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Thread: how is possibill demonstring ?

  1. #1
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    how is possibill demonstring ?

    - let n greater or equal 3 from the set of real numbers R,
    - let a and b graters or equal 1 from the set of real numbers R,
    - we have the equation : n=a+b+1

    - what will be one demonstration that for every value of number n always will be,always there two numbers a and b for this equation is true ?

  2. #2
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    Re: how is possibill demonstring ?

    Could you ask a friend who knows English to help you out posting this clearly?
    Thank you.
    I'm just an imagination of your figment !

  3. #3
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    Re: how is possibill demonstring ?

    - if n is a number greater or equal 3 from the set of real numbers R,
    - a and b are graters or equals with 1 from R,
    - and we have the equation n=a+b+1

    - what we have to demonstrate is that always,for every value of n will be one number a and one number b so that the equation will be true

  4. #4
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    Re: how is possibill demonstring ?

    - if n is greater or equal 3 ,we can writing n-1 greater or equal with 2
    - if a and b are graters or equals with 1 ,we can writing that a+b is greater or equal with 2
    - from this results that n-1=a+b ie where n=a+b+1

    - this demonstration may be accepted ?

  5. #5
    JeffM
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    Re: how is possibill demonstring ?

    PROVE
    If n is a real number greater than or equal to 3,
    Then there exists at least one pair of real numbers, a and b, each greater than or equal to 1, such that (a + b + 1) = n.

    Is that the problem?
    How have real numbers been defined for you? What can you assume?

    IF YOU ARE IN BEGINNING ALGEBRA I see nothing wrong with a proof like:
    Let a = n - 2.
    So, a is a real number and a >= (3 - 2).
    So, a is a real number >= 1.
    Let b = 1.
    So, b is a real number >= 1.
    So, (a + b + 1) = (a + 1 + 1) = (a + 2) = (n - 2) + 2 = n.
    (n - 2) and 1 are a pair of real numbers that meet the requirements.
    So there is at least one such pair of real numbers.

    IF YOU ARE IN SOME ADVANCED MATH COURSE AND MUST PROVE THAT THE REAL NUMBERS ARE CLOSED UNDER ADDITION, YOU ARE WAY BEYOND ME AND NEED TO POST THIS PROBLEM UNDER ADVANCED MATH.

  6. #6
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    Re: how is possibill demonstring ?

    - this equation can be demonstrated with complete mathematical induction ?
    - if so,how it can be demonstrated ?

  7. #7
    JeffM
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    Re: how is possibill demonstring ?

    Quote Originally Posted by jhonyy9
    - this equation can be demonstrated with complete mathematical induction ?
    - if so,how it can be demonstrated ?
    I gave you a constructive proof of what I THOUGHT was your problem. Clearly, I did not understand it. Sorry to be of no help.

    PS. As you have defined the problem SO FAR, n is to be a REAL number >= 3. The proof I gave was for any such real number. There is no need for mathematical induction in the problem as posed.

  8. #8
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    Re: how is possibill demonstring ?

    Quote Originally Posted by jhonyy9
    - let n greater or equal 3 from the set of real numbers R,
    - let a and b graters or equal 1 from the set of real numbers R,
    - we have the equation : n=a+b+1

    - what will be one demonstration that for every value of number n always will be,always there two numbers a and b for this equation is true ?
    n= a+b+1 is true for n=3

    assume that it is true for an arbitrary 'n'.

    n = p + q + 1

    Then:

    n+1 = (p+q+1) + 1 = p + (q+1) + 1 = a + b + 1

    This is what JeffM had presented. It is surprising that you could not rewrite that in this form!!!
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  9. #9
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    Re: how is possibill demonstring ?

    Quote Originally Posted by JeffM

    IF YOU ARE IN SOME ADVANCED MATH COURSE AND MUST PROVE THAT THE REAL NUMBERS ARE CLOSED UNDER ADDITION, YOU ARE WAY BEYOND ME AND NEED TO POST THIS PROBLEM UNDER ADVANCED MATH.
    You can state your message without yelling it by not using all caps, such as the above.
    I have been corrected on this same technique before.

  10. #10
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    Re: how is possibill demonstring ?

    Quote Originally Posted by lookagain
    Quote Originally Posted by JeffM
    IF YOU ARE IN SOME ADVANCED MATH COURSE AND MUST PROVE THAT THE REAL NUMBERS ARE CLOSED UNDER ADDITION, YOU ARE WAY BEYOND ME AND NEED TO POST THIS PROBLEM UNDER ADVANCED MATH.
    You can state your message without yelling it by not using all caps, such as the above.
    And the all caps also shows a poor response to your impatience and frustration with
    the user.
    Sir, please let the moderators do any reprimanding.
    It is the reason why they are moderators.
    We thank you for your cooperation.
    I'm just an imagination of your figment !

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