How can I prove that for all reals $x$, $2^x > x$? I can prove this for integers with induction, but I can't figure out how to prove it for reals. Perhaps you could say that since $2^x$ is strictly increasing, $2^x \ge 2^{floor(x)} > floor(x)$, but is this really the best approach?
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3what does mean $2^x$ when $x<0$ ? – Sebastien Apr 18 '14 at 16:03
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Perhaps you could try with $2^x-x$. Or if you really want to use induction, start by proving it for rationals, e.g. let $x=\frac {p}{q}$, $p,q\ \epsilon Z$ etc. – Shubham Apr 18 '14 at 16:04
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2@Sebastien: Presumably $\dfrac{1}{2^{-x}}$. – Najib Idrissi Apr 18 '14 at 18:45
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Since you have proven it for all integers, the key step has been done. Now note that we are only to show the inequality for $x \in \mathbb R^+$ because the inequality becomes trivial for $x \in \mathbb R^-$ and when $x=0$, the inequality is obviously true. So now we note that $2^n>n$ $\forall n$ $\geq 0$. In other words we have $2^n \geq n+1$ $\forall n$ $\geq 0$. Now choose any $\epsilon$ such that $n<\epsilon<n+1$. We then get $2^n$$^+$$^1$ $>$ $2^\epsilon$ $>$ $2^n$ $\geq$ $n+1$ $>$ $\epsilon$ $>$ $n$. Hence proved for all positive reals.
Note that I have claimed that the result is proved for all positive reals but has not shown that why it must be. Prove this and the result will be completely proved.
William Hilbert
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"Multiplying we get, $2^n$$^+$$^\epsilon$ $>$ $n+\epsilon$." I don't understand this. Wouldn't you get $2^{n+\epsilon} > n\epsilon$? I don't think this argument works since then we could use a similar argument (replacing 2 with 1.1 and setting $n=1$, $\epsilon = 0.5$) to prove $(1.1)^{1.5} > 1.5$, which by numerical computation is false. – Eric M. Schmidt Apr 19 '14 at 05:13
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@EricM.Schmidt: Thanks for your comment. I have mistaken. The proof is now edited. – William Hilbert Apr 21 '14 at 13:01
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Hint
The derivative of the function $f(x) = 2^x-x$ is $$f'(x) = \ln(2) \cdot 2^x - 1$$ which has only one zero (let's say at $x_0$). Check that this is a minimum point and make sure that $f(x_0)>0$, and you are done, since $f(x)$ is continous.
naslundx
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$$\log 2\cdot 2^x-1>0\iff 2^x>\frac1{\log 2}$$ and since $;2^x\xrightarrow[x\to-\infty]{}0;$ , the above inequality cannot be true for all real $;x;$ ... – DonAntonio Apr 18 '14 at 16:11
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My bad @DonAntonio - thanks for pointing that out. My updated answer is now valid. – naslundx Apr 18 '14 at 16:16
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Indeed so, @naslundx...though showing the minimal value (at not least than $;-\frac{\log\log2}{\log2}...!;$) is positive is not that easy... – DonAntonio Apr 18 '14 at 16:18
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1Proving that the minimum value is positive: If $c = -\frac{\log \log 2}{\log 2}$, then $2^c = \exp(c\log 2) = \exp(- \log \log 2) = 1/(\log 2)$. So, $f(c) = \frac{(\log \log 2) + 1}{\log 2}$. Thus (as $\log 2 > 0$), we just need to show $\log \log 2 > -1$, or 2 > $e^{1/e}$. Since $2 < e < 3$, it suffices to show $2 > 3^{1/2}$, or $4 > 3$. – Eric M. Schmidt Apr 19 '14 at 04:50
