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Tricky induction

  1. Aug 30, 2010 #1
    Q.Show that any integral power of [tex](\sqrt{2}-1)^k[/tex] can be wriitten as the [tex]\sqrt{N_k}-\sqrt{N_k-1}[/tex] for N a positive integer .
    ..............................................................................................................................................................................................................................
    I tried using induction to solve the problem but when looked at the book's solution they seem to suggest an extra condition that the following equation has to satisfy :
    [tex](\sqrt{2}-1)^k=\sqrt{N_k}-\sqrt{N_k-1}[/tex] for N_k a positive integer satisfying [tex]\sqrt{2}\sqrt{N_k}\sqrt{N_k-1} \in \mathbb{Z}[/tex] .

    Why is this last condition required to apparently "complete" induction argument..which I dont get why this would be necessary ?

    I managed to show the above is true for n+1 , but why would I have to show the above condition is true?

    Thanks
     
  2. jcsd
  3. Aug 30, 2010 #2

    Office_Shredder

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    Certainly it isn't necessary, if the statement is true for a positive integer satisfying your extra requirement, then it's true for just some positive integer (namely, the one satisfying your requirement). Likely what happened was in the proof they happened to show that [tex]\sqrt{2N_k (N_k -1)}[/tex] is an integer
     
  4. Aug 31, 2010 #3
    just mmanaged to find out why the condition is true

    true for k=1,

    Assume k=j
    [tex](\sqrt{2}-1)^j=\sqrt{N_j}-\sqrt{N_j-1}[/tex]

    need to prove
    [tex](\sqrt{2}-1)^{j+1}=\sqrt{N_{j+1}}-\sqrt{N_{j+1}-1}[/tex]

    So ,
    [tex](\sqrt{2}-1)^{j+1}=(\sqrt{2}-1)(\sqrt{2}-1)^j[/tex]

    using assumption step we have ,
    [tex](\sqrt{N_j}-\sqrt{N_j-1})(\sqrt{2}-1)[/tex]



    rearranging ,
    [tex](\sqrt{N_j}\sqrt{2}+\sqrt{N_j-1})-(\sqrt{2}\sqrt{N_j-1}+\sqrt{N_j})[/tex]

    Now noticing ,
    [tex](\sqrt{N_j}\sqrt{2}+\sqrt{N_j-1})^2=3N_j-1+2\sqrt{2}\sqrt{N_j}\sqrt{N_j-1}[/tex]

    let [tex]K=3N_j-1+2\sqrt{2}\sqrt{N_j}\sqrt{N_j-1} , K\in Z[/tex]

    [tex](\sqrt{2}\sqrt{N_j-1}+\sqrt{N_j})^2=3N_j-2+2\sqrt{2}\sqrt{N_j}\sqrt{N_j-1}=K-1[/tex]

    In my last line I assumed [tex]2\sqrt{2}\sqrt{N_j}\sqrt{N_j-1}[/tex] is an integer which was stupid of me.
     
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