Difference of Any Odd Int Minus Any Even Int is Odd?

[mr_coffee]

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!

 

[StatusX]

b - a = (2s+1) - 2r = 2s+1-2r = 2(s-r+1)-1.

Why did you add this one inside the parantheses? Without it you have something in the form you want.

 

[mr_coffee]

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!