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Math Help - Proof by Induction

  1. #1
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    Proof by Induction

    Using Mathematical Induction:

    Prove the sum of cubes of 3 consecutive pos. integers is divisble by 9.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Ideasman View Post
    Using Mathematical Induction:

    Prove the sum of cubes of 3 consecutive pos. integers is divisble by 9.
    The sum of the cubes of 1, 2, 3 is 1+8+27=36 which is divisible by 9.

    Now suppose that for some k: C(k)=k^3+(k+1)^3+(k+2)^3 is divisible by 9.

    Then:

    C(k+1)=C(k) - k^3 + (k+3)^3 = C(k) + 3 (k^2*3) + 3 (k*3^2) + 3^3

    ..............=C(k) + 9 [k^2 + 3k + 3]

    hence as by assumption C(k) is divisible by 9 so is C(k+1), and as we have
    established that C(1) is divisible by 9, we have established by mathematical
    induction that C(n) is divisible by 9 for all positive integers n.

    This proves (as it is the same thing in different words) that the sum of the
    cubes of three consecutive positive integers is divisible by 9.

    RonL
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  3. #3
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    Hello, Ideasman!

    This is Captain Black's proof . . . with my formatting.


    Using Mathematical Induction:
    Prove the sum of cubes of 3 consecutive pos. integers is divisble by 9.

    Verify S(1): .1 + 2 + 3 .= .36 . . . divisible by 9.


    Assume S(k): .k + (k+1) + (k+2) .= .9a . for some integer a.


    Add (k+3) - k to both sides:

    . . k + (k+1) + (k+2) + (k+3) - k .= . 9a + (k+3) - k

    . . (k+1) + (k+2) + (k+3) .= .9a + k - 9k + 27k - 27 - k

    . . (k+1) + (k+2) + (k+3) .= .9a - 9k + 27k - 27

    . . (k+1) + (k+2) + (k+3) .= .9(a - k + 3k - 3)


    The left side is the left side of S(k+1); the right side is a multiple of 9.
    . . The inductive proof is complete.


    Corrected my error . . . thanks, Captain!
    Last edited by Soroban; February 17th 2007 at 09:22 AM.
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    Grand Panjandrum
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    Quote Originally Posted by Soroban View Post
    Hello, Ideasman!

    This is Captain Black's proof . . . with my formatting.


    Verify S(1): .1 + 2 + 3 .= .36 . . . divisible by 9.


    Assume S(k): .k + (k+1) + (k+2) .= .3a . for some integer a.



    You need "k^3+(k+1)^3+(k+2)^3=9a" here, which is what you use later.

    RonL
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    Just a quick question-

    I thought mathematicians don't like solving two solves of the equation at the same time to prove something... I thought you work with one side (IE: the (k + 1) side to show the other side; according to my math professor, it's bad to try solve a proof like this. I may be wrong..your thoughts?
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  6. #6
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    Quote Originally Posted by Ideasman View Post
    Just a quick question-

    I thought mathematicians don't like solving two solves of the equation at the same time to prove something... I thought you work with one side (IE: the (k + 1) side to show the other side; according to my math professor, it's bad to try solve a proof like this. I may be wrong..your thoughts?
    Corret, it is a logical error called taking the converse.
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