## this bigger than grahams number?

is this bigger than grahams number, i again "accidentaly" checked grahams number out and felt dumb then lost/forgot innovation for classical progressive/ progressive progressive versions of this model

when somebody takes a sum of 2^3^4^5^6^7^8^9 and then this sum will "^" itself, so much times as the sum's value is, is it at the end greater than grahams number? and if, then why?(if graham loses then because it ends on g64?)

i knew i shouldnt let myself influence myself by infesting my mind again with grahams number to "show off" that my number is bigger, now i cant think anything because im also mentally in hard times where i also forget vocabulary from foreign languages and to confirm something like this isnt what im now 100% capable of

i wont say why its concepted as it is, i only need confirmation

here is original script

 classical 2^3^4^5^6^7^8^9 = i0 (basis value number) i0^i0 = i1 i1^i1 = i2 . . . result of highest value of procedures in i0^result of highest value of procedures in i0 = . number (classical) progressive 9^9^9^9^9^9^9^9^9 = in9 (basis value number) in9^in9 = in91 in91^in91 = in92 . . . result of highest value of procedures in in9^result of highest value of procedures in in9 = number (progressive)
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 Do you mean is (2^3^4^5^6^7^8^9)^(2^3^4^5^6^7^8^9) greater than grahams number? I don't have a proof, but I know what Grahams number is and your number is so much smaller that there is no word to compare them... Grahams number is so huge, it's crazy.

 Quote by fat f... when somebody takes a > > sum < < of 2^3^4^5^6^7^8^9 and then this > > sum < < will "^" itself, so much times as the > > sum's < < value is, ...
Those are not sums.

For instance, if you mean 2^3^4 to be $$2^{3^4},$$

then you would use grouping symbols, such as in:

2^(3^4) = 2^(81).

On my graphics TI-83 Plus calculator, 2^3^4 is evaluated as 4096,

or equivalently, it treats it as (2^3)^4.

So, your expression should be typed as (or similar to):

[2^(3^(4^(5^(6^(7^(8^9))))))]^[2^(3^(4^(5^(6^(7^(8^9))))))].

## this bigger than grahams number?

I always though with exponent towers you start at the top, not the bottom, I'm rather sure that's how you work with those... Regardless, your number is less than Grahams number.

 Quote by SumThePrimes 1) I always though with exponent towers you start at the top, not the bottom, I'm rather sure that's how you work with those... 2)Regardless, your number is less than Grahams number.
1) The Order of Operations would dictate work from left to right.

The use is inconsistent. For another example,

the TI-30XIIS with the display of "2^3^2" returns "64."

Therefore, to avoid any ambiguities, use grouping
symbols as I did.

2) It's not "my" number. It's a clarification of the OP's
intended number.

----------------------------------------------------------------

I'll give you a number to beat Graham's number:

Let Tier(n) = n raised to itself n times (n tiers)

Examples:

Tier(2) = $$2^2$$

Tier(3) = 3^(3^3) $$\ = \ 3^{3^3} \ = \ 3^{27}$$

.
.
.

Look at Tier(a googleplex**) to be Graham's number

____________________________________________

** where a googol = $$10^{100}$$

And a googleplex = $$10^{a \ googol}$$
 When I said your number I was referring to the original posts number, not necessarily your clarification of it. I thought I knew the syntax with these towers, I know tetration starts at the top, it probably is best to use grouping symbols, as you say. Although I am rather sure in general you start from the top... With grouping symbols you can't go wrong, though. Also, the new number, a googleplex to the googleplex a googleplex times, I think that might be smaller, although I can't do that in my head , is there any way to show which one is larger, I don't see that it is obvious.

 Quote by SumThePrimes ... I know tetration starts at the top, ... Although I am rather sure in general you start from the top... With grouping symbols you can't go wrong, though. Also, the new number, a googleplex to the googleplex a googleplex times, I think that might be smaller, although I can't do that in my head , is there any way to show which one is larger, I don't see that it is obvious.

$$2^{3^2} **\ is \ not \ ambiguous.$$

It is equal to $$2^{(3^2)} \ = \ 2^9.$$ So having the parentheses is
redundant in this case.

But, the grouping symbols in $$(2^3)^2$$are necessary to convey $$8^2.$$

** But this contrasts to the "different animal" of expressions typed out on one line,
which are not necessarily treated the same.

------------------------------

Agreed, to me it is not "obvious," and I don't use that word in mathematics anyway.

I don't even know if it's true.

 Quote by checkitagain I'll give you a number to beat Graham's number: Let Tier(n) = n raised to itself n times (n tiers) Examples: Tier(2) = $$2^2$$ Tier(3) = 3^(3^3) $$\ = \ 3^{3^3} \ = \ 3^{27}$$ . . . Look at Tier(a googleplex**) to be Graham's number ____________________________________________ ** where a googol = $$10^{100}$$ And a googleplex = $$10^{a \ googol}$$
Tier(googolplex) is MUCH less than Graham's number. For a good description of Graham's number, look here: http://frothygirlz.com/2010/01/14/big-numbers-part-2/

To understand Graham's number, you need to understand Knuth's up-arrow notation. Basically,

a^^b = a^a^a...^a with b a's
a^^^b = a^^a^^a...^^a with b a's
a^^^^b = a^^^a^^^a...^^^a with b a's

and so on. Note that operations are taken from left to right.

So Tier(googolplex) is googolplex^^googolplex. Tier(googolplex) is much less than 3^^^4, as

3^^^2 = 3^^3 = 3^3^3 = 3^27 = 7,625,597,484,987
3^^^3 = 3^^3^^3 = 3^^7,625,597,484,987 > Tier (7.6 trillion)
3^^^4 = 3^^3^^3^^3 = 3^^3^^7,625,597,484,987 > Tier(Tier(7.6 trillion))

and since Tier(7.6 trillion) much larger than googolplex, 3^^^4 > Tier (Tier (7.6 trillion)) is much larger than Tier(googolplex)

3^^^^3 = 3^^^(3^^^3) is MUCH larger than 3^^^4, and hence MUCH larger than Tier(googolplex). And yet this is just G1, the first stage in constructing Graham's number.

G2 = 3^^^...^^^3 where there are G1 arrows.
G3 = 3^^^...^^^3 where there are G2 arrows.
...
G64 = 3^^^...^^^3 where there are G63 arrows.

Graham's number is G64.

 Quote by fat f... i wasnt happy with how grahams number looked and i didnt understand why it was as it is (ends on g64)
It ends after 64 stages because that was how many stages Graham needed to yield a sufficiently large enough number of dimensions in his proof of the Graham-Leeb-Rothschild theorem. In other words, he was actually picking the smallest number that he needed, not the largest. Still, Graham's number is far larger than most people's attempts at creating large numbers.

 and i felt to prove a number most logically as possible which would make sense, also motivation was to this is that i have personal issues with numbers, i wanted to publish it long ago. im numerologist not mathematician thus i use maybe wrong terminology http://youtube.com/watch?v=rC_po1PMRWA here (video is aimed at average audience). there is also a text form of it if u dont want to watch the video, the problem to be solved is at the end of it.
It's not the biggest number ever, as you can just add one a create a bigger number. In fact, it's MUCH smaller than Graham's number. I don't know what you mean by the largest "logical" number.

 http://simple.wikipedia.org/wiki/Graham%27s_number and this version is different from http://en.wikipedia.org/wiki/Graham%27s_number which one is actually right?
Unsurprisingly, the regular wikipedia version is the accurate one, and the more thorough one at that. The simple wikipedia version has several errors, but the basic description is the same.

 but anyways, is classical-infinity at least bigger than graham's?
Nope. Recall the definition of ^^ and ^^^ from my previous post:

a^^b = a^a^a...^a with b a's
a^^^b = a^^a^^a...^^a with b a's

So i0 = 2^3^4^5^6^7^8^9 < 9^9^9^9^9^9^9^9^9 = 9^^9. (In fact, i0 < 9^(9^9) if you do the operations in calculator order.)

You then take i0 to the power of itself i0 times. This is similar to taking an exponential tower of i0 i0's; to be precise, classical-infinity will be between i0^^(i0+1) and i0^^(i0+2). So we have

classical-infinity < 9^^(9^^9) = 9^^^3.

9^^^3 is less than 3^^^4, which is MUCH less than 3^^^^3 = 3^^^(3^^^3). This last number is just G1, the first stage in constructing Graham's number.

So classical infinity is much less than G1, to say nothing of G64.

 he was actually picking the smallest number that he needed
it had something to do with hypercube or something, therefore it was a 3

 It's not the biggest number ever, as you can just add one a create a bigger number. In fact, it's MUCH smaller than Graham's number. I don't know what you mean by the largest "logical" number.
i said there also that it isnt the biggest number,but its the biggest number which made sense until now, in the sense of trying to find out the biggest number among numbers from 1-9

i started to doubt the result already while i tried to compare my number with g64, thus i got a problem to which i havent found answer yet because its about finding a perfect formula in the mysteries system of numbers which will then answer and without doubt will be the biggest number which will be used as an evidence and will make sense(in same sense as that number which i invented or discovered), its just that i dont know where to start finding the formula, its a precise formula which makes sense that i didnt discover yet, i know its mostly about 1 number because other numbers arent mysterious as the highest number in decimal system(it would cause problems if other systems will be considered but decimal system is more accepted as standard)

i couldnt let it stand that some retard made out of necessity the biggest number which i met although he doesnt respect numbers as i, i consider numbers mysteriously, mostly the 9(or only the 9) because i found out by myself the nature of this number, due to its nature it became my favorite number and the most special number out of all(except that 1 has also special nature to it), i consider numbers as something like tools of god, i wouldnt stand it if people would consider g64 as biggest number forever or that graham sometime just made this number and its considered until now by all people as the number, i could beat this number just by changing 3 to 9(and it would be legitimate in some sense) but it would be cheap and wouldnt prove anything as i also could 3 to value of 9^9^9^9^9^9^9^9^9, it has no reason which number there should be, i search for this mysterious formula which would make perfect sense, he had knowledge but he didnt understand numbers and he had time to study it all, i have no knowledge and im not mathematician, im numerologist and graham isnt in my category and im not in his(because i neither understand the problem for which he invented this number, but it looks like it was unimportant boring problem(not that it would be worldshaking the solution))

is 3^^^^3 = 7,625,597,484,987^7,625,597,484,987 ? or result of 3^7,625,597,484,987 ^ 3^7,625,597,484,987 ?

and why did he stop at 3^^^^3 and didnt made 3^^^^^3?

 this is actually ******** whats highlighted in magenta as i now found out, dont know why this 8 is stuck there(probably because 1 cant be exponentiated thus only 8 digits remain = 8 or 2+3+4+5+6+7+8+9=8, also needs to be solved)
 I'm new to the Forum and have always been interested in large numbers. I recently became aware of Graham's number and have seen various descriptions of it so I have a cloudy appreciation of how truly large it is (I emphasize 'cloudy'). I also recently came upon the website http://users.skynet.be/nizgorur/very_big_numbers.htm where it describes very large numbers with formal names attached (not sure of the naming validity but they seem legitimate). Near the bottom, it describes "fugagargantugoogolplex" as: gargantugoogolplex raised to the gargantugoogolplex power a gargantugoogolplex times. Question... assuming Graham's number is larger (?) than fugagargantugoogolplex, can anyone state at approximately which number (Gsub3, Gsub10, or whatever point along that path to Gsub64) Graham's number EXCEEDS fugagargantugoogolplex? I hope I've stated my question correctly, and I'm hoping to put both numbers in relative context with eachother in order to better appreciate both. Thank you!

A gargantugoogolplex is googolplexplexplex, or 10^10^10^10^100. (Apparently, there is some question about this, according to this website, which cites the original source and suggests that it should be smaller. But we will assume the larger value.)

fuga(n) = (...((n^n)^n)...^n)^n with n n's = n^(n^(n-1)) < n^(n^n).

So fugagargantugoogolplex < gargantugoogolplex ^ (gargantugoogolplex ^ gargantugoogolplex)
= 10^10^10^10^100^(10^10^10^10^100^(10^10^10^10^100))
= 10^10^10^10^100^(10^10^(10^10^100 + 10^10^10^100))
= 10^10^(10^10^100 + 10^(10^10^100 + 10^10^10^100))
< 10^10^10^(1 + 10^10^100 + 10^10^10^100))
< 10^10^10^10^10^10^101
< 3^3^3^3^3^3^3^3^3
= 3^^9,

where a^^b = a^(a^(a^...(a^a)...)) with b a's. See Knuth's_up-arrow_notation.

3^^^3 = 3^^(3^^3) = 3^^7625597484987, which is MUCH more than 3^^9, and therefore much more than fugagargantugoogolplex. And 3^^^3 is nothing compared to G(1), which is 3^^^^3. So G(1) is already much bigger than fugagargantugoogolplex.

@fat f:
 is 3^^^^3 = 7,625,597,484,987^7,625,597,484,987 ? or result of 3^7,625,597,484,987 ^ 3^7,625,597,484,987 ? and why did he stop at 3^^^^3 and didnt made 3^^^^^3?
If you read this, sorry for not responding to you. Neither of those numbers are anywhere near 3^^^^3. (This should be clear, as I showed previously that 3^^^^3 was greater than your classical-infinity and progressive-infinity numbers.) 3^7,625,597,484,987 ^ 3^7,625,597,484,987 = (3^3^3^3)^(3^3^3^3) < 3^3^3^3^3^3 = 3^^6, and 3^^^3 is already 3^^7,625,597,484,987, which is 3^3^3...3^3 with 7,625,597,484,987 3's. 3^^^^3 is much bigger, too big to be described as an exponential tower. The most we can simplify it is

3^^^^3 = 3^^^(3^^^3) = 3^^^(3^^7,625,597,484,987)
= 3^^3^^3^^3...3^^3 with 3^^7,625,597,484,987 3's

where each "3^^" means "take the previous number n and create an exponential tower with n 3's".

As to why Graham used 3^^^^3 instead of 3^^^^^3, he simply when as far as he needed to create an upper bound. Note that he doesn't stop at 3^^^^3; that is just G(1). He goes on to G(2), which is 3^^^...^^^3 with 3^^^^3 ^'s, and so on up to G(64).

I don't understand what you are saying with you "don't know why 8 is stuck in there".

 Quote by Need To Know Near the bottom, it describes "fugagargantugoogolplex" as: gargantugoogolplex raised to the gargantugoogolplex power a gargantugoogolplex times.
Well it would be nice if he actually defined everything properly.
According to the Googology Wiki, a gargantugoogolplex is $$10^{10^{10^{10^100}}}$$
So this is less than $10^^5$ and our fugagargantugoogolplex is less than
$(10^^5)^^(10^^5) < 10^^10^^10^^10 = 10^^^4 < 3^^^^3$
which is G1.

I hope I got that right.
 Thanks Deedlit and pwsnafu for your great replies! My appreciation for how truly large Graham's number is has now become (a bit) less cloudy. Also, thanks Deedlit for the link. That really helped to explain it in a very informative and entertaining way. Thanks again..