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Thread: Explain this!

  1. #1
    MHF Contributor Bruno J.'s Avatar
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    Explain this!

    In a column, write the integer part of the integer multiples of $\displaystyle \sqrt{2}$. Then, in the second column, write the missing integers in increasing order. In the third column, write the difference between the second and first columns. You get :

    1 3 2
    2 6 4
    4 10 6
    5 13 8
    7 17 10
    8 20 12
    9 23 14
    11 27 16
    12 30 18
    14 34 20
    15 37 22
    16 40 24
    18 44 26
    19 47 28
    21 51 30
    22 54 32
    24 58 34
    25 61 36
    26 64 38
    28 68 40
    29 71 42

    ...

    Why does the third column consist of the even integers?
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  2. #2
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    Hmmm...do you have a proof that the third column is in fact the set of positive even integers? Any easy argument would seem to require that the first column follows a pattern - at first glance it is unclear to me if this is true.
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by DrSteve View Post
    Hmmm...do you have a proof that the third column is in fact the set of positive even integers? Any easy argument would seem to require that the first column follows a pattern - at first glance it is unclear to me if this is true.
    Yes, it's true, and I'm challenging you to find the proof!
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  4. #4
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    HI , i think you are asking us to show that the form $\displaystyle 2n + [n\sqrt{2}] ~,~ n \in \mathbb{N} $ represents all the missing integers . I am now going to prove this fact .

    Before the proof , let me call the integers which cannot be expressed as $\displaystyle [k\sqrt{2}] $ happy integers . It is to avoid typing so much !


    Proof :

    Observation 1 :

    Let $\displaystyle m $ be a positive integer , then $\displaystyle m $ is happy if and only if $\displaystyle \{ \frac{m}{\sqrt{2}} \} < 1 - \frac{1}{\sqrt{2}} $ .

    Proof : It is easy to check that $\displaystyle m $ is happy if and only if there is no real number of the form $\displaystyle k\sqrt{2}$ lying between $\displaystyle m $ and $\displaystyle m + 1 $ , since the common difference of the sequence $\displaystyle k\sqrt{2}$ is $\displaystyle \sqrt{2} $ , we can see that there must be such a number lying between $\displaystyle m - ( \sqrt{2} - 1) $ and $\displaystyle m $ . In other words , there exists $\displaystyle k \in \mathbb{N} $ such that $\displaystyle m - ( \sqrt{2} - 1) < k\sqrt{2} < m $ or $\displaystyle \frac{m}{\sqrt{2}} - (1 - \frac{1}{\sqrt{2}}) < k < \frac{m}{\sqrt{2}} $ . Therefore , $\displaystyle \{ \frac{m}{\sqrt{2}} \} < 1 - \frac{1}{\sqrt{2}} $

    Now i am going to show all the integers of the form $\displaystyle 2n + [n\sqrt{2}]$ are happy :

    As $\displaystyle \frac{ 2n + [n\sqrt{2}] }{\sqrt{2}} = n \sqrt{2} + \frac{[n\sqrt{2}] }{\sqrt{2}} $

    $\displaystyle = [n\sqrt{2}] + \{ n\sqrt{2}\} + \frac{ n\sqrt{2} - \{ n\sqrt{2} \} }{\sqrt{2}} $

    $\displaystyle = n + [n\sqrt{2}] + \{ n\sqrt{2}\} ( 1 - \frac{1}{\sqrt{2}} ) $

    Since $\displaystyle \{ n\sqrt{2}\} ( 1 - \frac{1}{\sqrt{2}} ) < 1 $ we have

    $\displaystyle \left\{ \frac{ 2n + [n\sqrt{2}] }{\sqrt{2}} \right\} = \{ n\sqrt{2}\} ( 1 - \frac{1}{\sqrt{2}} ) < 1 - \frac{1}{\sqrt{2}} $ .

    Therefore , $\displaystyle 2n + [n\sqrt{2}] $ is happy for all positive integers $\displaystyle n $ .


    Observation 2 :

    If $\displaystyle m $ is happy , the next happy integer is greater than $\displaystyle m+2 $ .

    Proof :

    $\displaystyle \{ \frac{m+1}{\sqrt{2}} \} = \{ \frac{m}{\sqrt{2}} \} + \frac{1}{\sqrt{2}}$ ( since $\displaystyle \{ \frac{m}{\sqrt{2}} \} + \frac{1}{\sqrt{2}} < 1 $ )

    $\displaystyle > \frac{1}{\sqrt{2}} > 1 - \frac{1}{\sqrt{2}} $ so $\displaystyle m + 1 $ is not happy .

    $\displaystyle \{ \frac{m+2}{\sqrt{2}} \} = \{ \frac{m}{\sqrt{2}} \} + \sqrt{2} - 1 $ ( since $\displaystyle 1 < \{ \frac{m}{\sqrt{2}} \} + \sqrt{2} < 2 $ )

    $\displaystyle > \sqrt{2} - 1> \frac{\sqrt{2} - 1}{\sqrt{2}} = 1 - \frac{1}{\sqrt{2}} $ so $\displaystyle m+2 $ is not happy !

    We can see that $\displaystyle 3 = 2 + [\sqrt{2}] $ is the first happy integer which is of this form . Moreover , $\displaystyle 2(n+1) + [(n+1)\sqrt{2}] - (2n + [n\sqrt{2}] ) \leq 4 $ . The conclusion is : there doesn't exist a happy integer lying between $\displaystyle 2n + [n\sqrt{2}] $ and $\displaystyle 2(n+1) + [(n+1)\sqrt{2}] $ exclusively as from the second observation , the separation between them will be greater than 4 which is impossible .

    Therefore , $\displaystyle 2n + [n\sqrt{2}] $ represents all the happy integers !
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  5. #5
    MHF Contributor Bruno J.'s Avatar
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    Good job simplependulum, you rock as always!

    This fact is a consequence of the following more general and quite surprising theorem (a theorem which, obviously, you weren't expected to know off the top of your head) : the sets $\displaystyle \{\lfloor\alpha n\rfloor : n \in \mathbb{N}\}, \{\lfloor\beta n\rfloor : n \in \mathbb{N}\}$ form a partition of the integers if and only if $\displaystyle \alpha>1$ is irrational and $\displaystyle \frac{1}{\alpha}+\frac{1}{\beta}=1$.

    So with $\displaystyle \alpha = \sqrt 2$, we find that $\displaystyle \beta = 2+\sqrt 2$, which solves the problem!
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