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If a and b are two odd positive integers such that a $>$ b, then prove that one of the two numbers $\frac{a\ +\ b}{2}$ and $\frac{a\ -\ b}{2}$ is odd and the other is even.
Given: a and b are two odd positive numbers such that a $>$ b.
To do: Here we have to prove that one of the two numbers $\frac{a\ +\ b}{2}$ and $\frac{a\ -\ b}{2}$ is odd and the other is even.
Solution:
We know that;
$a$ and $b$ are two odd positive integers such that $a\ >\ b$.
Also,
Odd numbers are in the form of $2n\ +\ 1$ and $2n\ +\ 3$, where $n$ is an integer.
As $a\ >\ b$,
$a\ =\ 2n\ +\ 3$ and $b\ =\ 2n\ +\ 1$
Now,
Calculating value of ($\frac{a\ +\ b}{2}$):
$\frac{a\ +\ b}{2}\ =\ \frac{2n\ +\ 3\ +\ 2n\ +\ 1}{2}$
$\frac{a\ +\ b}{2}\ =\ \frac{4n\ +\ 4}{2}$
$\frac{a\ +\ b}{2}\ =\ 2n\ +\ 2$
$\frac{a\ +\ b}{2}\ =\ 2(n\ +\ 1)$
Any number multiplied by 2 is even. So, $2(n\ +\ 1)$ is even.
Therefore,
$\mathbf{\frac{a\ +\ b}{2}}$ is even.
Calculating value of ($\frac{a\ -\ b}{2}$):
$\frac{a\ -\ b}{2}\ =\ \frac{2n\ +\ 3\ -\ 2n\ -\ 1}{2}$
$\frac{a\ -\ b}{2}\ =\ \frac{2}{2}$
$\frac{a\ -\ b}{2}\ =\ 1$
Which is odd.
Therefore,
$\mathbf{\frac{a\ -\ b}{2}}$ is odd.
Hence, we can see that one number is odd and other is even.
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