Wikipedia:Reference desk/Archives/Mathematics/2023 November 4

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

November 4

The presence of at least 4 of something proves that there are at least any number of something

Do you think 4 is a lot?? Now, let's say that you have 4 coins and you want to pick 2. How many ways are there to do this?? The answer is 6. Now, can you find out how many ways there are to name 3 ways to pick 2 coins when there are 4 coins total?? The answer is 20. The next question is how many ways there are to name 10 ways to do this. The answer is 184756. The next number after that is way too big. If you understand this carefully, you'll see that if there's anything that there are at least 4 of, then there's something that there are at least any number of. This is contrary to the statement that "there isn't a googol of anything". (If possible, feel free to include the number of digits in the "way too big" number I talked about is; it is specifically $184756!/(92378!*92378!)$ .) Georgia guy (talk) 15:52, 4 November 2023 (UTC)

Sorry, GG, can't follow this at all. Can you try to restate more clearly? --Trovatore (talk) 16:51, 4 November 2023 (UTC)

User:Trovatore, let's start by taking 4 coins and calculating how many ways there are to pick exactly 2 of these coins. There are 6. Now let's play with the 6 ways of picking 2 coins when there are 4 total. Specifically, how many ways are there to name 3 of these ways?? The answer is 20. As for the 20 ways of naming 3 ways of picking 2 coins when there are 4 total, how many ways are there to pick 10 of these?? The answer is 184756. Do you understand?? Georgia guy (talk) 16:57, 4 November 2023 (UTC)

I'm pretty sure this is the central binomial coefficients iterated with a a division by 2. So a(2) = choose(4, 2) = 6, a(a(2)) = choose(6, 3) = 20, a(a(a(2))) = choose(20, 10) = 184756, a(a(a(a(2)))) = choose( 184756, 92378) = very large. I think the statement "there isn't a googol of anything" refers to actual things, not combinations of things; it's not really a mathematical statement. --RDBury (talk) 17:13, 4 November 2023 (UTC)

I'm not sure why you need to start as high as 4 or why you need anything stronger than adding 1.

ObSMBC

That said, there is a possible reason to use something stronger than adding 1. There are certain very weak fragments of Peano arithmetic (such as IΔ₀, which I think means you restrict the induction schema to formulas with bounded quantifiers and a primitive-recursive matrix, or roughly that anyway) that cannot prove that the exponentiation function is total. That is, they can't prove that for every $n$, $2^{n}$ exists. That's in spite of the fact that they do prove that for every $n$, $n+1$ exists. --Trovatore (talk) 17:14, 4 November 2023 (UTC)

I think the problem is the word 'som,ething'. Is the number of ways of arranging articles an actual physical thing? At most it is a new identifiable arrangement every nanosecond or so, Should I even consider an argument that there is a new arrangement convincing without actually being able to identify it. NadVolum (talk) 17:19, 4 November 2023 (UTC)

The number of ways to order just two standard decks of cards (104 cards) is much larger than a googol. CodeTalker (talk) 17:47, 4 November 2023 (UTC)

If you have only one 54-card deck but distinguish whether cards are vertical or horizontal in a laid-out arrangement such as

🃑 🃖 🃆 🂵 🂥 ···

the number of arrangements also exceeds one googol.  --Lambiam 20:45, 4 November 2023 (UTC)

@Lambiam: Very impressive cards rendering! --CiaPan (talk) 21:19, 4 November 2023 (UTC)

"there isn't a googol of anything" refers to physical objects but we don't know whether it's actually true. A googol is 10¹⁰⁰. Orders of magnitude (numbers)#1042 to 10100 says:

• Cosmology: Various sources estimate the total number of fundamental particles in the observable universe to be within the range of 10⁸⁰ to 10⁸⁵. However, these estimates are generally regarded as guesswork. (Compare the Eddington number, the estimated total number of protons in the observable universe.)

References

I guess the phrase was coined with this in mind. Mathematics is full of large numbers. We don't need combinatorics but can for example just start with something large. There are more than a googol prime numbers below 10¹⁰⁰⁰. PrimeHunter (talk) 22:24, 4 November 2023 (UTC)

The googolth prime number is something like 5.262...×10²³⁵, so there are quite a few more.  --Lambiam 23:21, 4 November 2023 (UTC)

It's actually only around 2.35×10¹⁰². PrimeHunter (talk) 01:59, 5 November 2023 (UTC)

Yes, I don't understand where that came from. The last thing I did (according to a log I kept) before I published the reply was to verify that the number $x$ I had found produced $x/(\log x)=9.999999999999769\times 10^{99}$ and print its value as being $x=2.357211588756798\times 10^{102}.$ That was the last line of the log. The incorrect number I published seems to be $10^{10^{{-}100}x}.$  --Lambiam 14:07, 6 November 2023 (UTC)

If you want significantly better accuracy of the prime-counting function with a slightly more complicated formula then use $x/(\log x-1)$. It predicts 2.34717×10¹⁰². Based on the logarithmic integral, the real value rounds to 2.34713×10¹⁰². If f is the function for which $x/(\log x-f(x))$ gives the exact count then f(x) tends to 1 when x tends to infinite. The logarithmic integral is still far better but few devices and programs offer it. PrimeHunter (talk) 21:05, 6 November 2023 (UTC)