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Math Help - Demonstration 'reductio ad absurdum'.

  1. #1
    Newbie jhonyy9's Avatar
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    Demonstration 'reductio ad absurdum'.

    Help me please with a demonstration 'reductio ad absurdum'!

    1. - substitution
    - be a and b,two numbers greater or equal than 1 from the set of real numbers R,
    - be n a number greater or equal than 3 from the set of real numbers R,
    2. - conclusion
    - there is always a and b such that equation is true: n=a+b+1
    3. - demonstration
    - ???

    Thank you very much your all helps !
    Last edited by mr fantastic; March 25th 2011 at 04:24 AM. Reason: Re-titled.
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  2. #2
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    This very confusing. First you say "let a and b be two numbers greater or equal than 1 from the set of real numbers R"
    But then you say "there is always a and b such that equation is true: n=a+b+1". You cannot select a and b first (which is what your first sentence means) and then say "there exist a and b such that".

    I think what you mean is "If n is any real number larger than 3 then there exist real numbers a and b, larger than 1, such that n= a+ b+ 1".

    That's obviously true but I can see no reason to use "reductio ad absurdum" (proof by contradiction). If n> 3, then n- 1> 2. Let a= b= (n-1)/2> 1. Then n-1= a+ b so n= a+ b+ 1.

    (Of course, there exist an infnite number of choices for a and b. To show "there exists" you only need to demonstrate one.)
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  3. #3
    Newbie jhonyy9's Avatar
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    one demonstration,please

    can you help me,please with one acceptable demonstration that in this condition for every n there always numbers a and b for the ecuation n=a+b+1 is true

    thank you very much
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  4. #4
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    in a reducto ad absurdum argument, you assume the opposite is true. in this case, that would be:

    for some real number n > 3 there are no real numbers a >1, b >1 with a+b+1 = n.

    we don't know what n may be (except that it is greater than 3).

    since n > 3, n - 1 > 2.

    since n - 1 > 2, (n - 1)/2 > 1.

    (n - 1)/2 + (n - 1)/2 + 1 = (n - 1) + 1 = n.

    but if n is a real number, so is n - 1, and so is (n - 1)/2.

    therefore we have proved that for this n > 3, the real numbers (n - 1)/2 don't exist. this is absurd.

    (the logical conclusion therefore being that no such real number > 3 actually exists, so our assumption was wrong,

    so EVERY real number > 3 has such an a and b > 1 with a+b+1 = n).
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  5. #5
    Newbie jhonyy9's Avatar
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    this prove is correct too - when those numbers are natural numbers ?

    HI Deveno ! can you tel me please if those numbers are not real numbers ,they are natural numbers - so in this case this prove is correct too ?

    - Thank you very much for your reply !
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