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Math Help - 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 \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  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  [k\sqrt{2}] happy integers . It is to avoid typing so much !


    Proof :

    Observation 1 :

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

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

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

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

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

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

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

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

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


    Observation 2 :

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

    Proof :

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

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

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

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

    We can see that  3 = 2 + [\sqrt{2}] is the first happy integer which is of this form . Moreover ,  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 2n + [n\sqrt{2}] and  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 ,  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 \{\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 \alpha>1 is irrational and \frac{1}{\alpha}+\frac{1}{\beta}=1.

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