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Suppose $f$ is a real-valued function and $f(x) \geq 1$ for every real $x \in [0,1]$.

Why we can always find a unique positive integer $n$ such that $2^{n} \leq f(x) < 2^{n+1}$ ?

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@User: Take $f(x)=e^{x}$. Clearly $f(x) \geq 1$ for every $x \in [0,1]$. But as $x \to \infty$, $e^{x} \to \infty$. –  anonymous Nov 17 '10 at 23:12
    
Do you mean for any given $x$? Or for all $x$ at once? The second statement is false; the first is true because $f(x)\ge 1=2^0$ and $\{[2^n,2^{n+1})\}$ partitions $[1,\infty)$. –  Paul VanKoughnett Nov 17 '10 at 23:15
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2 Answers

The intervals $[2^n,2^{n+1})$ cover the entire real half (existence) line $y\geq 1$ and no two of them intersect (uniqueness). So any point on the graph of the function must intersect one of these since $f(x)\geq 1$. You don't need any hypothesis on the domain of the function.

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You can't do it for $f(x)=3x+1$.

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@Paul: I meant for a given x. How do you prove that $[2^n,2^{n+1})$ partitions $[1,\infty)$ ? –  student Nov 17 '10 at 23:19
    
Got it now, thanks! –  student Nov 17 '10 at 23:25
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