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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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    $\begingroup$ what does mean $2^x$ when $x<0$ ? $\endgroup$
    – Sebastien
    Apr 18, 2014 at 16:03
  • $\begingroup$ 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. $\endgroup$
    – Shubham
    Apr 18, 2014 at 16:04
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    $\begingroup$ @Sebastien: Presumably $\dfrac{1}{2^{-x}}$. $\endgroup$ Apr 18, 2014 at 18:45

4 Answers 4

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Can you show that $x\log2>\log x$, $\, \forall\, x>0$?

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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.

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  • $\begingroup$ "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. $\endgroup$ Apr 19, 2014 at 5:13
  • $\begingroup$ @EricM.Schmidt: Thanks for your comment. I have mistaken. The proof is now edited. $\endgroup$ Apr 21, 2014 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.

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  • $\begingroup$ $$\log 2\cdot 2^x-1>0\iff 2^x>\frac1{\log 2}$$ and since $\;2^x\xrightarrow[x\to-\infty]{}0\;$ , the above inequality cannot be true for all real $\;x\;$ ... $\endgroup$
    – DonAntonio
    Apr 18, 2014 at 16:11
  • $\begingroup$ My bad @DonAntonio - thanks for pointing that out. My updated answer is now valid. $\endgroup$
    – naslundx
    Apr 18, 2014 at 16:16
  • $\begingroup$ Indeed so, @naslundx...though showing the minimal value (at not least than $\;-\frac{\log\log2}{\log2}...!\;$) is positive is not that easy... $\endgroup$
    – DonAntonio
    Apr 18, 2014 at 16:18
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    $\begingroup$ Proving 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 $4 > 3$. $\endgroup$ Apr 19, 2014 at 4:50
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Hint

Take the $^2\log$ on both sides

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  • $\begingroup$ A bit of a cheat... how do you know $\log$ is increasing? $\endgroup$
    – Jack M
    Apr 21, 2014 at 13:33

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