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Math Help - More Irrational Number ??>

  1. #1
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    Question More Irrational Number ??>

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    Can you get a rational nonzero number adding or subtracting 2 irrational numbers? If so, do you have examples?
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  2. #2
    Super Member Rebesques's Avatar
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    Well, if you substract the irrationals 3+\sqrt{2} , 2+\sqrt{2}, you 'll get (the very rational) 1
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  3. #3
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    you didn't use 2 irrational numbers, 3 is rational
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  4. #4
    Super Member Rebesques's Avatar
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    Do not judge too fast:

     (3+\sqrt{2})-(2+\sqrt{2}) = 1

    Cheap!!??
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  5. #5
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    but you need to add the irrationals together
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  6. #6
    Super Member Rebesques's Avatar
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    e?

    Just did

    Sorry, but I see kinda stress here...

    2+\sqrt{2} is very irrational, and so is the other fellow. Tell me what you do not understand.
    Last edited by MathGuru; July 27th 2005 at 01:54 PM.
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  7. #7
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    in the number model 2 + sqrt 2, only sqrt 2 is irrational. 2 as an addend is rational.
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  8. #8
    Super Member Rebesques's Avatar
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    ...So? The number  2+\sqrt{2} is irrational, still. What troubles you?
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  9. #9
    Site Founder MathGuru's Avatar
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    the point is you are simply saying that  \sqrt{2} - \sqrt{2} = 0
    I don't think that is the point of the question.
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  10. #10
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    the point of the question is to use two addends, both irrational, and get a sum that is rational.
    can it be done?
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  11. #11
    Site Founder MathGuru's Avatar
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    e^pi*i

    I don't know if you can do

    irrational + irrational = rational

    but you can do

    (irrational)^(irrational*i)=rational

    for example

    e^{\pi*i} = -1
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  12. #12
    Super Member Rebesques's Avatar
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    e?

    I think I get the idea. The lad asks whether two -irrelevant- irrationals, can add up to a rational. The general answer is no. Only with trickery can this happen (like the example with 2+\sqrt{2}, 3+\sqrt{2}).

    To see this, we remember that a number is rational, if and only if it has a terminating or a periodical decimal expansion. Therefore, if we have two irrationals A=\alpha.\alpha_0\alpha_1\alpha_2... , B=\beta.\beta_0\beta_1\beta_2... , then for the sum A+B we have the following possibilities:

    ------------------------------------------------------------------
    a) A+B is non-terminating and unperiodical.

    Then, it is irrational.


    --------------------------------------------------------------------

    b) A+B is terminating.

    Then, there exists a minimum index k, such that  \alpha_n =-\beta_n, \ \forall  \ n\geq k. This means that

    A+B=(\alpha+\beta).(\alpha_0+\beta_0)(\alpha_1+ \beta_1)... =  <br />
\gamma.\gamma_0\gamma_1\gamma_2...\gamma_{k-1}=C,

    this last number C being rational (as terminating). Then, A=-B+C, and so the two irrationals were related at first hand!



    (Example: A=1+\sqrt{2}=2.141421..., B=1-\sqrt{2}=-0.141421.... Then A+B=2.141421... -0.141421... =2.)


    ---------------------------------------------------------------------

    c) A+B is periodical.

    Then, there exists a minimum index k and a maximum m, such that the decimal expansion

    A+B= \gamma.\gamma_0\gamma_1\gamma_2...=\gamma.\gamma_0  \gamma_1...\gamma_{k-1}\gamma_k...\gamma_m\gamma_k...\gamma_m....

    This means that for all naturals t, \gamma_k=\gamma_{tm-k} or \alpha_k+\beta_k=\alpha_{tm-k}+\beta_{tm-k}, and so
    \alpha_k -\alpha_{tm-k}=-(\beta_k -\beta_{tm-k}), for all t. This (again) sais that A and B are forehand related, as promised.




    (Example: Consider the numbers A=0.1212212221... and B=0.1010010001... They are unperiodical by construction, and so irrational. If A=0.\alpha_0\alpha_1... and B=0.\beta_0\beta_1..., then \alpha_0=\beta_0=\alpha_2=\beta_2=\alpha_5=\beta_5  ...=1, that is they are quite related. And, as if by magic, A+B= 0.222..., which is periodical, and so very rational.)


    ------------------------------------------------------------------------


    Sorry for the long answer, hope it was worth it.
    Last edited by Rebesques; July 28th 2005 at 02:34 AM.
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  13. #13
    Super Member Rebesques's Avatar
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    e?

    Nobody agrees/disagrees?
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  14. #14
    Site Founder MathGuru's Avatar
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    I like the unperiodical by construction trick. But it is still an artificial construction.

    So I agree, and I appreciate the trick, and you have answered the question well.
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