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Math Help - Proof of checkerboard pieces

  1. #1
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    Proof of checkerboard pieces

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    Last edited by chillerbros17; February 28th 2007 at 02:38 PM.
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  2. #2
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    Quote Originally Posted by chillerbros17 View Post
    discrete2.doc

    Problem is in attachment.
    B) It is imposible as part b has two white squares adjacent, which
    never occurs on a real chess board.

    RonL
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  3. #3
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    Quote Originally Posted by chillerbros17 View Post
    discrete2.doc

    Problem is in attachment.
    6. Prove the following theorems. State the method of proof that you are using in parts (a) and (c).





    (a):
    Thm: The sum of two prime numbers, each greater than 2, is never a prime. Restatement: For all integers p, q, if p is a prime greater than 2 and q is a prime greater than 2, then p + q is not prime.


    Proof by Contradiction (Reductio ad absurdum)

    If p and q are primes greater than 2, then p and q are both odd, and may
    be written p=2a+1, q=2b+1, for some positive integers a, and b.

    Now suppose p+q is prime, then as it is greater than 2, it must be odd.
    But:

    p+q=2a+1+2b+1=2(a+b+1).

    Thus p+q is divisible by 2, and hence not prime (as p+q !=2), which
    is a contradiction, hence our assumption is false, and p+q is not prime.

    RonL
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  4. #4
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    discrete2.doc

    Here's is b with the right pattern. Sorry for the mistake.
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  5. #5
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    Quote Originally Posted by chillerbros17 View Post
    discrete2.doc

    Here's is b with the right pattern. Sorry for the mistake.
    See the attachment. This shows how to fit two of each shape together to make a 4x4 section of the chess board pattern. Four of these fitted together
    will make the entire board.

    RonL
    Attached Thumbnails Attached Thumbnails Proof of checkerboard pieces-gash.jpg  
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