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MTGAP
Daedalian Member

 Posted: Fri Oct 10, 2008 2:31 am    Post subject: 1 I have been looking at [url="http://en.wikipedia.org/wiki/Conway_chained_arrow_notation"]Chained Arrow Notation[/url], and something about it seems strange. It seems like 2 -> 2 => n will always be 4. (I have no proof, it just seems that way.) 2 -> 2 -> -1 (addition) = 4. 2 -> 2 -> 0 (2*2), 2-> 2-> 1 (2^2), 2-> 2-> 2 (2^2^2...^2^2 with two 2's, which is 2^2), etc. But according to chained arrow notation, 2-> 2-> 2 is 256, and 2-> 2-> 3 is unimaginably huge. It's 4->(4->(4->....->4)->4)->4, where there are 65536 4's on each side. Why is that? Shouldn't it be 4? That would seem more "natural". It seems like there are 2 general rules about that sort of thing: 1 [operation] 1 is always 1, unless it's 1 + 1. And 2 [operation] 2 is always 4, unless it's no operation. The first one works in chained arrow notation, but the second one doesn't, and it seems like it should. Why doesn't it work? Maybe I'm calculating it wrong..._________________This statement is false.
MTGAP
Daedalian Member

 Posted: Fri Oct 10, 2008 2:45 am    Post subject: 2 Aha! On the Large Numbers page of Wikipedia, I found this, which is in contradiction of my (probably false) calculations: "( 2 → 2 → Y ) = 4 for any subchain Y" I probably calculated incorrectly. Here's how I calculated 2-> 2-> 2: 2->2->2 = 2^2 -> 2^2 -> 1 - (2^2)^(2^2) = 4^4 = 256_________________This statement is false.
Bicho the Inhaler
Daedalian Member

 Posted: Fri Oct 10, 2008 4:47 am    Post subject: 3 Not sure what you did there... 2 -> 2 -> 2 2 -> 2 -> 1 2 -> 2 2 ^ 2 4
MTGAP
Daedalian Member

Posted: Fri Oct 10, 2008 5:41 pm    Post subject: 4

 Bicho the Inhaler wrote: Not sure what you did there... 2 -> 2 -> 2 2 -> 2 -> 1 2 -> 2 2 ^ 2 4

I'm not sure either. I did it again, and got 4.
_________________
This statement is false.
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