Editing Conway chained arrow notation (section)

==Graham's number==
[[Graham's number]] cannot be expressed in Conway chained arrow notation, but it is bounded by the following:

<math>3 \rightarrow 3 \rightarrow 64 \rightarrow 2 < G < 3 \rightarrow 3 \rightarrow 65 \rightarrow 2</math>

'''Proof:''' We first define the intermediate function <math>f(n) = 3 \rightarrow 3 \rightarrow n =  \begin{matrix}
    3\underbrace{\uparrow \uparrow \cdots \uparrow}3 \\
    \text{n arrows}
 \end{matrix}</math>, which can be used to define Graham's number as <math>G = f^{64}(4)</math>. (The superscript 64 denotes a [[Function composition#Functional powers|functional power]].)

By applying rule 2 and rule 4 backwards, we simplify:

<math>f^{64}(1)</math>
:<math>= 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (\cdots (3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow 1))\cdots ))</math> (with 64 <math>3 \rightarrow 3</math>'s)
:<math>= 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (\cdots (3 \rightarrow 3 \rightarrow (3 \rightarrow 3) \rightarrow 1) \cdots ) \rightarrow 1) \rightarrow 1</math>
:<math>= 3 \rightarrow 3 \rightarrow 64 \rightarrow 2;</math>
<math display="block">
\left.
 \begin{matrix}
=   &3\underbrace{\uparrow \uparrow \cdots\cdots\cdots\cdot \uparrow}3 \\
    &3\underbrace{\uparrow \uparrow \cdots\cdots\cdots \uparrow}3 \\
    &\underbrace{\qquad\;\; \vdots \qquad\;\;} \\
    &3\underbrace{\uparrow \uparrow \cdots\cdot \uparrow}3 \\
    &3\uparrow 3
 \end{matrix}
\right\} \text{64 layers}
</math>

<math>f^{64}(4) = G;</math>
:<math>= 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (\cdots (3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow 4))\cdots ))</math> (with 64 <math>3 \rightarrow 3</math>'s)
<math display="block">
\left.
 \begin{matrix}
=   &3\underbrace{\uparrow \uparrow \cdots\cdots\cdots\cdot \uparrow}3 \\
    &3\underbrace{\uparrow \uparrow \cdots\cdots\cdots \uparrow}3 \\
    &\underbrace{\qquad\;\; \vdots \qquad\;\;} \\
    &3\underbrace{\uparrow \uparrow \cdots\cdot \uparrow}3 \\
    &3\uparrow \uparrow \uparrow \uparrow 3
 \end{matrix}
\right\} \text{64 layers}
</math>

<math>f^{64}(27)</math>
:<math>= 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (\cdots (3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow 27))\cdots ))</math> (with 64 <math>3 \rightarrow 3</math>'s)
:<math>= 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (\cdots (3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (3 \rightarrow 3)))\cdots ))</math> (with 65 <math>3 \rightarrow 3</math>'s)
:<math>= 3 \rightarrow 3 \rightarrow 65 \rightarrow 2</math> (computing as above).
:<math>= f^{65}(1)</math> 
<math display="block">
\left.
 \begin{matrix}
=   &3\underbrace{\uparrow \uparrow \cdots\cdots\cdots\cdot \uparrow}3 \\
    &3\underbrace{\uparrow \uparrow \cdots\cdots\cdots \uparrow}3 \\
    &\underbrace{\qquad\;\; \vdots \qquad\;\;} \\
    &3\underbrace{\uparrow \uparrow \cdots\cdot \uparrow}3 \\
    &3\uparrow 3
 \end{matrix}
\right\} \text{65 layers}
</math>

Since ''f'' is [[strictly increasing]],
:<math>f^{64}(1) < f^{64}(4) < f^{64}(27)</math>
which is the given inequality.

With chained arrows, it is very easy to specify a number much greater than Graham's number, for example, <math> 3 \rightarrow 3 \rightarrow 3 \rightarrow 3 </math>.

<math> 3 \rightarrow 3 \rightarrow 3 \rightarrow 3</math>
:<math>= 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow 27 \rightarrow 2) \rightarrow 2\, </math>
:<math>= f^ {3 \rightarrow 3 \rightarrow 27 \rightarrow 2}(1) </math>
:<math>= f^{f^{27}(1)}(1)  </math>
<math display="block">
\left.
 \begin{matrix}
=   &3\underbrace{\uparrow \uparrow \cdots\cdots\cdots\cdot\cdot \uparrow}3 \\
    &3\underbrace{\uparrow \uparrow \cdots\cdots\cdots\cdot \uparrow}3 \\
    &3\underbrace{\uparrow \uparrow \cdots\cdots\cdots \uparrow}3 \\
    &\underbrace{\qquad\;\; \vdots \qquad\;\;} \\
    &3\underbrace{\uparrow \uparrow \cdots\cdot \uparrow}3 \\
    &3\uparrow 3
 \end{matrix}
\right\}
\left.
 \begin{matrix}
    3\underbrace{\uparrow \uparrow \cdots\cdots\cdots\cdot \uparrow}3 \\
    3\underbrace{\uparrow \uparrow \cdots\cdots\cdots \uparrow}3 \\
    \underbrace{\qquad\;\; \vdots \qquad\;\;} \\
    3\underbrace{\uparrow \uparrow \cdots\cdot \uparrow}3 \\
    3\uparrow 3
 \end{matrix}
\right\} \ 27
</math>
which is much greater than Graham's number, because the number <math>3 \rightarrow 3 \rightarrow 27 \rightarrow 2</math> <math>= f^{27}(1)</math> is much greater than <math>65</math>.