Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In class, we used the fact that $\lceil{a + b \rceil} \geq \lceil{a}\rceil + \lfloor{b}\rfloor$. However, we weren't given a proof of this statement.

I am interested to see how this works. Can anyone help? Thanks!

share|cite|improve this question
up vote 7 down vote accepted

By definition: $$b\geq \lfloor b\rfloor$$ Adding $a$ to both sides: $$a+b \geq a+ \lfloor b\rfloor$$ Taking the ceiling of both sides: $$\lceil a + b\rceil \geq \lceil a + \lfloor b\rfloor\rceil = \lceil a\rceil + \lfloor b \rfloor$$

This uses that if $n$ is an integer, then $$\lceil a + n\rceil = \lceil a \rceil + n$$ And if $x\geq y$ then $$\lceil x \rceil \geq \lceil y\rceil$$

share|cite|improve this answer
Oh wow, that turned out to be way easier than I thought. Thanks again! – Maria Nov 13 '12 at 19:46

By subtracting off integer parts, we can prove this for numbers in $[0,1)$. Unless both are $0$ the right side is $1$, and then the left is at least $1$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.