so what if 2^ω is actually log(ω1)?
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so what if 2^ω is actually log(ω1)?
i have decided that im just gonna go ahead and relearn apeirology, cuz i forgor.....
wish me luck
Number Tournament: FORTY-TWO vs GOOGOLPLEX
[link to all polls]
42 (forty-two)
seed: 14 (38 nominations)
class: cultural
definition: the answer to the ultimate question of life, the universe, and everything. (it turns out the question itself is "what do you get if you multiply six by nine?")
googolplex
seed: 51 (8 nominations)
class: googologism
definition: a one, followed by a googol zeroes (where a googol is a one followed by a hundred zeroes)
which is the best number?
forty-two
googolplex
how do we know the last digit of Graham's number but not the first??? i am not a mathemagician and cannot come up with a way to calculate the last digit of anything without calculating the entire number. please explain I'm desperate
hehe really big number with 69s and 420s and such
the number in brackets is the hyperoperation. (1=add, 2=multiply, 3=exponent) 3[3]3=27 4[4]4 is made with tetration, which makes the number so big, that you can't even write it with with exponents, as 3[4]4 is some number with trillions of digits. F1=360[69420[420[1337]420]69420]360 F2=F1^F1[F1^F1[F1^F1[F1^F1[F1^F1[F1^F1]F1^F1]F1^F1]F1^F1]F1^F1]F1^F1 F3 is the same equation as F2, but replace F1 with F2. continue until F694203601337.
i like making up really big numbers
Eiper He notation (WRITTEN COMPLETELY WRONG, REGARD THE NEW VERSION)
Eiper He notation will be used in multiple equations involving the sphere/hypersphere of influence of Orb Thaija.
The notation can alternatively be written like this:
HЄ(n, x)
for platforms that do not allow super/subscript.
(OUTDATED)
What makes a number big?
What makes a number big is how dense it is compared to the information required to describe it.
What is the biggest math? Lambda calculus seems pretty big. You might need to include the definitions of the operators there to get full coverage. So if you want to use a symbol, you have to define it. Lambda calculus still looks to win here.
here's a fun concept:
imagine a function f(x) where f(x) = x↑[x↑[x↑[x...]x]x]x. where the number of nested ↑[x↑x] structures (number of up arrows = x↑x) equals x.
f(1) = 1↑[1↑1]1 (1↑...↑1, number of up arrows = 1↑1) = 1↑1 = 1
f(2) = 2↑[2↑[2↑2]2]2 = 2↑[2↑[4]2]2 = 2↑[2↑↑↑↑2]2 = 2↑...↑2, number of up arrows = 2↑↑↑↑2)
etc.