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I'm trying to prove a statement with M.I. Here is my statement:

There exist an $n_{0}\epsilon\mathbb{N}$ such that $2^{n}\geq n^{4}$ for all $n\geq n_{0}$

Well, when I start to prove:

Initial step: Let n = $n_{0}$

I get $2^{n_0}\geq n_{0}^{4}$ . But I realize this equation is just true when $n_{0}$ equal to $0$, $1$.

So I mean I can't start because I get a problem at the initial step. Any advice? How can I solve this so I can continue to prove.

Thanks in advance.

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First, you have to work out what $n_0$ is. This you do by trying various values of $n$ until you are convinced that you have found a suitable $n_0$. Also, there seems to be some confusion as to whether you are trying to show $2^n\ge n^2$ or $2^n\ge n^4$. –  Gerry Myerson Dec 17 '12 at 4:50
    
Thx for warning :) –  Tiro Dec 17 '12 at 4:51
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1 Answer

Your inequality holds for $n \geq 16$. Hence, choose your $n_0 = 16$.

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