Talk:Graham's number

Digits

I rather like the listing of the last ten digits; it makes the number seem more concrete somehow. Are there any similar tricks we could do to get the first 10 digits? Yours hopefully, Agentsoo 09:34, 29 July 2005 (UTC)

Not by any method I know of, but then again, I don't know much. A method to calculate the first ten digits may exist, but if it did, it would certainly be a lot more complicated. Calculating the last digits of the number is quite simple (as the article says, elementary number theory) using modular arithmetic. The powers of the number can go higher and higher but the final digits will cycle, so it is possible to work out. Calculating the first digits would be much more difficult, if possible at all. -- Daverocks 10:12, 8 August 2005 (UTC)

Number of digits

How many digits does this large number have?? 66.32.246.115

This large number has a large number of digits. ;) Okay, take the common logarithm of Graham's number, round it down to the nearest whole number, and then add 1. That's the number of digits. I'd write it out in full for you, but there aren't enough particles in the known universe... :) -- Oliver P. 00:26, 9 Mar 2004 (UTC)

So how about the number of digits of THAT number (namely, the number you are saying you'd write down)??

How many times would you have to take the logarithm of the number acquired in the previous step (starting with Graham's number) before you get a number that can be written using the particles in the known universe? :) Fredrik 00:37, 9 Mar 2004 (UTC)

Even the number of times you take the logarithm, to get a writeable numbers, is too large to write. That is, you don't just take a logarithm once, or twice, or even a million times; if you took the logarithm $2^{10^{80}}$ times (about as many as you could write with all the particles in the universe) you still wouldn't be even in the ball park of being able to write out the result - even if you had any particles left over to do it 8^) In fact, even the number of times you had to take the logarithm, is so large that the number of times you had to take it's logarithm, is too large to write - and so on, and so on, 65 layers deep... Securiger 16:49, 3 May 2004 (UTC)

I don't see any way to visualise the magnitude of the number.... my only conclusion is 'whoa this is large'. Can't we find some kind of magnitude comparison to help out to fathom G? — Sverdrup 00:42, 9 Mar 2004 (UTC)

Well, if you'll allow me to define numbers in O notation, then Graham's number is O(A⁶⁴). This grows faster than exponentiation, iterated exponentiation, iterated iteration of exponentiation, etc. So Graham's number simply cannot be written this way, or any related way. Conway's chained arrow notation, however, is sufficient to write Graham's number. --67.172.99.160 02:35, 19 March 2006 (UTC)

An amusing exercise for the student

Obviously Graham's number is divisable by three, and hence not prime. And since all powers of three are odd, obviously Graham's number minus one is even, and hence not prime. Question: is Graham's number minus two prime? Prove your result. Securiger 16:49, 3 May 2004 (UTC)

• Securiger, I know enough to know that Graham's number minus 2 is not prime. Given that it is a number of the form 3^3^3^3^3^3^3^3... with at least 2 3's, its final digit is a 7. A number ending in 7 will end in 5 if you subtract 2, and apart from the single digit 5, a number ending in 5 is not prime because it has 5 as a factor. Therefore, Graham's number minus 2 is not prime. And, Graham's number minus 3, which is even, is not prime. Georgia guy 22:26, 8 Mar 2005 (UTC)

Wow

I tried figuring out just how big this number is, and my mind blew a gasket when it got to the 4th Knuth iteration. Yikes. I'm glad there's only 1e87 particles in the universe, otherwise some wacko would already be trying to write it out. --Golbez 06:53, Sep 3, 2004 (UTC)

Operation needed for Graham's number

Recall from the above the Graham's number is to large to visualize simply with the help of tetration. Try the question "What operation is needed to visualize it??" Please highlight it in the list below, defining each operation as being related to the previous one the way multiplication is related to addition.

1. Addition
2. Multiplication
3. Exponentiation
4. Tetration
5. Pentation
6. Hexation
7. Heptation
8. Oktation
9. Enneation
10. Dekation
11. Hendekation or above

66.245.74.65 23:51, 30 Oct 2004 (UTC)

None of these are even close to sufficient. Regular exponentiation requires 1 Knuth arrow to be represented, and dekation requires 8. Dekation in itself is a very powerful operator that produces large numbers. But for every layer of the 65 layers that Graham's number goes further into, what happens is that the number of arrows is equal to the result of the previous step. $3\uparrow \uparrow 3=3^{3^{3}}=3^{27}=7625597484987$, so we have 7625597484987 arrows already. Which is much, much larger than dekation. The next step will have $3^{7625597484987}$ arrows, a gargantuan number. This is just the number of arrows. And we're doing this 65 times. -- Daverocks 07:42, 19 August 2005 (UTC)

Sorry, I have to correct a fact in my own statement here. You start with $3\uparrow \uparrow \uparrow \uparrow 3$, and this is the number of arrows required for the next step, and so on and so on (one does not start with $3\uparrow \uparrow 3$ as I stated above). -- Daverocks 12:12, 30 August 2005 (UTC)

Comparison

Is it bigger than googol? Googolplex?

Yes. It is even bigger than googolduplex, googoltriplex, and even googolcentuplex, defining googol(n)-plex as the number 10^10^10^10^...10^10^10^10^googol where there are a total of n 10's. 66.245.110.114 23:55, 6 Dec 2004 (UTC)

Adding to that, one can easily find numbers larger than a googolplex, even though it is also unconceiveably large. However, $9^{9^{9^{9^{9^{9}}}}}$ is larger, and can be written with fewer symbols. Remember that a googolplex is $10^{10^{100}}$, which requires 7 symbols to be written down. The above-mentioned number only requires 6. -- Daverocks 10:30, 8 August 2005 (UTC)

BTW, isn't $9^{9^{9^{9}}}$ already larger than $10^{10^{100}}$ ? --FvdP 20:56, 19 August 2005 (UTC)

Actually, FvdP, you're probably right. -- Daverocks 12:05, 30 August 2005 (UTC)

Definately right in fact; stacking new power layers is much more powerful than incresing the numbers of lower layers. (which is why the procedure for G gets so very large very fast.) For example, $9^{9^{9^{9}}}=9^{9^{387,420,489}}$ which is clearly larger than $10^{10^{100}}$

Still informally but with more precision, here is another way to see it, by taking twice the logarithm on each side: $\log 9^{9^{9^{9}}}=9^{9^{9}}\log 9$ so $\log \log 9^{9^{9^{9}}}=9^{9}\log 9+\log \log 9\approx 9^{9}\log 9$, while similarly, $\log \log 10^{10^{100}}\approx 100\log 10$. There is (almost) no more dispute which is greater ;-) You see here how the actual numbers in the lower exponentiation layers just seem to evaporate and play no significant role in the proof. --FvdP 22:08, 23 December 2005 (UTC)

notation

Are you saying that even if you used the notation like 10^10^100^100000000 etc, there isnt enough ink or hdd space in the universe to express graham's #? 69.142.21.24 05:20, 5 September 2005 (UTC)

That is correct. Graham's number is far too large to simply be expressed in powers of powers of powers. Even if one takes the exponentiation operator to a higher level, such as tetration or many (actually, an incomprehensible number of) hyper operators above this, it would still not be even close to sufficient to expressing Graham's number with all the matter in the known universe. -- Daverocks 12:32, 19 September 2005 (UTC)

Largest number with an article

I assume from the above discussion that this is the largest number to have a serious Wikipedia article devoted to it. Is that correct? CDThieme 03:02, 8 October 2005 (UTC)

Well, the greatest integer. We have plenty of articles about infinite numbers, and Graham's number is finite. Factitious 19:51, 10 October 2005 (UTC)

Can numbers be infinite? I don't think so, a number is a number. Infinite is a concept. Rbarreira 19:24, 20 December 2005 (UTC)

See Aleph-null, for instance. Factitious 22:32, 21 December 2005 (UTC)

Years in Hell

Imagine being in Hell for a Graham's Number of years. (It's said that people stay in hell for all eternity...) If someone was taken out of Hell after that many years, what could they say and how would they feel? Think about and imagine this. --Shultz 00:15, 2 January 2006 (UTC)

Nonsense. The universe is only 13,700,000,000 years old, a number that can be written in comma format with only 4 periods of digits. Graham's number is far too large to write in comma form with even millions of such periods. Georgia guy 01:10, 2 January 2006 (UTC)

Letter

Since this is obviously too large a number to write down, is there a letter to represent it in common usage?

Upper bound of something

You know with this nummer being so large, how on earth was it proven to be the upper bound of that question?