Talk:Graham's number

Too technical?

Pending some discussion of the issues here, I've removed the recently-added tag that said the article is too technical for most readers to understand. When the tag was added, the edit-comment asked "how can lay readers understand how large graham's number is?". Here are some points:

• The focus of the present article is "Graham's number" G, as originally published by Gardner in terms of Knuth's up-arrow notation. Any difficulties in understanding this notation should be resolved at the linked article, rather than addressing them in this one. I think the section Magnitude of Graham's number that's already in the article is not "too technical" for anyone who understands the up-arrow notation.
• The focus of the present article is not "Graham's problem", whose solution is bounded above by G. Although some discussion of this problem is needed for the historical context of Graham's number, Graham's problem involves a degree of technical detail that's totally irrelevant to understanding the definition of G. (Previous attempts to "simplify" the statement of Graham's problem have in my opinion backfired.)
• Graham's number G is regarded as incomprehensibly large, and no amount or kind of explanation is likely to overcome this. Additional attempts to convey how large it is will not make the incomprehensible comprehensible.
  — r.e.s. (talk) 17:49, 2 March 2011 (UTC)

Very good points. I agree. This article seems to be as non-technical as possible given that the subject is a number so large, technical notations are necessary to define it. Cheers, — sligocki (talk) 01:31, 5 March 2011 (UTC)

Hi everyone. I put up the original "too technical" tag. I understand the idea of an article being as "non-technical" as it can be. But I would encourage editors here to think harder about it. It's not really as non-technical as it can be. Rather, it's as non-technical as you're willing to make it. Now, don't get me wrong, there is real value in describing a topic in technical detail. Furthermore, I know that many topics can only really be described adequately if technical language is used. However, if the lay reader is not really understanding the topic to even a reasonable degree, then I think we ought to try harder.

As with many mathematics articles on Wikipedia, this article is difficult, if not impossible to truly understand without a degree in mathematics. Almost all of the information below the lede is beyond the "understandibility" of a lay person. One needs to know perhaps obscure (to the lay person at least) mathematical notation and concepts. And the lede itself tells the lay person only that Graham's number is unimaginably large. I understand the pitfalls of describing something that is so large that it can never even be written out. I also understand the pitfalls of explaining something that can only be described in a circumspect manner -- such as an unimaginably large number. But I think that lay readers still deserve a better basic explanation than they are receiving here.

These are just some thoughts. I'd love to hear the thoughts of some of you who have contributed to and are invested in this article. Cheers, ask123 (talk) 20:27, 5 March 2011 (UTC)

Also, in response to the previous informative post by r.e.s., the section Magnitude of Graham's number is definitely too technical for the lay reader to understand without outside references. The mathematical notation used in that section does not magically become digestible to most readers simply because the concept of Graham's Number is rather incomprehensible to begin with. I get it that the number is incomprehensibly large. All I'm saying is that 4/5ths of this article is made up of jargon and notation that most people don't get. Cheers, ask123 (talk) 20:35, 5 March 2011 (UTC)

You make it sound like there is a lot of jargon, but the only jargon is Knuth's up-arrow notation. And the up-arrow notation is simply enough defined:

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

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

a^^^^b = a^^^(a^^^(a^^^(...(a^^^a)...))) with b a's,

and so on.

I'm not sure what there is to understand. How would you make the above definitions more acessible to a lay audience? Deedlit11 (talk) 10:58, 13 April 2011 (UTC)

First, I disagree that the up-arrow notation is simply enough defined. It is defined through the writing out of mathematical notation, which is not adequate for most readers. Second, jargon was the wrong word -- the wrong way to describe it. I mean mathematical notation and jargon. Most people cannot readily understand the mathematics in that section. I know there may be no other way to express it, but, given that, at least provide adequate linkage (as has been done with the word "tetration") to other wiki articles that will help readers get through it. And, also, it would be helpful to offer more definitional explanations -- be more verbose so that readers can understand through language. Again, the word tetration is a good example of a term that can quickly be defined in a brief appositive. ask123 (talk) 19:27, 18 June 2011 (UTC)

I can't comprehend why a layperson would feel the need to understand this number. Outside of the field of pure math, there really is no value to this information. Nobody will ever need to use it to measure anything. I would imagine that studying and understanding the relevant information (which appears to be adequately Wikilinked in the article) is the only way to appreciate the topic. I don't want to tell anyone what they can and can't learn, but if you don't have a math-inclined mind, you really have no reason to be reading this. Some things just inherently can only be dumbed down so far. InedibleHulk (talk) 04:36, 18 June 2011 (UTC)

Your point is taken, Hulk, but wikipedia is not designed as an advanced encyclopedia. It is meant as an encyclopedia for all. Ideally, there is not supposed to be any topic herein that is beyond the understandability of the average reader. The reader doesn't need to become an expert, but at least understand the topic adequately. If you are interested in advanced and technical encyclopedias, there are other wikis for that. ask123 (talk) 19:19, 18 June 2011 (UTC)

There are a lot of wikipedia math and physics articles that are full of notations and are extremely useful for those studying math and physics. If you are not interested in subjects like this in general then you shouldn't be reading this article. Besides, if you really want to understand how big this number is, image that number of particles in the universe, (10^80), for each particle, put a universe inside, now you have a lot more particles. For each of the particles inside the universe that's in another particle, put another universe in there. Repeat the process the number of times that equals the number of particles in the universe. Then repeat this repetition the number of times that equals the number of times in the universe. And even this construction wouldn't be large enough to compare to Graham's number. Can you think of way to simplify this description? — Preceding unsigned comment added by 99.47.69.214 (talk) 11:59, 4 December 2011 (UTC)

The underlying reason why any "ordinary" process of comparison is bound to fall short (e.g., repeatedly replacing particles by "ordinary" universes of particles an "ordinary number" of times, then repeating all of that a similar number of times), is that it repeats at each stage no faster-growing operations than those of ordinary arithmetic (add, multiply, exponentiate). In contrast, Graham's number involves two levels of "acceleration into the large": (1) The ordinary operations of arithmetic (+, *, ↑) are extended into a sequence of ever-more-rapidly-growing operations (using Knuth up-arrows: 3↑n, 3↑↑n, 3↑↑↑n, ...), and (2) a procedure called diagonalization is used to leap over this sequence into one that's a version of the (diagonalized) Ackermann function (i.e., 3↑ⁿ3). See my earlier comments just below, in which f_k(n) = 3↑^kn, and f_ω(n) = 3↑ⁿ3.
— r.e.s. (talk) 18:46, 4 December 2011 (UTC)

--
In the article, Graham's number G is defined in terms of the g_k-sequence, which involves iterating a certain function defined in terms of "base-3" up-arrow notation, and this up-arrow notation can itself be neatly defined using iterated-function notation. So it might be more to the point to just "cut to the chase" and define G directly in terms of iterated function notation alone. Here's the equivalent definition of G that begins by iterating 3× (i.e. iterated multiplication by 3, otherwise known as base-3 exponentiation):

For any integer n > 0,
f1(n) ::= (3×)n-1(3)  ( = 3n)
fk(n) ::= fk-1n-1(3) for any integer k > 1
fω(n) ::= fn(3)

Graham's number G ::= fω64(4).

There is really only one simple standard notation that a reader needs to grasp here: for any function name f, fⁿ(x) means f(...f(f(x))...) with n fs. (I used the function name f_ω only to show this as a kind of fast-growing hierarchy — it might be less confusing to use a function name like g instead of f_ω. Note that f_ω^k(4) is what the article calls g_k.)
— r.e.s. (talk) 16:54, 29 June 2011 (UTC)

PS: If even the use of superscript notation is considered too confusing, then that too can be replaced (as in Deedlit's comment above) with more explicit but more cumbersome forms as follows:

For any integer n > 0,
f1(n) ::= 3×(...3×(3×(3))...)  with n-1 "3×"s   ( = 3n)
fk(n) ::= fk-1(...fk-1(fk-1(3))...)  with n-1  "fk-1"s, for any integer k > 1
fω(n) ::= fn(3)

Graham's number G ::= fω(...fω(fω(4))...)  with 64 "fω"s

Whatever notational system is used to express them, the essential mathematical concepts here are (1) recursion of functions (i.e., starting with one basic function and then defining an ordered collection of functions in which each is defined in terms of those previously defined), and (2) iteration of functions (i.e. repeating the application of a function a specified number of times).
— r.e.s. (talk) 15:53, 30 June 2011 (UTC)

Scale: Comparisons of Graham's Number, Skewes' Number, Moser's Number, and Googolplex

On space-related Wikipedia articles, the authors often provide the reader with analogies to make the magnitude of the article subject more readily understandable. So, for example, in the Naos article, the authors state that the star is so luminous that the Earth would need to be 450 AU away to maintain traditional temperature ranges, etc. The VY Canis Majoris and Betelgeuse articles also do a great job of comparing their sizes with the Sun. (I believe that the Betelgeuse article compares the Sun to an orange, Sirius to a soccer ball, the Earth to a pearl, and Betelgeuse to a stadium.)

Can someone provide a similar analogy between these four numbers? If a googolplex were represented as '1', how would the other numbers compare? Would Skewes' Number be 1,000? 10,000,000? Another googolplex? I imagine this would go a long way towards making this number more comprehensible to the reader. 70.138.217.107 (talk) 20:41, 25 November 2011 (UTC)

It's possible to do this with Googolplex and Skewes' Number, but not with truly large numbers like the other two. The ratio between Graham's Number and Moser's Number is, for all practical purposes, roughly equal to Graham's Number. The same will hold true even if you took, say, the logarithm of either or both numbers. Any comparison that is not, itself, built on recursion is doomed to be meaningless when dealing with such numbers. And again, comparing either of them with Skewes' or with Googolplex is like comparing them with '1'. Owen× ☎ 20:50, 25 November 2011 (UTC)

Thanks for the perspective! Although there doesn't seem to be a simple way to compare the two larger numbers with a googolplex or with Skewes' number, I still found your explanation to be really helpful. 70.138.217.107 (talk) 02:26, 1 December 2011 (UTC)

Smallest factor of G-128

Is there any project anywhere online that reveals people are searching for the smallest prime factor of G-128?? For example, there's http://www.oddperfect.org where people are searching for odd perfect number requirements. (Note: G means Graham's number.) Georgia guy (talk) 14:45, 28 December 2011 (UTC)

By "G-128" do you mean "Graham's number minus 128"? If so, why 128? If not, what's the definition? — r.e.s. (talk) 02:46, 29 December 2011 (UTC)

Yes, that's what I mean. In an archived part of this talk page, it reveals that G-H has known prime factors for all 0 <= H < 128. Georgia guy (talk) 13:52, 29 December 2011 (UTC)

Ah, I'd forgotten about that archived section. I'm unaware of any projects of that kind, though. — r.e.s. (talk) 16:28, 29 December 2011 (UTC)

Rightmost (frozen) digits calculation

In my book (unfortunately in Italian) on tetration and others hyperoperators, it is described the "frozen digits" distribution for every natural base. A preview (introduction) of the book can be found here: http://www.uni-service.it/images/stories/product_uploaded_file/Coda%20Preview.pdf So, it's possible to calculate the p-adic convergence range of Graham's number too [p. 13].

In this case (n:=3), the convergence speed is equal to the height of the iterated tetration tower minus one (say [hyper-4 exponent]-1). Let this height be "k+1", for example, G(mod 10^k)==G^^G(mod 10^k). So "k", the number of "frozen" (calculable) digits have to satisfy the following inequality:

g(63) < < k < g(64)

Could I have the permission to add this info/reference to the Graham's number page?

Best regards, Marco Ripà — Preceding unsigned comment added by Marcokrt (talk • contribs) 15:47, 19 January 2012 (UTC)

I suggest we limit to 500, because:

• We don't want Wikipedia articles to be huge.
• Somewhere deep along the line, the digits will start to mismatch the infinite 3^3^3^3^3...3^3^3^3^3 sequence; we don't have a way to tell where this point is.

Georgia guy (talk) 15:50, 19 January 2012 (UTC)

Marco:

What you're calling "frozen digits" are evidently the ones I've called "stable digits", e.g. in the recently-archived section Rightmost digits of G (cont'd) that was at the top of this very talk-page when you posted. Note that the upper bound in your inequality can be made tighter, as explained there. (I think "Georgia guy" may have misunderstood your intention, if I'm correct that you merely want to include in the article such a bounding inequality on the number of stable/frozen digits.) I haven't yet looked at your publication, so I have no comment about whether it is adequate as a source for citing such a result.
— r.e.s. (talk) 16:13, 19 January 2012 (UTC)

Hello again, My book is about the randomness of the tetrates digits too (i.e. caos theory and so on)... not only about perfect convergence (a very small amount of info can be found here http://math.eretrandre.org/tetrationforum/showthread.php?tid=720). BTW, IMHO, we could only add the point that the stable digits correspond (exactly) to k, where k+1 is the height of the equivalent hyper-4 hyperexponent value. This is a specific natural number, not only an upper bound. Let me know if this info could be found interesting or not for the page.

Marcokrt (talk) 18:26, 19 January 2012 (UTC) — Preceding unsigned comment added by Marcokrt (talk • contribs) 16:27, 19 January 2012 (UTC)

Visualizing Graham's number

The number of times you have to perform the "How many Knuth arrows are needed??" operation on Graham's number is finally small enough to write out; it is only 64. But have numbers for which the number of times you have to perform the "How many Knuth arrows are needed??" operation on a number too large to write out ever been used?? There have been a few references to numbers larger than Graham's number in some sections of the talk page. Do any of these numbers meet this criterion?? Georgia guy (talk) 16:52, 19 January 2012 (UTC)

For example, my BOX_M~ (a specific natural number involving "human problems" solutions) satisfies this criterion: http://www.scribd.com/doc/77714896/The-largest-number-ever-il-numero-dei-record The busy beaver function "Σ" is even faster (non computable functions). BTW I don't find this outcome very interesting for this section :)

Marcokrt (talk) 18:24, 19 January 2012 (UTC) — Preceding unsigned comment added by Marcokrt (talk • contribs) 18:19, 19 January 2012 (UTC)

Georgia guy:

In the fast-growing hierarchy we have f_ω⁶⁴(6) > G, where f_ω is a version of Ackermann's function and corresponds to the operation you're asking about. If we look at a much larger ordinal index, say ε₀ instead of ω, then we'll encounter numbers like f_ε0(2), which certainly meets your criterion -- even with the small argument of 2, it requires f₈(8) > 2↑⁷8 iterations of f_ω.

Marko:

I think it's not hard to show that your number satisfies $Box_{-}\widetilde{M} \ll f_{\epsilon_0}(3)\,\!$, where f_ε0 is as just described (linked) above.

(Please sign your posts by typing four tildes (~~~~) at the very end.)
— r.e.s. (talk) 18:30, 20 January 2012 (UTC)

Ok r.e.s., even if probably $Box_{-}\widetilde{M} > f_{\epsilon_0}(3)\,\!$, it's easily possible to surpass its size, for example considering the one at the bottom of this page: http://www.polytope.net/hedrondude/array.htm My number is based on a set of philosophical/empiric assumptions and its definition could (abstracly) be useful to construct upper limits for particular kinds of "human condition related problems"... it's quite similar to a universal outlet, even if far less interesting ;)

Best regards Marcokrt (talk) 00:08, 21 January 2012 (UTC)