How to convey the largest possible number with conventional symbols?
August 7, 2018 11:08 AM   Subscribe

The short version: I'm curious about how to represent the largest possible number in an arbitrarily-limited physical space (think, say, a sheet of paper with room for only so many legible characters), using only standard, commonly-recognized numbers and mathematical symbols.

This is dumb and weird, but: I had a dream the other night that a friend and I had worked up a new dispute-settling mechanism where whoever could name the higher number on an identical piece of paper would win the dispute. I then woke up and spent the rest of the night half-asleep, half-awake, thinking about this. My 2 a.m. conclusion was that the best bet would be basically 9^999999999.... where the first 9 is regular-sized and all of the 9s in the exponent were smaller. But in the cold light of day, that doesn't seem right. My math-heavy undergrad days are far in the past now, but I know things get pretty huge when you throw factorials in.
posted by the phlegmatic king to Grab Bag (15 answers total) 6 users marked this as a favorite


(You will need to turn the page sideways.)
posted by Huffy Puffy at 11:12 AM on August 7, 2018 [10 favorites]

Response by poster: Meant to add: infinity is explicitly excluded, since it's not a specific number.
posted by the phlegmatic king at 11:14 AM on August 7, 2018

posted by adamrice at 11:21 AM on August 7, 2018 [4 favorites]


Where "Bob" is used as the name of your competitor?
posted by turkeybrain at 11:22 AM on August 7, 2018 [2 favorites]

Knuth's up arrow notation allows you to represent extremely large numbers. I don't know if that is commonly recognized enough for you.

I think repeated factorials gives you a bigger number than your 9^999..., especially if you're talking about handwriting and can fit in more !s than 9s. 9!! is already bigger than 9^9.
posted by cruelfood at 11:30 AM on August 7, 2018 [6 favorites]

My mathematician also suggests up-arrows.
posted by LobsterMitten at 11:33 AM on August 7, 2018

Use base 62 (represented by 0..9,a..z,A..Z) number system. Then Follow adamrice's method of Z^Z^Z^Z^... If you have to use actual super script instead of the '^' character, than it would be a separate equation to see for the particular base number system if it's a win to do giant lines of Z^ZZZZZZZZZZ vs. fewer (because of more vertical space needed) lines of Z^Z^Z .

If you don't want to use letters, you could use a radix to represent a larger base. So again, if you can use standard characters instead of super (and super super (and super super super ...)) script,

(10)(subscript)99999 ^ (10)(subscript)99999 ^ (10)(subscript)99999

If the subscript involves less width than the standard characters, this would involve less space than writing:


If the font is non-fixed width, (11)(subscript)111111 instead will be larger and take up less space. However I think 9^9^9^9^... will actually be larger, so one is best off using the largest single-value digit. So 9 for base-10, but a larger base system with explicit character definition would be prefered.

9^9^9 is 3.7*10e8 decimal digits.

9!! is only 9.1*10e4

9!!! however is 6.0*10e1859932 digits

and 9!!!! (same number of characters as 9^9^9) is large enough that wolfram alpha won't give me the same values. Instead I get 10^10^10^10^6.2694 ...

However I note that Wolfram alpha is making me represent this as ((9!)!)! for instance to get 9!!! , so that eats into charactes/space.

If you can have a non-fixed width font, and don't need to surrond with parenthesis, than largest singledigit base followed by factorial markers looks to be pretty darn dense.
posted by nobeagle at 12:02 PM on August 7, 2018 [1 favorite]

Wow, I've missed hearing of up arrow notation until now. Assuming that character is allowed, filling a page with uparraws with a 9 in the front and end I feel would beat factorials for most fonts. I'm having difficulty working with this with Wolfram Alpha so am misssing some scaling comparing against "small" values like 9!!!! .
posted by nobeagle at 12:16 PM on August 7, 2018

Aaronson's paper, linked by lantius, is a must read.

Sorta related is the Bignum Bakeoff where you had to submit a computer program for a hypothetical computer that could handle arbitrarily large integers and had infinite memory. The program had to compute a number, biggest number won. Obviously you couldn't actually run the programs, so a certain amount of analysis was done to determine the winner.

Warning: Mega-epic nerdy. More so than usual.
posted by It's Never Lurgi at 12:40 PM on August 7, 2018 [1 favorite]

Hey, just as a heads-up, x!! doesn’t mean (x!)!, but rather means “multiply every other number”. So it’s less than a regular factorial.

I of course learned this by getting it wrong in trivia.
posted by Huffy Puffy at 1:02 PM on August 7, 2018 [2 favorites]

Best answer: I think if up arrows are allowed, then nobeagle's suggestion of two 9's (or something bigger if you're allowed to use another base than base 10) sandwiching as many arrows as allowed is probably the way to go.

If not, then on an operation-by-operation basis, iterated factorials are better than iterated exponentiation. To see this, suppose we already have some large n, and we want to obtain a larger number by either taking n!, or taking k^n, where k is some constant base (presumably the largest digit we can write in our chosen base--9 in base 10).

By Stirling's approximation, log(n!) ~ n*log(n) - n (discarding the O(log n) term). On the other hand, log(k^n) = n*log k. So as long as log(n) - 1 > log(k), which will be true for any sufficiently large n, we will get more out of writing down another factorial than writing down another exponentiation.

(Of course, how many times you can repeat each operation will be affected by the details of your notation and constraints, such as whether or not you need a new set of parentheses for each operation. That may change the answer.)
posted by egregious theorem at 2:03 PM on August 7, 2018 [1 favorite]

Best answer: Some exploring on Wikipedia led me to the page on iterated exponentiation. That would give you a compact way to represent ridiculously large values.
posted by adamrice at 3:03 PM on August 7, 2018

I had a dream the other night that a friend and I had worked up a new dispute-settling mechanism where whoever could name the higher number on an identical piece of paper would win the dispute.

MIT and Princeton got there ahead of you.

The trouble with this mechanism is that once the procedural specifications for deriving the numbers start to involve uncomputable functions, legitimate disputes can arise over which of two numbers arrived at by different methods actually is bigger. So you might want to add a computability restriction to the existing one that prohibits infinities.

Introduction to Googology
posted by flabdablet at 8:46 PM on August 7, 2018 [1 favorite]

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