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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 (a + b)/2 and (a – b)/2 is odd and the other is even.

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Question

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

Theorem
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Solution

We know that any odd positive integer is of the form 4q + 1 or 4q + 3 for some whole number q.

Now that it’s given a > b

So, we can choose a = 4q + 3 and b = 4q + 1.

∴ `((a + b))/2 = ([(4q + 3)+(4q + 1)])/2`

`=> ((a + b))/2 = ((8q + 4))/2`

`=> ((a + b))/2 = 4q + 2 = 2(2q + 1)` which is clearly an even number.

Now, doing `((a - b))/2`

`=> ((a - b))/2 = ([(4q + 3)-(4q + 1)])/2`

`=> ((a - b))/2 = ((4q + 3 - 4q - 1))/2`

`=> ((a - b))/2 = ((2))/2`

`=> ((a - b))/2 = 1` which is an odd number.

Hence, one of the two numbers `((a + b))/2` and `((a - b))/2` is odd and the other is even.

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Chapter 1: Real Numbers - EXERCISE 1.1 [Page 1.9]

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R.D. Sharma Mathematics [English] Class 10
Chapter 1 Real Numbers
EXERCISE 1.1 | Q 14. | Page 1.9
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