Given:
a and b are two odd positive integers such that a > b.
To prove:
Out of the numbers \(\frac{a+b}{2}\), \(\frac{a-b}{2}\) one is odd and other is even.
Proof:
a and b both are odd positive integers,
Since,
\((\frac{a+b}{2})\) + \((\frac{a-b}{2})\) = \((\frac{a+b+a-b}{2})\) = a
Which is an odd number as "a" is given to be a odd number.
Hence one of the number out of \(\frac{a+b}{2}\), \(\frac{a-b}{2}\) must be even and other must be odd because adding two even numbers gives an even number and adding two odd numbers gives an even number.
Here,
\(\frac{a+b}{2}\) will give an odd number and \(\frac{a-b}{2}\) will give an even number.
EXAMPLE:
Take a = 7 and b = 3 such that a > b
Now,
\(\frac{a+b}{2}\) = \(\frac{7+3}{2}\) = \(\frac{10}{2}\) = 5 which is odd
And \(\frac{a-b}{2}\) = \(\frac{7-3}{2}\) = \(\frac{4}{2}\) = 2 which is even
Conclusion:
Out of out of the numbers \(\frac{a+b}{2}\), \(\frac{a-b}{2}\), \(\frac{a+b}{2}\) is an odd number and \(\frac{a-b}{2}\) is an even number.
