Ceiling[(2x + 1) / 2)] - ceiling[(2x + 1) / 4] + floor[(2x + 1) / 4]

  • Thread starter Thread starter Ash L
  • Start date Start date
Click For Summary
SUMMARY

The expression ceiling[(2x + 1) / 2] - ceiling[(2x + 1) / 4] + floor[(2x + 1) / 4] consistently evaluates to either floor(x) or ceiling(x) for any real number x. The proof involves analyzing the behavior of the ceiling and floor functions based on the divisibility of 2x + 1 by 2 and 4. Specific cases arise depending on whether x is an integer or not, particularly focusing on values such as 0, 1/4, and 2/4.

PREREQUISITES
  • Understanding of floor and ceiling functions
  • Familiarity with real number properties
  • Basic algebraic manipulation skills
  • Knowledge of piecewise functions
NEXT STEPS
  • Study the properties of floor and ceiling functions in detail
  • Explore piecewise function definitions and their applications
  • Investigate the implications of divisibility in mathematical expressions
  • Learn about mathematical proofs by cases and their methodologies
USEFUL FOR

Mathematics students, educators, and anyone interested in understanding the behavior of floor and ceiling functions in real analysis.

Ash L
Messages
18
Reaction score
0

Homework Statement



Show that for any real number x, ceiling[(2x + 1) / 2)] - ceiling[(2x + 1) / 4] + floor[(2x + 1) / 4] always equals to floor(x) or ceiling(x). In what circumstances does each case arise?


Homework Equations


I know that:

floor(x) = largest integer ≤ x
ceiling(x) = smallest integer ≥ x
I also looked at other equations in wikipedia, but don't know where to begin.

I tried converting it to another form:
ceil(x+0.5) - ceil(x/2+0.25) + floor(x/2+0.25)

Then I tried to prove it by cases by assuming that x \notin Z, then x/2+0.25 \notin Z as well. But then I got stuck with ceil(x+0.5) and don't know what to do.

ceil(x+0.5) - ceil(x/2+0.25) + floor(x/2+0.25)
= ceil(x+0.5) - [2floor(x/2 + 0.25) + 1]

Thanks.
 
Physics news on Phys.org
Those individual parts change only if 2x+1 is divisible by 2 or 4. You could consider values like 0, 1/4, 2/4, ... and the intervals between them.
 

Similar threads

Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 105 ·
4
Replies
105
Views
7K
Replies
5
Views
954
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
2
Views
4K