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Prove by Contradiction: For all integers x greater than 11

  1. Aug 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove by Contradiction: For all integers x greater than 11, x equals the sum of two composite numbers.


    2. Relevant equations
    A composite number is any number that isn't prime
    To prove by contradiction implies that if you use a statement's as a negation, a contradiction arises

    3. The attempt at a solution
    The negation of the original statement is:
    There exists an integer x such that if x > 11, then x does not equal the sum of two composite numbers.

    I'm really stuck on this one, I tried substituting in values using the quotient remainder theorem (a number can be represented as 2r, or 2r + 1.. or alternatively 3r, 3r + 1, or 3x + 2, but I wasn't getting anywhere with it). Some direction would be appreciated!
     
  2. jcsd
  3. Aug 6, 2012 #2

    micromass

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    Given that x is not the sum of two composite numbers, what can you say about things like

    x-4
    x-6
    x-8
    x-9
     
  4. Aug 7, 2012 #3
    All even numbers are composite (besides 2!), and since 4 is composite as well, any even number greater than 4 can be written as the sum of two composite numbers. This leaves only odd ones left.

    Can you apply a similar logic for odd numbers?
     
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