Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Proving the diff of any odd int minus any even int is odd. I got 2t-1, will it work

  1. Sep 17, 2006 #1
    Hello everyone.
    I'm so close to this proof, that i think it might be right. But it doesn't follow the definition exactly, or does it? The definition of an odd number is: n is odd <=> There exists an integer k such that n = 2k + 1. My conclusion came out with 2k-1. Here is my proof.

    26. The difference of any odd integer minus any even integer is odd.

    proof: suppose a is an even integer and b is an odd integer. [we must show b-a is odd]. By definition of even and odd, a = 2r and b = 2s+1 for some integers r and s. By subsitution and algebra, b - a = (2s+1) - 2r = 2s+1-2r = 2(s-r+1)-1. Let t = s-r+1. Then t is an integer, because sums and differences of integers are integers. Thus b-a = 2t-1, where t is an integer, and so, by definition of odd, b-a is odd.

    Because 2t-1 is not 2t+1, is this proof showing that the difference of any odd integer minus any even integer is not odd? or did i screw up somewhere? Thanks!
    :biggrin:
     
  2. jcsd
  3. Sep 17, 2006 #2

    StatusX

    User Avatar
    Homework Helper

    Why did you add this one inside the parantheses? Without it you have something in the form you want.
     
  4. Sep 17, 2006 #3
    Ahh my bad! thank u for catching that! Now its:
    26. The difference of any odd integer minus any even integer is odd.

    proof: suppose a is an even integer and b is an odd integer. [we must show b-a is odd]. By definition of even and odd, a = 2r and b = 2s+1 for some integers r and s. By subsitution and algebra, b - a = (2s+1) - 2r = 2s+1-2r = 2(s-r) +1. Let t = s-r. Then t is an integer, because sums and differences of integers are integers. Thus b-a = 2t+1, where t is an integer, and so, by definition of odd, b-a is odd.

    Thanks again!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook