combinatorics pigeonhole-principle







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  • 2




    $begingroup$
    Divide the numbers into 123, 4567, 8 to 15, 16 to 31, etc and apply pidgeonhole principle.
    $endgroup$
    – didgogns
    Dec 19 '18 at 2:49










  • $begingroup$
    If the numbers don't have to be distinct, this is trivial.
    $endgroup$
    – Matt Samuel
    Dec 19 '18 at 3:03
















1












$begingroup$


Suppose we have a subset of $2n-1$ numbers from the set ${1, 2, 3, ..., 2^n-2}$. Prove that there exits $x, y, z$ such that $frac {1}{2} leq frac {x^2}{yz} leq 2$



I'm fairly new to this topic, and i honestly have no idea how to prove this. Any hint on how can i start the proof?



Thanks in advance.






combinatorics pigeonhole-principle







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Divide the numbers into 123, 4567, 8 to 15, 16 to 31, etc and apply pidgeonhole principle.
    $endgroup$
    – didgogns
    Dec 19 '18 at 2:49










  • $begingroup$
    If the numbers don't have to be distinct, this is trivial.
    $endgroup$
    – Matt Samuel
    Dec 19 '18 at 3:03














1












1








1


1



$begingroup$


Suppose we have a subset of $2n-1$ numbers from the set ${1, 2, 3, ..., 2^n-2}$. Prove that there exits $x, y, z$ such that $frac {1}{2} leq frac {x^2}{yz} leq 2$



I'm fairly new to this topic, and i honestly have no idea how to prove this. Any hint on how can i start the proof?



Thanks in advance.






combinatorics pigeonhole-principle







share|cite|improve this question









$endgroup$




Suppose we have a subset of $2n-1$ numbers from the set ${1, 2, 3, ..., 2^n-2}$. Prove that there exits $x, y, z$ such that $frac {1}{2} leq frac {x^2}{yz} leq 2$



I'm fairly new to this topic, and i honestly have no idea how to prove this. Any hint on how can i start the proof?



Thanks in advance.






combinatorics pigeonhole-principle




combinatorics pigeonhole-principle






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 19 '18 at 2:44









AnthonyAnthony

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324








  • 2




    $begingroup$
    Divide the numbers into 123, 4567, 8 to 15, 16 to 31, etc and apply pidgeonhole principle.
    $endgroup$
    – didgogns
    Dec 19 '18 at 2:49










  • $begingroup$
    If the numbers don't have to be distinct, this is trivial.
    $endgroup$
    – Matt Samuel
    Dec 19 '18 at 3:03














  • 2




    $begingroup$
    Divide the numbers into 123, 4567, 8 to 15, 16 to 31, etc and apply pidgeonhole principle.
    $endgroup$
    – didgogns
    Dec 19 '18 at 2:49










  • $begingroup$
    If the numbers don't have to be distinct, this is trivial.
    $endgroup$
    – Matt Samuel
    Dec 19 '18 at 3:03








2




2




$begingroup$
Divide the numbers into 123, 4567, 8 to 15, 16 to 31, etc and apply pidgeonhole principle.
$endgroup$
– didgogns
Dec 19 '18 at 2:49




$begingroup$
Divide the numbers into 123, 4567, 8 to 15, 16 to 31, etc and apply pidgeonhole principle.
$endgroup$
– didgogns
Dec 19 '18 at 2:49












$begingroup$
If the numbers don't have to be distinct, this is trivial.
$endgroup$
– Matt Samuel
Dec 19 '18 at 3:03




$begingroup$
If the numbers don't have to be distinct, this is trivial.
$endgroup$
– Matt Samuel
Dec 19 '18 at 3:03










1 Answer
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HINT.- Consider the rational function $f(X)=dfrac{(X-2)^2}{(X-3)(X-4)}$.



It is easy to prove that $$begin{cases}f(X)le2text{ when } Xgt7\f(X)gedfrac12text{ when } Xgt5end{cases}$$
Making now $X=2^x$ the inequalities above prove that for $nge3$ (in which case $Xge8$) a solution is given by
$$x=2^n-2\y=2^n-3\z=2^n-4$$ It remains to solve the values $n=1$ and $n=2$ in which cases $(x,y,z)=(1,1,1),(1,1,2)$ are trivial solutions.






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    $begingroup$

    HINT.- Consider the rational function $f(X)=dfrac{(X-2)^2}{(X-3)(X-4)}$.



    It is easy to prove that $$begin{cases}f(X)le2text{ when } Xgt7\f(X)gedfrac12text{ when } Xgt5end{cases}$$
    Making now $X=2^x$ the inequalities above prove that for $nge3$ (in which case $Xge8$) a solution is given by
    $$x=2^n-2\y=2^n-3\z=2^n-4$$ It remains to solve the values $n=1$ and $n=2$ in which cases $(x,y,z)=(1,1,1),(1,1,2)$ are trivial solutions.






    share|cite|improve this answer









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      2












      $begingroup$

      HINT.- Consider the rational function $f(X)=dfrac{(X-2)^2}{(X-3)(X-4)}$.



      It is easy to prove that $$begin{cases}f(X)le2text{ when } Xgt7\f(X)gedfrac12text{ when } Xgt5end{cases}$$
      Making now $X=2^x$ the inequalities above prove that for $nge3$ (in which case $Xge8$) a solution is given by
      $$x=2^n-2\y=2^n-3\z=2^n-4$$ It remains to solve the values $n=1$ and $n=2$ in which cases $(x,y,z)=(1,1,1),(1,1,2)$ are trivial solutions.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        HINT.- Consider the rational function $f(X)=dfrac{(X-2)^2}{(X-3)(X-4)}$.



        It is easy to prove that $$begin{cases}f(X)le2text{ when } Xgt7\f(X)gedfrac12text{ when } Xgt5end{cases}$$
        Making now $X=2^x$ the inequalities above prove that for $nge3$ (in which case $Xge8$) a solution is given by
        $$x=2^n-2\y=2^n-3\z=2^n-4$$ It remains to solve the values $n=1$ and $n=2$ in which cases $(x,y,z)=(1,1,1),(1,1,2)$ are trivial solutions.






        share|cite|improve this answer









        $endgroup$



        HINT.- Consider the rational function $f(X)=dfrac{(X-2)^2}{(X-3)(X-4)}$.



        It is easy to prove that $$begin{cases}f(X)le2text{ when } Xgt7\f(X)gedfrac12text{ when } Xgt5end{cases}$$
        Making now $X=2^x$ the inequalities above prove that for $nge3$ (in which case $Xge8$) a solution is given by
        $$x=2^n-2\y=2^n-3\z=2^n-4$$ It remains to solve the values $n=1$ and $n=2$ in which cases $(x,y,z)=(1,1,1),(1,1,2)$ are trivial solutions.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 25 '18 at 22:49









        PiquitoPiquito

        18.1k31539




        18.1k31539






























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