real numbers: If a > b are 2 positive numbers, then prove that one of...

[kbp.deepa]

if a & b are two positive numbers a>b then prove that one of a+b/2 is even then the other a-b/2 is odd

 

[Ishuda]

if a & b are two positive numbers a>b then prove that one of a+b/2 is even then the other a-b/2 is odd

What are your thoughts? What have you done so far? Please show us your work even if you feel that it is wrong so we may try to help you. You might also read
http://www.freemathhelp.com/forum/threads/78006-Read-Before-Posting

Hint:
a = 2 n + i
b = 2 m + j
where i and j are either, individually, 0 or 1 and
2 (n-m) + i - j > 0

Oh, and also note that you possibly should have used grouping symbols, i.e. (a+b)/2

 

[stapel]

if a & b are two positive numbers a>b then prove that one of a+b/2 is even then the other a-b/2 is odd

When you posted this to the "Arithematic" category, you originally titled this only "real numbers". But the concepts of "odd" and "even" apply only to the naturals. So should "positive real numbers" be changed to "natural numbers"? Also, does the exercise say something more along the lines of "exactly" (or "at least") "one of (a + b)/2 and (a - b)/2 is even"?

When you reply with a clear statement of the actual exercise, kindly please include a clear listing of your efforts so far. Thank you!

 

[pka]

But the concepts of "odd" and "even" apply only to the naturals.

I do think that parity applies to all integers. Yes?