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Proof of different parities

  1. Oct 29, 2012 #1
    Hello all I've been practicing proofs and would like to know if I'm on the right track. Here it is:

    If the sum of 3x + 3y is an odd number then x and y are different parities.

    Proof: Let x and y be two integers with opposite parity. Without loss of generality, suppose x is even and y is odd:

    x = 2m
    y = 2n + 1

    Then:

    3x + 3y = 3(2m) + 3(2n + 1) = 6m + 6n + 1 = 2(3m + 3n) + 1

    Since 2(3m + 3n) has a factor of 2 it is even. When 1 is added, 2(3m + 3n) + 1 is odd. Therefore, 3x + 3y is odd and x and y have opposite parities.

    Is this enough or do I need more?
     
  2. jcsd
  3. Oct 29, 2012 #2
    You've only proved that, if x and y are opposite parities, then 3x + 3y is odd. How do you know that it can't be odd in other circumstances?
     
  4. Oct 29, 2012 #3
    So would I have to show two more cases: one where both were even and one where both were odd?
     
  5. Oct 29, 2012 #4
    Something like this:

    Assume x and y are of the same parity. Then we have two cases: (1) both are even, and (2) both are odd.
    (1) Both are even; let:

    x = 2a
    y = 2b

    3x + 3y = 3(2a) + 3(2b) = 2(3)(a+b)
    Since there is a factor of 2 the sum is even.

    (2) Both are odd; let:

    x = 2a + 1
    y = 2b + 1

    3x + 3y = 3(2a+1) + 3(2b+1) = 6a + 3 + 6a + 3 = 12a + 6 = 6(2a) = 2(3)(2a)
    Again, since there is a factor of 2 the sum is even.

    Would this complete the proof?
     
  6. Oct 29, 2012 #5
    That looks complete, yes.
    For questions like these, it's useful to keep in mind the contrapositive: If x and y are not of different parities, then 3x + 3y is not odd (remember that an implication and its contrapositive mean the same thing; it's often easier to prove one than the other)
     
  7. Oct 29, 2012 #6
    So, since my last post proves the statement by contrapositive, would the proof in my first post be unnecessary?
     
  8. Oct 30, 2012 #7
    My math was wrong in my last proof in this line:

    3x + 3y = 3(2a+1) + 3(2b+1) = 6a + 3 + 6a + 3 = 12a + 6 = 6(2a) = 2(3)(2a)

    It should be:

    3x + 3y = 3(2a+1) + 3(2b+1) = 6a + 3 + 6b + 3 = 6a + 6b + 6 = 6(a + b) = 2(3)(x + y)
     
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