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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!

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up vote 4 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 $$\lceil x \rceil \geq \lceil y\rceil$$

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Oh wow, that turned out to be way easier than I thought. Thanks again! –  Maria Nov 13 '12 at 19:46
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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 $1$.

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