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Thread: how is possibill demonstring ?

  1. #11
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    Re: how is possibill demonstring ?

    Quote Originally Posted by Subhotosh Khan

    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 = [tex]\text{ Error here:}[/tex]a + b + 1
    [tex]n = p + q + 1[/tex]

    [tex]\text{Therefore}[/tex]

    [tex]n + 1 =[/tex]

    [tex](p + q + 1) + 1 =[/tex]

    [tex]p + (q + 1) + 1 =[/tex]


    [tex]\text{What did you intend for your conclusion?}[/tex]

  2. #12
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    Re: how is possibill demonstring ?

    Quote Originally Posted by lookagain
    Quote Originally Posted by Subhotosh Khan

    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 = [tex]\text{ Error here:}[/tex]a + b + 1
    [tex]n = p + q + 1[/tex]

    [tex]\text{Therefore}[/tex]

    [tex]n + 1 =[/tex]

    [tex](p + q + 1) + 1 =[/tex]

    [tex]p + (q + 1) + 1 =[/tex]


    [tex]\text{What did you intend for your conclusion?}[/tex]

    a = p
    b = q+1

    thus

    n+1 = a + b + 1

    Thus if (n+1) can be expressed as a+b+1.

    Thus anunumber can be expressed as a+b+1

    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #13
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    Re: how is possibill demonstring ?

    Quote Originally Posted by Subhotosh Khan

    a = p
    b = q+1

    thus

    n+1 = a + b + 1

    Thus if (n+1) can be expressed as a+b+1.

    Thus anunumber can be expressed as a+b+1[tex]You \ can't \ use \ the \ same \ variables[/tex]
    [tex]here \ as \ were \ given \ in \ the \ problem.[/tex]

    But (n + 1) cannot be expressed as (a + b + 1), because n already equals (a + b + 1).

    You have to use at least one different variable (~ letter) for your point.

    n and (n + 1) cannot both be equal to (a + b + 1).


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  4. #14
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    Re: how is possibill demonstring ?

    Thank you for the correction.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  5. #15
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    Re: how is possibill demonstring ?

    - this equation can be demonstrated with complete mathematical induction ?

    - n=a+b+1 --- 3=1+1+1 --- is true
    --- 4=2+1+1 --- is true
    --- 5=2+2+1 --- is true
    - if n=m --- m=a+b+1 --- suppose is true,
    then m+1=a+b+1+1
    - m+1=a+1+b+1 --- is true -- commutativity
    - m+1=a+(b+1)+1 --- is true --- associativity

    - results that this equation n=a+b+1 -- is true

    - this demonstration may be accepted ?

  6. #16
    JeffM
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    Re: how is possibill demonstring ?

    jhonny9

    In your first post, you said "let n greater or equal 3 from the set of real numbers R."

    The proof you give in your last post is of a type that is valid for integers. If it is true for some integer k, it is true for the integer (k + 1). It is true for some integer m. Therefore it is true for any integer n [tex]\geq \[/tex] m. That precise form of proof is NOT valid for real numbers.

    If we understand what you are being asked to prove (please see my first post under JeffM), it can be proved for any real number WITHOUT mathematical induction. If you are required to prove the statement for a real number using some form of mathematical induction, please let us know. I am sure that someone other than me can show you how that is done. We need you to tell us what is to be proved and by what method.

    Again a valid CONSTRUCTIVE proof (treating as axiomatic that the system of real numbers is closed under addition) of what I THINK you are being asked is:

    Let n be an ARBITRARY real number [tex]\geq \[/tex] 3.
    Let a = (n - 2).
    So a is a real number [tex]\geq \[/tex]1.
    Let b = 1.
    So, b is a real number [tex]\geq \[/tex] 1.
    (a + b + 1) = (a + 1 + 1) = (a + 2) = [(n - 2) + 2] = n.
    So there is at least one pair of real numbers a and b such that a [tex]\geq \[/tex] 1 [tex]\leq \[/tex] b and (a + b + 1) = n for any real n [tex]\geq \[/tex] 3.

  7. #17
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    Re: how is possibill demonstring ?

    ... this demonstration is like wrong ... or why dont reply me nobody nothing ???

  8. #18
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    Re: how is possibill demonstring ?

    sorry but i now i have got that this my problem have two pages ... so this demonstration with math complete induction what i have wrote on below , on end of the first page write me please this demonstration it is acceptable or ... what is your opinion,please ?

    for JefM what i have asked is how is possible writing a demonstration that for every value of n,when n is greater or equal 3 from R will be always one number a and one number b , every two graters or equals 1 from R,so that the equation n=a+b+1 is true
    - for example : 3=1+1+1 , 4=2+1+1 , 5=2+2+1 ,...
    - check please my last demonstration ... and i wait your reply !

    thank you very much !

  9. #19
    JeffM
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    Re: how is possibill demonstring ?

    Quote Originally Posted by jhonyy9
    - this equation can be demonstrated with complete mathematical induction ?

    - n=a+b+1 --- 3=1+1+1 --- is true
    --- 4=2+1+1 --- is true
    --- 5=2+2+1 --- is true
    - if n=m --- m=a+b+1 --- suppose is true,
    then m+1=a+b+1+1
    - m+1=a+1+b+1 --- is true -- commutativity
    - m+1=a+(b+1)+1 --- is true --- associativity

    - results that this equation n=a+b+1 -- is true

    - this demonstration may be accepted ?
    In my opinion, your demonstration is not acceptable. You are effectively assuming that n is an integer, but you said that n is a real number.

    As I have said twice now, there is a demonstration for any real n [tex]\geq\[/tex] 3 that does not require induction in its proof. Are you required to give a proof by induction?

  10. #20
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    Re: how is possibill demonstring ?

    - sorry - yes this is true ... - so how you think this ,please,will be true if a,b,n are numbers from the set of natural numbers N ?
    - in this conditions - if a,b,n are from N - this prove can may be accepted ?

    - thank you for your reply

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