Proof by Induction - New to proofs












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totally new to proofs and found this challenge problem and struggling a bit. Any help would be appreciated!



There are some real numbers $x$ such that $x+frac{1}{x}$ is an
integer. For example, $2+sqrt{3}+frac{1}{2+sqrt{3}}=4$,
$1+frac{1}{1}=2$, and $2sqrt{6}-5+frac{1}{2sqrt{6}-5}=-10$.



Prove for all $xinmathbb{R}$ that if $x+frac{1}{x}$ is an integer,
then $x^n +frac{1}{x^n}$ also is an integer for all $ninmathbb{N}$.






induction







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  • $begingroup$
    I recommend this answer as a good start
    $endgroup$
    – Ross Millikan
    2 hours ago










  • $begingroup$
    math.stackexchange.com/questions/936479/…
    $endgroup$
    – lab bhattacharjee
    2 hours ago
















2












$begingroup$


totally new to proofs and found this challenge problem and struggling a bit. Any help would be appreciated!



There are some real numbers $x$ such that $x+frac{1}{x}$ is an
integer. For example, $2+sqrt{3}+frac{1}{2+sqrt{3}}=4$,
$1+frac{1}{1}=2$, and $2sqrt{6}-5+frac{1}{2sqrt{6}-5}=-10$.



Prove for all $xinmathbb{R}$ that if $x+frac{1}{x}$ is an integer,
then $x^n +frac{1}{x^n}$ also is an integer for all $ninmathbb{N}$.






induction







share|cite|improve this question







New contributor




Robin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    I recommend this answer as a good start
    $endgroup$
    – Ross Millikan
    2 hours ago










  • $begingroup$
    math.stackexchange.com/questions/936479/…
    $endgroup$
    – lab bhattacharjee
    2 hours ago














2












2








2





$begingroup$


totally new to proofs and found this challenge problem and struggling a bit. Any help would be appreciated!



There are some real numbers $x$ such that $x+frac{1}{x}$ is an
integer. For example, $2+sqrt{3}+frac{1}{2+sqrt{3}}=4$,
$1+frac{1}{1}=2$, and $2sqrt{6}-5+frac{1}{2sqrt{6}-5}=-10$.



Prove for all $xinmathbb{R}$ that if $x+frac{1}{x}$ is an integer,
then $x^n +frac{1}{x^n}$ also is an integer for all $ninmathbb{N}$.






induction







share|cite|improve this question







New contributor




Robin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




totally new to proofs and found this challenge problem and struggling a bit. Any help would be appreciated!



There are some real numbers $x$ such that $x+frac{1}{x}$ is an
integer. For example, $2+sqrt{3}+frac{1}{2+sqrt{3}}=4$,
$1+frac{1}{1}=2$, and $2sqrt{6}-5+frac{1}{2sqrt{6}-5}=-10$.



Prove for all $xinmathbb{R}$ that if $x+frac{1}{x}$ is an integer,
then $x^n +frac{1}{x^n}$ also is an integer for all $ninmathbb{N}$.






induction




induction






share|cite|improve this question







New contributor




Robin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Robin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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asked 2 hours ago









RobinRobin

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  • $begingroup$
    I recommend this answer as a good start
    $endgroup$
    – Ross Millikan
    2 hours ago










  • $begingroup$
    math.stackexchange.com/questions/936479/…
    $endgroup$
    – lab bhattacharjee
    2 hours ago


















  • $begingroup$
    I recommend this answer as a good start
    $endgroup$
    – Ross Millikan
    2 hours ago










  • $begingroup$
    math.stackexchange.com/questions/936479/…
    $endgroup$
    – lab bhattacharjee
    2 hours ago
















$begingroup$
I recommend this answer as a good start
$endgroup$
– Ross Millikan
2 hours ago




$begingroup$
I recommend this answer as a good start
$endgroup$
– Ross Millikan
2 hours ago












$begingroup$
math.stackexchange.com/questions/936479/…
$endgroup$
– lab bhattacharjee
2 hours ago




$begingroup$
math.stackexchange.com/questions/936479/…
$endgroup$
– lab bhattacharjee
2 hours ago










1 Answer
1






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9












$begingroup$

HINT: Note that for $ngeq1$ you have
$$left(x^n+frac{1}{x^n}right)left(x+frac{1}{x}right)=left(x^{n+1}+frac{1}{x^{n+1}}right)+left(x^{n-1}+frac{1}{x^{n-1}}right).$$



For more details, hover over the the block below:




The equation above can be rewritten to get
$$x^{n+1}+frac{1}{x^{n+1}}=left(x^n+frac{1}{x^n}right)left(x+frac{1}{x}right)-left(x^{n-1}+frac{1}{x^{n-1}}right).$$
If the three terms in parentheses on the right hand side are integers, then so is the left hand side. Now to use induction, all you need is that $x^n+frac{1}{x^n}$ is an integer for $n=0$ and $n=1$.







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    9












    $begingroup$

    HINT: Note that for $ngeq1$ you have
    $$left(x^n+frac{1}{x^n}right)left(x+frac{1}{x}right)=left(x^{n+1}+frac{1}{x^{n+1}}right)+left(x^{n-1}+frac{1}{x^{n-1}}right).$$



    For more details, hover over the the block below:




    The equation above can be rewritten to get
    $$x^{n+1}+frac{1}{x^{n+1}}=left(x^n+frac{1}{x^n}right)left(x+frac{1}{x}right)-left(x^{n-1}+frac{1}{x^{n-1}}right).$$
    If the three terms in parentheses on the right hand side are integers, then so is the left hand side. Now to use induction, all you need is that $x^n+frac{1}{x^n}$ is an integer for $n=0$ and $n=1$.







    share|cite|improve this answer









    $endgroup$


















      9












      $begingroup$

      HINT: Note that for $ngeq1$ you have
      $$left(x^n+frac{1}{x^n}right)left(x+frac{1}{x}right)=left(x^{n+1}+frac{1}{x^{n+1}}right)+left(x^{n-1}+frac{1}{x^{n-1}}right).$$



      For more details, hover over the the block below:




      The equation above can be rewritten to get
      $$x^{n+1}+frac{1}{x^{n+1}}=left(x^n+frac{1}{x^n}right)left(x+frac{1}{x}right)-left(x^{n-1}+frac{1}{x^{n-1}}right).$$
      If the three terms in parentheses on the right hand side are integers, then so is the left hand side. Now to use induction, all you need is that $x^n+frac{1}{x^n}$ is an integer for $n=0$ and $n=1$.







      share|cite|improve this answer









      $endgroup$
















        9












        9








        9





        $begingroup$

        HINT: Note that for $ngeq1$ you have
        $$left(x^n+frac{1}{x^n}right)left(x+frac{1}{x}right)=left(x^{n+1}+frac{1}{x^{n+1}}right)+left(x^{n-1}+frac{1}{x^{n-1}}right).$$



        For more details, hover over the the block below:




        The equation above can be rewritten to get
        $$x^{n+1}+frac{1}{x^{n+1}}=left(x^n+frac{1}{x^n}right)left(x+frac{1}{x}right)-left(x^{n-1}+frac{1}{x^{n-1}}right).$$
        If the three terms in parentheses on the right hand side are integers, then so is the left hand side. Now to use induction, all you need is that $x^n+frac{1}{x^n}$ is an integer for $n=0$ and $n=1$.







        share|cite|improve this answer









        $endgroup$



        HINT: Note that for $ngeq1$ you have
        $$left(x^n+frac{1}{x^n}right)left(x+frac{1}{x}right)=left(x^{n+1}+frac{1}{x^{n+1}}right)+left(x^{n-1}+frac{1}{x^{n-1}}right).$$



        For more details, hover over the the block below:




        The equation above can be rewritten to get
        $$x^{n+1}+frac{1}{x^{n+1}}=left(x^n+frac{1}{x^n}right)left(x+frac{1}{x}right)-left(x^{n-1}+frac{1}{x^{n-1}}right).$$
        If the three terms in parentheses on the right hand side are integers, then so is the left hand side. Now to use induction, all you need is that $x^n+frac{1}{x^n}$ is an integer for $n=0$ and $n=1$.








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        ServaesServaes

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