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

  • Thread starter Ash L
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  • #1
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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 [itex]\notin[/itex] Z, then x/2+0.25 [itex]\notin[/itex] 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.
 

Answers and Replies

  • #2
34,976
11,164
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.
 

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