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I have the following statement that I'm trying to prove:

Assume that $f,g: \mathbb{N} \rightarrow \mathbb{R}^{\ge0}$.

If $f(n) \ge g(n)$ then $\lceil f(n) \rceil \ge \lceil g(n) \rceil $.

I have a kind of proof. But it is ugly (by using $f(n) - g(n) \ge 0$ and this will also mean $\lceil f(n) - g(n) \rceil \ge 0$, then check for different conditions, e.g. if the difference less than or equal to 1 etc.) and I'm sure there is a much simpler and more elegant way to prove this. But I couldn't figure out and seeking some help.

Edit: I'm not sure if it will help for elegant proof but f and g are also eventually non-decreasing function.

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2 Answers 2

up vote 1 down vote accepted

For any such function $h$, $\lceil h(n)\rceil -1 < h(n)\leq \lceil h(n)\rceil$, with equality only when $f(n)$ is an integer. So $\lceil f(n)\rceil<\lceil g(n)\rceil$ implies that $$f(n)\leq \lceil f(n)\rceil\leq \lceil g(n)\rceil-1<g(n)\leq\lceil g(n)\rceil$$ which proves the contrapositive.

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great, that was kind of a proof that I was looking for. –  mathamania Jan 6 '13 at 0:09

$a=\lceil g(n)\rceil$ is the smallest integer which is greater or equal than $g(n)$. Since $b=\lceil f(n)\rceil$ is an integer which is greater or equal than $f(n)$, and the latter is greater or equal than $g(n)$, we see that $b\geq a$.

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thanks for your answer, sorry if I was not clear enough in my question. But I was looking for a more formal approach similar to the Alexander pointed out. –  mathamania Jan 6 '13 at 0:12
    
My argument is completely formal! –  Mariano Suárez-Alvarez Jan 6 '13 at 0:19

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