If 'a' and 'b' are two odd positive integers such that a>b, then how do we prove that one of the two numbers (a+b)/2 and (a-b)/2 is odd and the other is even?

First we can easily verify that \(\frac{a+b}{2}\) and \(\frac{a-b}{2}\) are positive integers since the sum of two odd numbers is always even and, the difference of two odd numbers is always even respectively. This implies that on division by \(2\) we we will have a positive integer.

Let \(x=\frac{a+b}{2}+\frac{a-b}{2}\)

\(\Rightarrow x=a\)

Therefore, we have that \(x\) is an odd positive integer.
We know that the sum of two even or sum of two odd numbers is never odd.
Thus, it follows that \(\frac{a+b}{2}\) is even when \(\frac{a-b}{2}\) is odd and vice-versa.

Take a as 2k+1 and b as 2m+1, now consider all the possible cases such that, either k is odd or m is odd or both are odd or neither of them. when you will observe all four of these cases u will automatically get the result.

\( \frac{a+b}{2} + \frac{a-b}{2} = a \). We know \(a\) is an odd integer, and so, when addition of two numbers is odd, they have to be of opposite parity.

but 'a' and 'b' are odd positive integers, so if we add up both of them and divide it by 2....the result has to be even, it can't be odd, so i don't agree with your last two lines.

I would also like to add that your question asks as I perceive is that when one of the given number is odd the other should be even. Thus, when \(\frac{a+b}{2}\) is even \(\frac{a-b}{2}\) is odd.(See my solution)

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TopNewestFirst we can easily verify that \(\frac{a+b}{2}\) and \(\frac{a-b}{2}\) are positive integers since the sum of two odd numbers is always even and, the difference of two odd numbers is always even respectively. This implies that on division by \(2\) we we will have a positive integer.

Let \(x=\frac{a+b}{2}+\frac{a-b}{2}\)

\(\Rightarrow x=a\)

Therefore, we have that \(x\) is an odd positive integer. We know that the sum of two even or sum of two odd numbers is never odd. Thus, it follows that \(\frac{a+b}{2}\) is even when \(\frac{a-b}{2}\) is odd and vice-versa.

Hence proved.

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thanks for the answers

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Take a as 2k+1 and b as 2m+1, now consider all the possible cases such that, either k is odd or m is odd or both are odd or neither of them. when you will observe all four of these cases u will automatically get the result.

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\( \frac{a+b}{2} + \frac{a-b}{2} = a \). We know \(a\) is an odd integer, and so, when addition of two numbers is odd, they have to be of opposite parity.

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but 'a' and 'b' are odd positive integers, so if we add up both of them and divide it by 2....the result has to be even, it can't be odd, so i don't agree with your last two lines.

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Not necessarily. As an explicit example take \(a=5\) and \(b=1\). The result will be odd. \(\frac{5+1}{2}=3\).

EDIT: To make things clearer, we can say that \(a=b+2k\) for some positive integer k since \(a>b\).

Let y=\(\frac{a+b}{2}\)

\(y=\frac{2b+2k}{2}\)

\(y=b+k\)

This means that y is only even when k is odd, i.e \(\frac{a+b}{2}\) is even only when \(a-b=2k\).

This again implies that when k is odd \(\frac{a-b}{2}\) is also odd, which can be another proof for your question.

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I would also like to add that your question asks as I perceive is that when one of the given number is odd the other should be even. Thus, when \(\frac{a+b}{2}\) is even \(\frac{a-b}{2}\) is odd.(See my solution)

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