Given:
a > 0, b > 0 and a, b are Integers
a - b and a/b are both even intergers
==> \(a - b = 2m_1\)
\(\frac{a}{b} = 2m_2\) ==> \(a = (b)(2m_2)\)
==> a & b are both even, otherwise the above conditions won't be satisfied.
Now lets go with options one by one ...
A) a/2
==> can be even or odd ( a = 6 or a = 8)
B) b/2
==> Can be either even or odd ( b = 6 or b = 8)
C) (a+b) /2
==> Since, \(a = (b)(2m_2)\), \(\frac{a + b}{2}\) =\(\frac {(b)(2m_2) + b}{2}\)
==> \(\frac{a + b}{2}\) =\(\frac {(b)(2m_2 + 1)}{2}\)
==> Can be even or odd
D) (a+2)/2
==> \(\frac{a + b}{2}\) =\(\frac{2bm_2 + 2}{2}\)
==> \(\frac{a + b}{2}\) =\(bm_2 + 1\)
==> Since, b is even, \(bm_2\) is even and \(bm_2 + 1\) is odd
alwaysE) (b+2)/2
==> \(\frac{b+2}{2} = 1 + \frac{b}{2}\)
==> Can be even or odd
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Cheers!
Ravi
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