## Large Numbers, Part 2: Graham and Conway

Direct continuation of: Large Numbers, Part 1: Magnitude and Simple Functions

When we last left off, we had discussed exponentiation and Knuth up-arrows. Remember that Knuth up-arrows looked like this:
$a\uparrow^n b= \left\{ \begin{matrix} a^b, & \mbox{if }n=1; \\ 1, & \mbox{if }b=0; \\ a\uparrow^{n-1}(a\uparrow^n(b-1)), & \mbox{otherwise} \end{matrix} \right.$

Now let’s move on to something a little bit bigger: Graham’s Number, perhaps the most famous “Large Number”, even though it’s quite small compared to what we will do by the end of this post. The number actually means something and isn’t just made “for the fun of it”, but what it means is something rather complicated that I don’t personally understand. That’s what Wikipedia is for, anyway. (It’s the 1980 Guiness Book of World Records holder for largest number.)

So… we currently have a notation that can describe the amount of arrows to be used. What if we used arrows to describe the amount of arrows there are? What if we did this?

$a\uparrow^{c \uparrow^e d} b$

This is how we get to Graham’s Number. Graham’s number is found by continuing a pattern:
$g_1 = 3 \uparrow^{4} 3 = 3 \uparrow \uparrow \uparrow \uparrow 3$

$g_2 = 3 \uparrow^{g_1} 3 = \underbrace{3 \uparrow \cdots \uparrow 3}_{3 \uparrow \uparrow \uparrow \uparrow 3 \text{ arrows}}$

$g_3 = 3 \uparrow^{g_2} 3 = \underbrace{3 \uparrow \cdots \uparrow 3}_{\underbrace{3 \uparrow \cdots \uparrow 3}_{3 \uparrow \uparrow \uparrow \uparrow 3}}$

And eventually we get to Graham’s Number, also written as G, after 64 steps:
$\left. \begin{matrix} G &=& g_{64} &=& 3\underbrace{\uparrow \uparrow \cdots\cdots\cdots\cdots\cdots \uparrow}3 \\ & & & &3\underbrace{\uparrow \uparrow \cdots\cdots\cdots\cdots \uparrow}3 \\ & & & &\underbrace{\qquad\;\; \vdots \qquad\;\;} \\ & & & &3\underbrace{\uparrow \uparrow \cdots\cdot\cdot \uparrow}3 \\ & & & &3\uparrow \uparrow \uparrow \uparrow3 \end{matrix} \right \} \text{64 layers}$

## Graham’s Number in Hyper Notation

Graham’s number can also be represented by the Hyper Function. We know that the c of the Hyper Function is the same as the amount of arrows – 2, so:
g1 = hyper(3, 3, 6)
g2 = hyper(3, 3, g1+2) = hyper(3, 3, hyper(3, 3, 6)+2)
g3 = hyper(3, 3, g2+2) = hyper(3, 3, hyper(3, 3, g1+2)+2) = hyper(3, 3, hyper(3, 3, hyper(3, 3, 6)+2)+2)

$G = g_{64} = \underbrace{hyper(3, 3, hyper(3, 3, hyper( \cdots hyper(3, 3, 6)))))}_{64 \text{ nested functions}}$

## Conway Chained Arrow Notation

So we now can make larger numbers by using arrows to define the number of arrows, and then using arrows to define the number of arrows of the number arrows, etc.

John Conway summarized this in\to a single notation and then made the ability \to go even larger.

Conway’s notation uses “chains” of arrows that have a finite amount of left-arrows, looking like this: a → b → c → d → e

A single left-arrow is the same as a single exponent: a→b = a^b

Two left-arrows equal Knuth’s notation:
$a \to b \to c = a\uparrow^c b$

Three left-arrows then blow our current systems out of the water. In order \to understand a→b→c→d, we have to use the formal definition of Conway’s chained-arrow notation. This notation uses the following two rules:

1. {A chain of any length} → 1 → {Another chain} = {The first chain}
2. {A chain of any finite length} → y → z = {That chain} → [{That chain} → (y-1) → z] → z-1

#### Examples

6→9→3→8→1→4→5→3→3 = 6→9→3→8

5→5→5→5→5 = 5→5→5→(5→5→5→4→5)→4

3→3→2 = 3→(3→2→2)→1 = 3→(3→2→2) = 3→(3→(3→1→2)→1) = 3→(3→3) = 3→27 = 3^27 = 7625597484987

3→2→2→2 = 3→2→(3→2) = 3→3→9 = 3↑↑↑↑↑↑↑↑↑3

## Larger with Each Arrow

Not only is the number represented getting larger with the addition of each arrow (remember not to confuse conway right arrows with knuth up arrows), it’s getting larger at a much increasing magnitude. The gulf between 3 (0 arrows) and 27 (1 arrow) is much different between 27 (1 arrow) and 3 cubed, 7625597484987 times (2 arrows).

And of course the gulf between 2 arrows and 3 arrows is even larger! The difference between something representable by three Knuth up-arrows and something representable only by arrows larger than Graham’s Number.

Now consider the difference between 3 arrows and 4 arrows. It’s unimaginable by any other standard. Then we can keep going to 5 arrows, 6 arrows, 7 arrows… etc. Any number larger than three arrows is too large to be of any mathematical use, but in the FUN TIMES of making large numbers, we need more!

What if we defined the amount of arrows with Conway notation?

### Comparing Graham’s Number to Four Entries

Graham’s Number evaluates very much like a form of $a \to b \to c \to 2$, but no value of a can be both the 3 needed to be in front and at the end of the arrows, and the 4 for the initial amount of arrows. Thus that places Graham’s number between two Conway chains:

$\left. \underbrace{3 \uparrow \cdots \uparrow 3}_{ \underbrace{3 \uparrow \cdots \uparrow 3}_{ \underbrace{\vdots}_{\underbrace{3 \uparrow \cdots \uparrow 3}_{3}}}} \right\} 64 < G < \left. \underbrace{3 \uparrow \cdots \uparrow 3}_{ \underbrace{3 \uparrow \cdots \uparrow 3}_{ \underbrace{\vdots}_{\underbrace{3 \uparrow \cdots \uparrow 3}_{3}}}} \right\} 65$

Which means, that $3 \to 3 \to 64 \to 2 < G < 3 \to 3 \to 65 \to 2$

Note that $3 \to 3 \to 3 \to 3 = 3 \to 3 \to (3 \to 3 \to 2 \to 3) \to 2$, which is much larger than G, and $3 \to 3 \to 3 \to 4$ is much larger than even ggG.

## n Conway Right-Arrows

So we can then define any amount of Conway arrows with a new notation.

$a \to_2 b = \underbrace{a \to a \to \cdots \to a}_{b \text{ right-arrows}}$

And then we can apply the same definitions and rules of the old arrows \to the new arrows.

$a \to_2 b \to_2 c = a \to_2 (a \to_2 b-1 \to_2 c) \to_2 c-1$

For example, $3 \to_2 2 \to_2 2 = 3 \to_2 (3 \to_2 1 \to_2 2) \to_2 1$
$3 \to_2 (3 \to_2 1 \to_2 2) \to_2 1 = 3 \to_2 3 = 3 \to 3 \to 3$

That didn’t produce anything out of the ordinary, so here’s a larger example. …Just by increasing the last number by 1:

$3 \to_2 2 \to_2 3 = 3 \to_2 (3 \to_2 1 \to_2 3) \to_2 2$
$= 3 \to_2 3 \to_2 2 = 3 \to_2 (3 \to_2 2 \to_2 2) \to_2 1$
$= 3 \to_2 (3 \to_2 2 \to_2 2) = 3 \to_2 (3 \to_2 (3 \to_2 1 \to_2 2) \to_2 1)$
$= 3 \to_2 (3 \to_2 3) = 3 \to_2 27$
$= \underbrace{3 \to 3 \to \cdots \to 3}_{27 \text{ right-arrows}}$

We can also then use even larger second level chains to get even larger numbers — numbers so large that the can’t be expressed with regular Conway arrows.

$3 \to_2 3 \to_2 3 \to_2 3 = 3 \to_2 3 \to_2 (3 \to_2 3 \to_2 3 \to_2 2) \to_2 2$
etc.

## But Wait! There’s More!

We can then make even larger levels.

$a \to_3 b = \underbrace{a \to_2 a \to_2 \cdots \to_2 a}_{b \text{ right-arrows}}$
and
$a \to_3 b \to_3 c = a \to_3 (a \to_3 b-1 \to_3 c) \to_3 c-1$

## Any level of Conway

We can actually add a new rule to the previous two, and make this work for any subscript:

1. a →X b = a →X-1 a →X-1 a →…→X-1 a (b right-arrows)
2. {A chain of any length} →X 1 →X {Another chain} = {The first chain}
3. {A chain of any finite length} →X y →X z = {That chain} →X [{That chain} →X (y-1) →X z] →X z-1

#### And Some Examples

6→29→23→28→21→24→25→23→23 = 6→29→23→28

5→25→25→25→25 = 5→25→25→2(5→25→25→24→25)→24

3→23→22 = 3→2(3→22→22)→21 = 3→2(3→22→22) = 3→2(3→2(3→21→22)→21) = 3→2(3→23) = 3→2(3→3→3→3) = 3→3→3→…→3 (with 3→3→3→3 arrows)

## Extending the Extensions

You can probably see where this is going. If we extended the Knuth up-arrows by defining the number of up-arrows in up-arrows, and defining the number of Conway right-arrows with Conway right-arrows, then we can define the subscripts with subscripts.

C(a) = a→aa

Then we can build off of that:

C(a, 1) = a→C(a)a
C(a, b) = x→C(a, b-1)a

And then build off of that with three arguments:

C(a, 1, 1) = C(a, C(a, a))
C(a, b, 1) = C(a, C(a, b-1, 1))
C(a, 1, c) = C(a, C(a, a, c-1), c-1)
C(a, b, c) = C(a, C(a, b-1, c), c-1)

That’s probably good enough. In the next part, we’ll introduce what this ordinal thing is all about.

Continued in: Large Numbers, Part 3: Functions and Ordinals

-

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On 9 May 2011 in All, Mathematics. 7 Comments.

1. #1 Deedlit says:

One pedanatic correction:

g_2 = hyper (3,3,g_1+2)
g_3 = hyper (3,3,g_2+2)
etc.

(You’re missing the +2′s)

2. #2 Peter Hurford (author) says:

Fixed! Thanks!

3. #3 Louis Epstein says:

Also,in defining g sub 1,you are including g sub 4 instead of 4 for the number of arrows between the threes.

So Moser rates no mention?

4. #4 Peter Hurford (author) says:

Also,in defining g sub 1,you are including g sub 4 instead of 4 for the number of arrows between the threes.

Fixed, thanks.

So Moser rates no mention?

I do like Moser, but I found it a bit complicated to explain and a tad counter-intuitive, a bit hard to convert into a function, and not any larger than the other notations.

5. #5 Luiz says:

Any level of Conway

And Some Examples
part

…→(3→{2}3) not is …→(27)
…→(3→{2}3) is equal …→(3→3→3→3)

6. #6 Luiz says:

3→23→22 = … = 3→2(3→3→3→3)

unimaginable right arrows with 3

7. #7 Peter Hurford (author) says:

Fixed. Thanks! Let me know if there are any other errors.