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?
This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.
When posting on Brilliant:
Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.
Markdown
Appears as
*italics* or _italics_
italics
**bold** or __bold__
bold
- bulleted - list
bulleted
list
1. numbered 2. list
numbered
list
Note: you must add a full line of space before and after lists for them to show up correctly
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
Math
Appears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3
2×3
2^{34}
234
a_{i-1}
ai−1
\frac{2}{3}
32
\sqrt{2}
2
\sum_{i=1}^3
∑i=13
\sin \theta
sinθ
\boxed{123}
123
Comments
First we can easily verify that 2a+b and 2a−b 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=2a+b+2a−b
⇒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 2a+b is even when 2a−b 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.
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 2a+b is even 2a−b is odd.(See my solution)
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
First we can easily verify that 2a+b and 2a−b 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=2a+b+2a−b
⇒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 2a+b is even when 2a−b is odd and vice-versa.
Hence proved.
2a+b+2a−b=a. We know a is an odd integer, and so, when addition of two numbers is odd, they have to be of opposite parity.
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.
thanks for the answers
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.
Log in to reply
Not necessarily. As an explicit example take a=5 and b=1. The result will be odd. 25+1=3.
EDIT: To make things clearer, we can say that a=b+2k for some positive integer k since a>b.
Let y=2a+b
y=22b+2k
y=b+k
This means that y is only even when k is odd, i.e 2a+b is even only when a−b=2k.
This again implies that when k is odd 2a−b is also odd, which can be another proof for your question.
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 2a+b is even 2a−b is odd.(See my solution)