Numbers, Part 5: Extending Conway’s Chains

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Numbers, Part 5: Extending Conway’s Chains

5 February 2010, 12:00 am

“Numbers” Table of Contents

Part 1: Magnitude
Part 2: Exponentiation
Part 3: Hyper and Knuth Up Arrows
Part 4: Graham’s Number and Conway Arrows
Part 5: Extended Conway Arrows
Part 6: Limit Functions
Part 7: Larger Limit Functions
Part 8: Exponentiation Within Limit Functions

Essentially very small amount of Conway arrows can represent very large numbers, and the number represented get’s much larger with the addition of just one arrow.

$3 \to 3 = 3^3 = 27$

$3 \to 3 \to 3 = 3\uparrow\uparrow\uparrow3 = \underbrace{3^{3^{.^{.^{.{3}}}}}}_{3^{3^{3}}} = \underbrace{3^{3^{.^{.^{.{3}}}}}}_{7625597484987}$ or 3 cubed, 7625597484987 times.

$3 \to 3 \to 3 \to 3$ can be seen written out in Part 4, but, using Knuth up-arrows, it has an amount of arrows greater than Graham’s Number.

$3 \to 3 \to 3 \to 3 \to 3$ is indescribable in any notation we’ve discussed so far other than Conway’s. It’s equal to $3 \to 3 \to 3 \to (3 \to 3 \to 3 \to 2 \to 3) \to 2$ , which expands further to $3 \to 3 \to 3 \to (3 \to 3 \to 3 \to (3 \to 3 \to 3) \to 2) \to 2$ … It’s unfeasible to write-out this four-arrow expression as anything other than a four-arrow expression. It’s that big!

Larger with Each Arrow

Not only is the number represented getting larger with the addition of each arrow, 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!

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
$a \to_3 b \to_3 c = a \to_3 (a \to_3 b-1 \to_3 c) \to_3 c-1$

or just a formula for any level:

$a \to_X b = \underbrace{a \to_{X-1} a \to_{X-1} \cdots \to_{X-1} a}_{b \text{ right-arrows}}$

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

$(\text{Chain of any length}) \to_X 1 \to_X (\text{Another chain}) = (\text{The first chain})$

$(\text{Any Chain}) \to_X a \to_X b = (\text{That chain}) \to_X [(\text{That chain again}) \to_X b-1 \to_X c] \to_X c-1$

Next On Numbers!

Next: Limit Functions

But Wait! There's More!

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9 Comments

Hiato says:

Sunday, February 21, 2010 at 6:26 am

Heh, nice to see someone also taking a look at extending Conway chains. A while ago, Tricky and myself attempted to create our own takes on chaining. While his embedded (powerfully) conway chains in a funky new chain, I took a completely different approach to making veeeery big chains, possibly worth a look: http://echochamber.me/viewtopic.php?f=14&t=7469&sid=e36cf4ebbc8553053fec008b3fb25550&start=200#p1252045

But, in summary (for the *reduced* version which still pawns the extended conway chains tricky-style :P ):

a/b = a^b [used to be a->a->(b^a) or something]
a/b/c = a/(a/(a/ … (b/c)))) ; b/c times
a/b/c/d = a/b/(a/b/(a/b/ … (b/c/d))) ; b/c/d times
… and so on

then a//b = a/(a/(a/ .. (a/b))) ; a/b times {Special case: a//b = a/a/b}
a///b = a//(a//(a//…(a//b))) ; a//b times
and so on
with a/b//c = a/(a/ .. (b/c)); a/b/c times

I mean, 2/3/2 = 2/(2/(2/(2/(2/(2/(2/(2/(3/2)))))))) >> 2|^|^10
while 2/2/3/2 = 2/2/(2/2/ …(2/3/2) ) ; 2/3/2 times … sheesh. Then every term needs to be re-evaluated, the first collapse being 2/2/(2/3/2) = 2/2/((2|^|^9)^3) and so on
Louis Epstein says:

Wednesday, May 19, 2010 at 10:26 pm

What about Moser polygons,considered the step beyond Conway arrows at Susan Stepney’s site?…I’d like to see a good comparison between 3->3->3->3 and
3[3[3[3]]]] (i.e. 3[3[27]]])…both are way bigger than Moser’s Number but are they both bigger than Graham’s?
Peter says:

Saturday, May 22, 2010 at 12:52 am

What is the link to Susan Stepney’s site?

The Steinhaus-Moser notation in its original form at http://en.wikipedia.org/wiki/Steinhaus%E2%80%93Moser_notation generates numbers much smaller than what you can do with even simple non-extended Conway arrows.

3->3->3->3 is much bigger than a 3 in four triangles.
Louis Epstein says:

Sunday, June 20, 2010 at 8:07 pm

My linked website has a link to Susan Stepney’s.

You are misinterpreting her notation for Moser polygons…3[3[3[3]]] is NOT a 3 in 4 triangles…it is 3 in a 3-in-a-27-sided-polygon-sided-polygon!
Louis Epstein says:

Friday, June 25, 2010 at 10:13 pm

You haven’t responded to my correction,but further examination of Tim Chow’s proof indicates that 3[3[3[3]]] is less than 3->3->(3->3->49)->1 So it’s actually still less than Graham’s Number.
4[4[4[4]]] is less than 4->4->(4->4->(2(((256^256)^(256^256))^((256^256)^(256^256))-2)-1)->1 but I’m not sure how that chain would simplify!
Peter says:

Friday, June 25, 2010 at 10:20 pm

I’ll make a post for you within a few days or so. I’ve been focused on projects besides blogging recently.
Peter says:

Sunday, July 11, 2010 at 11:07 pm

I’ll make a post about it sometime before the end of the month. :/
Louis Epstein says:

Sunday, August 1, 2010 at 2:12 am

I thought you meant July…
Louis Epstein says:

Saturday, August 28, 2010 at 2:10 am

….will it be August?