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How can we prove that given that $x\leq 2$, then $2^x>x$?

I think that this seems to be intuitively correct but I don't know how to prove it. Can it be proven without calculus?

Xam
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TheNotoriousSc
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    Try to look at derivatives of both sides, prove that one is always less than the other – Andrei Kulunchakov Mar 02 '17 at 22:10
  • Split it into parts: prove it first when $1 \le x \le 2$, since the maximum value of $x$ on this interval equals the minimum value of $2^x$, and then argue separately for the interval $0 \le x \le 1$. – Chris Mar 02 '17 at 22:13
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    @Peter, this isn't really true. (See my comment.) Although admittedly the definition of $2^x$ usually (but not necessarily) comes from calculus. – Chris Mar 02 '17 at 22:14
  • @Peter: In particular we don't need calculus for case $x \le 0$. – hardmath Mar 02 '17 at 22:15
  • @Peter by definition we'll have $\frac{2^y}{2^x} = 2^{y-x}$; if $y > x$ then this says $\frac{2^y}{2^x} \ge 1$. – Chris Mar 02 '17 at 22:18
  • @Chris Can you use that rule of exponent if $x$ and $y$ are irrational, without first proving that rule (without calculus)? – imranfat Mar 02 '17 at 22:19
  • It suffices to prove it for rationals then use continuity. – eyeballfrog Mar 02 '17 at 22:20
  • @Peter Not necessarily. It's possible to define $a^x$ as the supremum of all numbers $a^{\frac{p}{q}}$, for $\frac{p}{q} \le x$. Using this definition, and without calculus per se, we can establish all the usual laws of exponents for real numbers. – Chris Mar 02 '17 at 22:20
  • I assumed no calculus to mean no derivatives. – eyeballfrog Mar 02 '17 at 22:22
  • For $x\geq 1$ you could use the generalised Bernoulli inequality:

    $$2^{x} , =, (1+1)^{x} ,\geq, 1+x ,>, x$$

     – AloneAndConfused Mar 02 '17 at 23:36

3 Answers3

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$$ \begin{align} \color{red}{0}\lt \color{blue}{x}\le\color{green}{1} \quad &\rightarrow\quad 2^\color{red}{0}\lt2^\color{blue}{x}\le2^\color{green}{1} \quad\Rightarrow\quad x\le2^0\lt2^x \\[2mm] \color{red}{1}\lt \color{blue}{x}\le\color{green}{2} \quad &\rightarrow\quad 2^\color{red}{1}\lt2^\color{blue}{x}\le2^\color{green}{2} \quad\Rightarrow\quad x\le2^1\lt2^x \\[2mm] \color{red}{2}\lt \color{blue}{x}\le\color{green}{4} \quad &\rightarrow\quad 2^\color{red}{2}\lt2^\color{blue}{x}\le2^\color{green}{4} \quad\Rightarrow\quad x\le2^2\lt2^x \\[2mm] \color{red}{4}\lt \color{blue}{x}\le\color{green}{16} \,\,\,\, &\rightarrow\quad 2^\color{red}{4}\lt2^\color{blue}{x}\le2^\color{green}{16} \,\,\,\Rightarrow\quad x\le2^4\lt2^x \\[4mm] \cdots\quad\text{etc}\quad &\rightarrow\quad \color{red}{x\lt2^x\quad\colon x\in\mathbb{R}} \end{align} $$

Hazem Orabi
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Note: Hazem Orabi's solution is really the one I prefer personally but since i had fun messing with a more convoluted and ultimately inconclusive approach I'd just like to share it.

Solution Let's do this for rational numbers $x = p/q$ and then just assume it extends properly to real numbers.

Let $p,q$ be relatively prime and let's just note that

$$0 < x < 2 \Rightarrow p < 2q$$

Now the number $2^{x} = 2^{p/q}$ is the solution to the equation $y^q = 2^p$ and the idea that $2^x > x$ can be rephrased as the implication

$$p < 2q \quad \text{and} \quad y^q = 2^p \quad \Rightarrow \quad qy > p \quad \text{(to be proven)}$$

The reason I phase it like this is that we're just dealing with integral powers the results of which are hopefully more intuitive (easier to prove) than rational or irrational powers.

Let us start by observing

$$(qy)^q = q^q y^q = q^q 2^p$$

And then divide our intention to two special cases $q < p$ and $p < q$

Case 1. $p < q$. In this case we use $q^q > p^q$

$$(qy)^q = q^q 2^p > p^q 2^p > p^q $$ and therefore $$(qy)^q > p^q \Leftrightarrow qy > p$$

Case 2. $q < p$. In this case we use $2^p > 2^q$

$$(qy)^q = q^q 2^p > q^q 2^q > (2q)^q > p^q $$

(where $2q > p$ was invoked in the final step) and therefore again $$(qy)^q > p^q \Leftrightarrow qy > p$$

(The principle invoked over and over again is that integral powers preserve order which while not entirely trivial is at least natural)

Squid
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The inequality is obvious for $x\le0$, so we can assume $x>0$.

Consider the function $f(x)=x\log 2-\log x$ (natural logarithm). It's easy to prove that this function has a minimum for $x=1/\log2$ and that $$ f(1/\log2)=1+\log\log2 $$ Since $2^2>e$, we have $2\log2>1$, so $\log2>1/2$ and $$ \log\log2>-\log2 $$ Therefore $1+\log\log2>1-\log2>0$, because $2<e$.

Since the minimum of $f$ is positive, we have $x\log2>\log x$, for every $x>0$, hence $2^x>x$.

egreg
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