# Prove that $2^{2^{\sqrt3}}>10$

With a computer or calculator, it is easy to show that $$2^{2^\sqrt{3}} = 10.000478 \ldots > 10.$$

How can we prove that $2^{2^{\sqrt3}}>10$ without a calculator?

• Considering that $2^{2^{\sqrt{3}}} \approx 10.000478\ldots$ is very close to $10$, this will be quite a challenge. – JimmyK4542 Dec 28 '14 at 7:22
• What does that even mean? It sounds to me ill-posed... the calculator doesn't do anything magical that I couldn't also do (laboriously) by hand: computing logarithms with proven error bounds using series, etc... – user7530 Dec 28 '14 at 7:24
• My comment wasn't meant to be taken as a solution. I was simply pointing out that this would be a difficult task. – JimmyK4542 Dec 28 '14 at 7:25
• What contest is this? – user7530 Jan 9 '15 at 6:28
• I didn't use a calculator in my answer! I think the meaning is clear, and the problem is entertaining. – copper.hat Jan 9 '15 at 22:17

I suspect any adequate answer to this question is going to be very computation-heavy. Here's one.

First we claim that $$\sqrt{3} > \frac{3691}{2131} \tag{1}$$ This fraction is not pulled out of a hat; rather, it is found by taking the continued fraction $\sqrt{3} = [1; 1, 2, 1, 2, 1, 2, \ldots]$ and carrying it out a while. Anyway, the above is proved directly by $$3 \cdot 2131^2 = 3 \cdot 4541161 = 13623483 > 13623481 = 3691^2.$$

We next claim that $$2^{{3691}/{2131}} > \frac{1465}{441} \tag{2}$$

To prove this, one needs to show $2^{3691} 441^{2131} > 1465^{2131}$. With repeated squaring, each term requires squaring about $12$ times, so this is doable$^{a}$ with a pencil and paper.

The last step is $$2^{1465/441} > 10 \tag{3}$$ Again one uses repeated squaring, and if one is working in base 10 one only needs to count the number of digits in $2^{1465}$. This should not take nearly as long as the previous computation.$^{b}$

From (1), (2), and (3), $$2^{2^{\sqrt{3}}} > 2^{2^{3691/2131}} > 2^{1465 / 441} > 10. \quad \square$$

$^{a}$ Honestly, this is quite a stretch. I can't guarantee it won't take like a year of work.

$^{b}$ As a rough estimate, perhaps a day or so.

• Why is it obvious that any adequate answer would be computation-heavy? Do you have a proof lying around somewhere? :-P – Aryabhata Jan 9 '15 at 2:39
• Well, then don't claim that $\pi + e$ isn't rational. – Aryabhata Jan 9 '15 at 2:44
• @Aryabhata I am not at all optomistic that a proof could be found. It would involve proving that any proof of $2^{2^{\sqrt{3}}}$ must be sufficiently long. And once you start talking about all proofs of a statement, you are venturing into territory where many things are provably undecidable. – 6005 Jan 9 '15 at 2:45
• You are missing the fact that this is a contest problem. If this was something someone just noticed while playing with a calculator, then there would be good reason to be skeptical. – Aryabhata Jan 9 '15 at 2:47
• All I am saying is, don't state your opinions as if they were facts :-P – Aryabhata Jan 9 '15 at 2:50

We want to prove $(\log_2 \log_2 10)^2 < 3$.

One way is to use the procedure outlined here to compute $\log_2$. The only mild complication is knowing when to stop. Since $\log_2$ is non-decreasing, it suffices to find $x\ge\log_2 10$ and $y \ge \log_2 x$ such that $y^2 <3$.

The procedure is straightforward, I am just repeating the parts necessary to see how an upper bound is found. Given $x>0$, we first compute the integer part of $\log_2 x$ by finding the smallest $n$ such that ${x \over 2^n} \in [1,2)$, then $\log_2 x = n + \log_2 {x \over 2^n}$. Then, supposing $x \in (1,2)$, we repeatedly square $x$ until $x^{2^n} \in [2,4)$. Then we have $\log_2 x = {1 \over 2^n}(1+ \log_2 {x^{2^n} \over 2})$, where ${x^{2^n} \over 2} \in [1,2)$, and so we can repeat ad nauseam.

For the purposes of this problem, we note that in the latter step, we always have $\log_2 x \le {1 \over 2^n}(1+ 1)$, since $\log_2 {x^{2^n} \over 2} \le 1$. So by replacing $\log_2 {x^{2^n} \over 2}$ by $1$ at any stage we obtain an upper bound. The error will be $\le {1 \over 2^{n_1}} \cdots {1 \over {2^{n_k}}}$, where the $n_1,...,n_k$ are the number of 'squarings' at each step.

Then it is a matter of trial and error to find suitable $x,y$:

$x = 3+{1 \over 4} + {1 \over 16} + {1 \over 128} + {1 \over 1024} + {1 \over 2048}+{1 \over 8192}+ {1 \over 65536}+ {1 \over 65536} = {217706 \over 65536} \ge \log_2 10$.

$y = 1+{1 \over 2} + {1 \over 8} +{1 \over 32} + {1 \over 128} + {1 \over 256} + {1 \over 1024} + {1 \over 2048} + {1 \over 16384} + {1 \over 65536}+ {1 \over 65536}= { 113510 \over 65536 } \ge\log_2 x$.

We have $y^2=({ 113510 \over 65536 })^2 < 3$.

No calculators or computers were harmed during this computation.

There is nothing magical about a calculator: any computation performed on a calculator can be done by hand (and perhaps with extra rigor). (*)

Most special functions have series approximations that are known to converge particularly quickly; but Taylor's theorem with remainder can be applied to almost all important special functions even without knowing these specialized series. If $f(x)$ is smooth on $(a-\epsilon, b+\epsilon)$ and all derivatives $f^{(n)}$ are bounded on that interval:

$$|f^{(n)}(x)| \leq K^n\qquad \forall x\in (a-\epsilon, b+\epsilon)$$

then Taylor's theorem guarantees that

$$|f(b) - F^i(b)| \leq \frac{K^i(b-a)^{i+1}}{(i+1)!}$$ where $F^i$ is the Taylor expansion to $i$th order $$F^i(b) = \sum_{j=0}^i f^{(j)}(a) \frac{(b-a)^j}{j!}.$$

In particular if $K^n$ can be chosen to decay sufficiently quickly in $n$, and $f^{(n)}(a)$ is easy to evaluate exactly at a special value $a$, the above can be used to approximate $f(b)$ to arbitrary precision by hand.

For this particular problem, we want to show that $$\sqrt{3}\log 2 \leq \log\log 10 - \log \log 2.$$

$\sqrt{x}$ and $\log x$ can be trivially estimated using the above. $\log \log x$ can be computed by composition (this will require computing $\log x$ to high precision) or directly using Taylor expansion about $x=e$ (NB the formulas for the higher-order derivatives are not particularly pleasant).

(*) However even when guaranteed accuracy is required, the advantages of hand calculation over computer algebra packages like Mathematica with arbitrary-precision support are rather dubious.

First take logs $$2^{\sqrt3}>\log_2 10$$ Then again $$\sqrt{3}>\log_2 \log_2 10$$ Now, $$\log_2(10)=\frac{\log 10}{\log 2}$$ With $\log 10\approx 2.302$ and $\log 2\approx 0.6931$, We find $$\log_2(10)\approx3.32131$$ And the $\log_2$ of that is $(\approx1.731)$ Now, $\sqrt 3\approx 1.732,$ and therefore we conclude that $$2^{2^{\sqrt3}}>10$$ Note that these values can be found with Newton's method and taylor series, though tedious to find and calculate.

• Unfortunately, $\sqrt 3 \approx 1.732$, so your argument fails. Also, how do you get $\log_2 3.3270 \approx 1.732$, though it is correct? – Ross Millikan Jan 9 '15 at 3:38
• I used a few terms of the taylor series of $e^\log 2 x$ and newtons method with $f(x)=2^x-3.3270$. And then I probably failed at memorizing sqrt 3 :/ I will redo my computations with more accuracy then. – Cyclohexanol. Jan 9 '15 at 5:16
• Fixed now. I got a better estimate of $\log 10$ and that was the problem. – Cyclohexanol. Jan 9 '15 at 5:24
• You really should be using inequalities in the correct directions instead of $\approx$. Especially since your approximations of $\log_2(10)$ and $\log_2(\log_2(10))$ are actually off in the wrong direction. (You'll need more decimal places in your approximation of $\sqrt{3}$ when you correct those.) – aes Jan 9 '15 at 5:30