Talk:Conway chained arrow notation

As it stands, this page is gibberish. I have no doubt that the equations are "correct" but they need explanation. Are the first 3 equations alternative schemes or are they all a single scheme defining one large number. -- SGBailey 22:07, 2003 Aug 27 (UTC)

They're supposed to make up a recursive definition. They're not correct, though. They don't cover the case with more than two arrows. Matthew Woodcraft

I understand Knuths up-arrow notation which is explained. This "X>p>1=X>p" and "p>q=p^q" (using > for right arrow and ^ for exponent) explains nothing. I am inclined to empty the article unless someone can explain why I shouldn't. -- SGBailey 09:57, 2003 Aug 28 (UTC)

Fixed up the first part, however I don't get the chaining of 3 arrows as below. Dysprosia 10:06, 28 Aug 2003 (UTC)

Ach. Didn't see this page before. I sorta expected email. The eqns were correct, but although i specified, i guess i didn't make clear what X, p, and q meant. Okay, I tried working three 4-element chains, two fizzles and one probably explodes. You try it! =) Kwantus

Microp&aeling;dia

(Some things which don't belong in a main page IMO but are useful to understanding.--Kwantus)

Every chain is equivalent to a shorter chain with the same head. That is, with any subchains X and Y, X→Y=X→n for some (usu very large) number n dependent in a nontrivial way on X and Y. This is a simple consequence of rules 1 & 2.
A chain beginning with two 2s stands for 4. By rules 3 and 2, 4=2→2=2→2→1. By rule 1, 2→2→(n+1)=2→2→n. By induction, 4=2→2→n for all n. By the result above, all chains 2→2→Y=2→2→n for some n.
A chain can be truncated ahead of any 1. By the result above, any chain X→1→Y=X→1→n for some n, and rule 1 instantly reduces that to X.