What is the formula for calculating the sum of ceiling values for two integers?

  • Thread starter rayukpant
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In summary, to find the value of ceiling[(n+m)/2] + ceiling[(n-m+1)/2], we must consider two cases separately: (1) when both n and m have the same parity (either both odd or both even), and (2) when n and m have different parities (one odd and one even). In both cases, we can explicitly find the ceilings in question by plugging in expressions for even and odd integers. This will allow us to evaluate the answer explicitly. Additionally, it is important to note that the sum or difference of two even numbers is even, the sum or difference of two odd numbers is even, and the sum or difference of one odd and one even number is even.
  • #1
rayukpant
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For integers 'n' and 'm', find the value of ceiling[(n+m)/2] + ceiling[(n-m+1)/2].
 
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  • #2
Consider the two cases separately, of
1. Either both even or both odd and
2. One odd the other even.
 
  • #3
Just split it into cases.

Either
(1) m and n have the same parity (either both odd or both even); or
(2) m and n have different parities (one odd and the other even).

In both cases, you can explicitly find the ceilings in question.
 
  • #4
can you please tell me more in detail as to how to consider those two separately with even and odd parities
 
  • #5
How can you express an even integer? Odd integer? Once you have expressions for each, go through each case and plug them in. After you do this, you can evaluate explicitly what the answer is.
 
  • #6
The sum of difference of two even numbers is even, the sum or difference of two odd numbers is even. The sum or difference of one odd and one even number is even. Fit those into your formula. The crucial point is that in any case one of (n+m)/2 and (n- m+ 1)/2 is an integer and the other is a half integer.
 

FAQ: What is the formula for calculating the sum of ceiling values for two integers?

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