Answer:
search-icon-header
Search for questions & chapters
search-icon-image
Question
Bookmark
If a and b are two odd positive integers such that a>b, then prove that one of the two numbers
2
a+b
and
a−b
is odd and the other is even.
Medium
Solution
verified
Verified by Toppr
We have
a and b are two odd positive integers such that a & b
but we know that odd numbers are in the form of 2n+1 and 2n+3 where n is integer.
so, a=2n+3, b=2n+1, n∈1
Given ⇒ a>b
now, According to given question
Case I:
=
2n+3+2n+1
4n+4
=2n+2=2(n+1)
put let m=2n+1 then,
=2m ⇒ even number.
Case II:
2n+3−2n−1
=1 ⇒ odd number.
Hence we can see that, one is odd and other is even.
This is required solutions.
Step-by-step explanation:
please mark me as brilliant
Copyright © 2024 EHUB.TIPS team's - All rights reserved.
Answers & Comments
Answer:
search-icon-header
Search for questions & chapters
search-icon-image
Question
Bookmark
If a and b are two odd positive integers such that a>b, then prove that one of the two numbers
2
a+b
and
2
a−b
is odd and the other is even.
Medium
Solution
verified
Verified by Toppr
We have
a and b are two odd positive integers such that a & b
but we know that odd numbers are in the form of 2n+1 and 2n+3 where n is integer.
so, a=2n+3, b=2n+1, n∈1
Given ⇒ a>b
now, According to given question
Case I:
2
a+b
=
2
2n+3+2n+1
=
2
4n+4
=2n+2=2(n+1)
put let m=2n+1 then,
2
a+b
=2m ⇒ even number.
Case II:
2
a−b
=
2
2n+3−2n−1
2
2
=1 ⇒ odd number.
Hence we can see that, one is odd and other is even.
This is required solutions.
Step-by-step explanation:
please mark me as brilliant