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Place the following numbers by their size:

$$A=2^{4^{2^{.^{.^{.^{2^{4}}}}}}},B=4^{2^{4^{.^{.^{.{4^{2}}}}}}},C=2^{2^{2^{.^{.^{.^{2^{2}}}}}}}$$

In number $C$ there are $2000$ "$2$" digits, and in numbers $B,A$ there are $500$ "$2$" and $500$ "$4$" digits. It seems to me that $C>B>A$, but I can't give a proof. Any hints?

Here is the same problem in art of problem solving. I hope that it helps.

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Alternative and relatively easy solution without using logarithms. Let us set

$$h=2^{4^{2^4}}$$

$$j=4^{2^{4^2}}$$

$$k=2^{2^{2^{.^{.^{.^{2^{2}}}}}}}$$

where $h $ and $j $ only contain four numbers (two $2$ and two $4$), and $k $ contains sixteen $2$. Calculating the exponential towers starting from the top, we have

$$h=2^{4^{16}} = 2^{2^{32}} $$

$$j=4^{2^{16}} = 2^{2 \cdot 2^{16}} = 2^{ 2^{17}} $$

so that $h>j $ . Also, considering $k $ and solving its first two exponentials from the top, we get an exponential tower formed by fourteen elements where the first thirteen are $2$ and the last one is $16$. Therefore we clearly have $k>h>j$.

Now let us consider the numbers $A , B , C $ reported in the OP. Again starting to solve the exponentials from the top, we can rewrite them as

$$A=(A_1)^h $$ $$B=(B_1)^j $$ $$C=(C_1)^k $$

where $A_1$ and $B_1 $ are exponential towers similar to the initial towers $A $ and $B $, respectively, but with $500-4=496 \,\,\, $ elements instead of $500$; and where $C_1$ is an exponential tower similar to the initial $C $ but with $2000-16=1984 \,\,\, $ elements instead of $2000$.

Because $k>h>j \,\,\, $, if we could show that $C_1>A_1>B_1 \,\,\, $, then this would necessarily imply $C>A>B \,\,\, $. With this in mind, we can now repeat the same procedure above, applying it to $A_1, B_1, C_1 \,\,\, $. So we can set

$$A_1=(A_2)^h $$ $$B_1=(B_2)^j $$ $$C_1=(C_2)^k $$

where $A_2$ and $B_2 $ are again exponential towers similar to $A $ and $B$, respectively, but with $500-2\cdot 4 =492 \,\,\, $ elements instead of $500$; and where $C_2$ is an exponential tower similar to $C $ but with $2000-2 \cdot 16=1968 \,\,\, $ elements instead of $2000$. As above, because $k>h>j \,\,\, $, if we could show that $C_2>A_2>B_2$, then this would necessarily imply $C_1>A_1>B_1 \,\,\, $, and then $C>A>B \,\,\, $.

To generalize, after repeating this procedure $m $ times, we obtain two numbers $A_m$ and $B_m $ that are exponential towers similar to $A $ and $B $, respectively, but with $500-m\cdot 4 \,\,\,$ elements instead of $500$, and a third number $C_m$ that is an exponential tower similar to $C $ but with $2000-m \cdot 16 \,\,\, $ elements instead of $2000$. As above, because $k>h>j \,\,\, $, showing that $C_m>A_m>B_m \,\,\, \,\, $ necessarily implies $C_{m-1}>A_{m-1}>B_{m-1} \,\, \,\,\, $, which in turn implies $C_{m-2}>A_{m-2}>B_{m-2} \,\,\, \,\, $ and so on, until $C>A>B \,\,\, \,\, $.

So, repeating this procedure $m=124 \,\,$ times, we arrive to a point where $A_{124} $ and $B_{124} $ are exponential towers with $500-124\cdot 4 =4 \,\,\, $ elements, and $C_{124}$ is an exponential tower with $2000-124 \cdot 16=16 \,\,\, $ elements. Since $ A_{124}=h \,\, \,\,\, $, $B_{124}=j \,\, \,\,\, $, and $C_{124}=k \,\, \,\,\, $, we have $ C_{124}>A_{124}>B_{124} \,\,\, $, which necessarily implies, as shown above, $C>A>B \,\,\, $.

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The problem can be re-formulated in algorithmic style as follows: $$ \begin{matrix} a_0 = 1 & b_0 = 1 & c_0 = 1 \\ & & c_1 = 2^{c_0} \\ a_1 = 4^{a_0} & b_1 = 2^{b_0} & c_2 = 2^{c_1} \\ & & c_3 = 2^{c_2} \\ a_2 = 2^{a_1} & b_2 = 4^{b_1} & c_4 = 2^{c_3} \\ \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot \\ a_k = 4^{a_{k-1}} & b_k = 2^{b_{k-1}} & c_{2k} = 2^{c_{2k-1}} \\ & & c_{2k+1} = 2^{c_{2k}} \\ a_{k+1} = 2^{a_k} & b_{k+1} = 4^{b_k} & c_{2k+2} = 2^{c_{2k+1}} \\ \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot \\ A = a_{1000} = 2^{a_{999}} & B = b_{1000} = 4^{b_{999}} & C = c_{2000} = 2^{c_{1999}} \end{matrix} $$ Now we know how to calculate these huge numbers - though very much in principle :-)

The logarithm with base $2$ may be defined formally as $\,\operatorname{lg}(x) = \ln(x)/\ln(2)$ .
But in the practice below its meaning is much simpler, as might become clear from a few examples: $\lg(2) = 1 \; ; \; \lg(4) = \lg\left(2^2\right) = 2 \; ; \; \lg(8) = \lg\left(2^3\right) = 3$ . In general : $\,\operatorname{lg}\left(2^x\right) = x$ .
We need a function to chop off the exponent from a power (tower) of two. We could have called this function "chopper", but the proper name for it in common mathematics is $2$-logarithm. Take it, or if you don't like logarithms: leave it. The whole employment of our "exponent chopper" is like in here: $2^x > 2^y \; \Longleftrightarrow \; x > y$ . So what's the problem?

We are going to employ mathematical induction. To that end, define subsequent "approximations" of $A,B,C$ as follows: $$ A_n = a_{2n} \quad ; \quad B_n = b_{2n} \quad ; \quad C_n = c_{4n} $$ As the first induction step, we calculate: $$ A_1 = a_2 = 2^4 = 16 \quad ; \quad B_1= b_2 = 4^2 = 16 \quad ; \quad C_1 = c_4 = 2^{2^{2^2}} = 65536 $$ That's not enough to establish an inequality between $A$ and $B$, so we take a second step for these: $$ A_2 = a_4 = 2^{4^{2^4}} \quad ; \quad B_2 = b_4 = 4^{2^{4^2}} = 2^{2\cdot{2^{4^2}}} $$ Logarithms base $2$ : $$ \operatorname{lg}\left(2^{4^{2^4}}\right) = 4^{2^4} = 2^{2\cdot{2^4}} \quad ; \quad \operatorname{lg}\left(4^{2^{4^2}}\right) = 2\cdot 2^{4^2} = 2^{1+4^2} $$ And again: $$ \operatorname{lg}\left(4^{2^4}\right) = 2\cdot 2^4 = 32\quad ; \quad \operatorname{lg}\left(2\cdot 2^{4^2}\right) = 1 + 4^2 = 17 $$ From $\,32 > 17\,$ it follows that $\;A_2 > B_2\,$ , namely: $\;2^{2^{32}} > 2^{2^{17}}$ .

Now assume that $\;C_n > A_n > B_n \gg 1\;$ and prove that $\;C_{n+1} > A_{n+1} > B_{n+1}$ . $$ \begin{matrix} A_n = a_{2n} & B_n = b_{2n} & C_n = c_{4n} \\ A_{n+1} = a_{2n+2} = 2^{4^{A_n}} & B_{n+1} = b_{2n+2} = 4^{2^{B_n}} & C_{n+1} = c_{4n+4} = 2^{2^{2^{2^{C_n}}}} \end{matrix} $$ Taking logarithms two times: $$ \begin{matrix} \operatorname{lg}\left(A_{n+1}\right) = 4^{A_n} & \operatorname{lg}\left(B_{n+1}\right) = 2\cdot 2^{B_n} & \operatorname{lg}\left(C_{n+1}\right) = 2^{2^{2^{C_n}}} \\ \operatorname{lg}\left(\operatorname{lg}\left(A_{n+1}\right)\right) = 2A_n & \operatorname{lg}\left(\operatorname{lg}\left(B_{n+1}\right)\right) = 1+B_n & \operatorname{lg}\left(\operatorname{lg}\left(C_{n+1}\right)\right) = 2^{2^{C_n}} \end{matrix} $$ From which it follows that $\;C_{n+1} > A_{n+1} > B_{n+1}\;$ as well. Especially: $$ C_2 > A_2 > B_2 $$ So for all $\,n \ge 2\,$ we have $\;C_n > A_n > B_n$ . Now specialize for $\,n = 500\,$ and you're done.
Conclusion : $\;C > A > B$ .   Suggestive picture (not pretending anything more):

enter image description here

BONUS. Somewhat more of a challenge is the following modification of the question:
in $C$ there are $1500$ numbers $2$ (instead of $2000$). Then we have: $$ A_n = a_{2n} \quad ; \quad B_n = b_{2n} \quad ; \quad C_n = c_{3n} $$ And the first induction step results in an equality $\;A_1=B_1=C_1$ : $$ A_1 = 2^4 = 16 \quad ; \quad B_1 = 4^2 = 16 \quad ; \quad C_1 = 2^{2^2} = 16 $$ A second step is needed to establish an inequality: $$ \operatorname{lg}(\operatorname{lg}(A_2)) = 32 \quad ; \quad \operatorname{lg}(\operatorname{lg}(B_2)) = 17 \\ \operatorname{lg}(\operatorname{lg}(C_2)) = \operatorname{lg}\left(\operatorname{lg}\left(2^{2^{2^{16}}}\right)\right) = \operatorname{lg}\left(2^{2^{16}}\right) = 2^{16} = 65536 $$ Finally resulting in the same as before: $\;C > A > B$ .

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  • $\begingroup$ But I set the bounty for an answer without using logarithem. $\endgroup$ – Taha Akbari Jan 14 '17 at 20:29
  • $\begingroup$ @TahaAkbari: Even if that answer without logarithms is wrong? I'm always trying to give an answer that is correct mathematically. And in the way I can comprehend it at best myself. If that answer is not according to the OP's wishes, then so be it. Your +100 points bounty is only one percent of my current reputation. It's second to the challenge of the question itself. $\endgroup$ – Han de Bruijn Jan 15 '17 at 8:53
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    $\begingroup$ I think it doesn't need to change with new edits even with using logarithems it is a nice answer. $\endgroup$ – Taha Akbari Jan 16 '17 at 16:47
  • $\begingroup$ Pft, who needs bounties. It's easy to earn +800 rep at least once a week in one day, amirite? $\endgroup$ – Simply Beautiful Art Jan 17 '17 at 0:45
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    $\begingroup$ @TakahuroWaki haha, I don't want to do that. And if you want to get a good income of rep, keep answering questions, fail a bit, but keep doing it. Practice will give you good answering skills. $\endgroup$ – Simply Beautiful Art Jan 19 '17 at 12:02
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One approach is to use that if $x > 3y$ we have $2^x > 3\cdot4^y$ if $y\ge2$. This is because:

$$2^x > 2^{3y}$$ $$4^y = 2^{2y}$$

so their ratio is:

$${2^x\over4^y} > {2^{3y}\over2^{2y}}=2^y$$

In similar way we have that $4^x > 3\cdot2^y$.

So if $y\ge2$ we have that the ratio is at least $4$ and definitely larger than $3$. So we can use this to get relative estimates. Starting at the tail of $A$ and $B$. Obviously $2^4=16=4^2$, but next we have $4^{2^4} = 4^{16} = 2^{17}$ and $2^{4^2}=2^{16}$. Next step we will have that $2^{4^{2^4}} > 3\cdot4^{2^{4^2}}$, so thereby we will have that result all the way until we've reached $A$ and $B$ so we will have $A>3B>B$ (I assume that they all contain the same number of numbers).

The relation between $A$ and $C$ can be seen in similar and somewhat easier way. To do this we will first make sure that the exponent chain is equally long. This is done by just replacing the tail of $C$. Obviously $2^{2^{\cdots{997}\cdots^2}} > 3\cdot4$, the rest follows - so $C>A$.

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  • $\begingroup$ sorry but we didn't learn logarithem yet. $\endgroup$ – Taha Akbari Jan 12 '17 at 15:00
  • $\begingroup$ Yes we have learned them.And I have to say there aren't $498$ numbers omitted there are $996$ because I told there are $500$, $2$s and $500$,$4$s But I can't understand your way!Now how is their order? $\endgroup$ – Taha Akbari Jan 13 '17 at 7:12
  • $\begingroup$ @HandeBruijn I've modified my answer using another aproach making the above comments redundant. $\endgroup$ – skyking Jan 13 '17 at 14:52
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    $\begingroup$ It's a bit unclear use brakets to clearfiy put the power in {}.And also in $C$ we have $2000$ numbers not $1000$. $\endgroup$ – Taha Akbari Jan 13 '17 at 17:08
  • $\begingroup$ Indeed, I find that $C > B$ and not $B > C$ . $\endgroup$ – Han de Bruijn Jan 14 '17 at 13:58
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This is an excellent exercise to practice how to deal with exponentiation. Remembering that $x\cdot x^y=x^{y+1}$ and that $(x^y)^z=x^{y\cdot z}$, let:

$a_1=2^4=16~~~~~~a_{i+1}=2^{4^{a_{i}}}=2^{(2^2)^{a_i}}=2^{2^{\mathbf{\left(2\cdot a_{i}\right)}}}$

$b_1=4^2=16~~~~~~b_{i+1}=4^{2^{b_{i}}}=(2^2)^{2^{b_i}}=2^{2\cdot 2^{b_i}}=2^{2^{\mathbf{\left(b_i+1\right)}}}$

$c_1=2^{2^{2^{2}}}=2^{16}~~~c_{i+1}=2^{2^{2^{2^{c_{i}}}}}=2^{2^{\mathbf{\left(2^{2^{c_{i}}}\right)}}}$

so that $A=a_{500}, B=b_{500}, C=c_{500}$.

Now, if $1<b_i\leq a_i\leq c_i$ then $\mathbf{(b_i+1)}\leq (a_i+1) < \mathbf{(2\cdot a_i)} \leq (2\cdot c_i) < 2^{c_i} < \mathbf{\left( 2^{2^{c_i}}\right)}$, so if we look at what is in parentheses "on top" of the $2^2$ in the last term of the expressions of $a_{i+1}$, $b_{i+1}$, $c_{i+1}$ we have:

$1<b_1=a_1<c_1 \implies 1<b_2<a_2<c_2 \implies\dots \implies 1<b_{500}<a_{500}<c_{500}$, i.e.

$B<A<C$.

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