24

I have heard that it is unknown whether or not $\Large \pi^{\pi^{\pi^\pi}}$ is an integer. How can this be? $\pi$ is known to many digits and it seems like only a matter of time on a computer to find the integer part or to find that there must be a decimal part to the number.

What makes it difficult to determine? I know that it is overwhelmingly likely that the constant is irrational but I am interested in why it is hard to show.

Another aspect of the question could involve answering why there is no proof that for $\alpha$ transcendental $\Large \alpha^{\alpha^{\alpha^\alpha}}$ is not an integer. I am not naive enough to think such a proof would be easy but I also don't know too much about it to know why it would be hard.

user157227
  • 2,024
  • 13
  • 30
  • 9
    How hard is it to figure out if an electron is a particle or a wave? I mean, it's just an electron. And how hard is it to figure out who killed JFK? There were only finitely many people in Dallas that day. And how hard is it to find out what gene causes what? There are only so few of them. – Asaf Karagila Jul 01 '14 at 17:11
  • 10
    I once heard this quote: "There only are two kinds of problems: the trivial ones, and the ones you can't solve". – Ivo Terek Jul 01 '14 at 17:16
  • 16
    Since $3$ is a lower bound for $\pi$, a lower bound for $\pi^{\pi^{\pi^{\pi}}}$ is given by $3^{3^{3^3}} = 3^{3^{27}} = 3^{7625597484987}$. That's kind of a big(ish) number. – Dustan Levenstein Jul 01 '14 at 17:19
  • 4
    The likely issue is the size of the number rather than any technical difficulty with the proof - if the number is (as is likely) not an integer, it should be possible to bound it away from an integer using sufficiently accurate estimates at each stage. – Mark Bennet Jul 01 '14 at 17:21
  • @AsafKaragila Brilliant! –  Jul 01 '14 at 17:24
  • 2
    For what it's worth, Wolfram|Alpha says $\frac{22}{7}^{\frac{22}{7}^{\frac{22}{7}^{\frac{22}{7}}}}$ is not an integer. http://www.wolframalpha.com/input/?i=is+%2822%2F7%29%5E%28%2822%2F7%29%5E%28%2822%2F7%29%5E%2822%2F7%29%29%29+an+integer%3F –  Jul 01 '14 at 17:30
  • It may be possible to prove its not an integer but if its very close to an integer, and may be one, wouldn't we need to the maths with infinite digits of precision? – Warren Hill Jul 01 '14 at 17:33
  • 3
    @AlonsodelArte: It says the same about $\pi^{\pi^{\pi^\pi}}$ but I doubt it has proved either. – Charles Jul 01 '14 at 17:33
  • 2
    Would Alfonzo's question with $\frac{22}{7}$ instead of $\pi$ be an easier or a harder thing to prove? – Robert Soupe Jul 02 '14 at 04:12
  • A slight variation a few places down will have a variation in the output. 3.1^3.1 = 33.35 but 3.2^3.2 = 41.35 so modification of 1 decimal point (.1%)yields a difference of 33%! To take this to three powers 3.1^3.1^3.1 = 24640259360223718.858126253459385 while 3.2^3.2^3.2 = 772972481920577459398.87465531998 so you can see huge differences occur. ... Hmm, though. Maybe we can calculate how many decimals to get the error within one digit. But then the we have to see that the est. is not within the error of a digit. And if the result is an intger we'll never disprove its not. – fleablood Feb 11 '16 at 02:09
  • There are countably infinite cases where $\alpha^{\alpha^{\alpha^{\alpha}}}$ is an integer. – fleablood Feb 11 '16 at 02:31
  • problem solved http://prntscr.com/t2wdhe – Sam Jun 19 '20 at 17:49

2 Answers2

13

That number has 666262452970848504 decimal places, so to determine if it's an integer you'd have to compute it with that precision. But this would take 270,000 TB, and we don't have many hard drives that large.

Charles
  • 32,122
  • How much storage would be needed to show that it has a fractional part? (assuming it isn't a near integer) – user157227 Jul 01 '14 at 17:22
  • 5
    You need to compute it to infinite precision if it actually is an integer – TROLLHUNTER Jul 01 '14 at 17:23
  • 1
    @user157227: About 276659495136553954 bytes should suffice. – Charles Jul 01 '14 at 17:26
  • @user1708: Right -- but the question assumed "it isn't a near integer". – Charles Jul 01 '14 at 17:31
  • How many decimal places of pi do we have to know to calculate the number within, say .25? Then if the answer is not in the range of x.25 through x.75 (as it has a 50% chance of not being) how many decimal places further to get it within 1/8? – fleablood Feb 11 '16 at 02:19
  • @fleablood: You'd need about 666262452970848504 to calculate the number to within 0.25, and 666262452970848505 should be enough to get within 1/8 = 0.125. – Charles Feb 11 '16 at 02:24
  • So if we knew pi to the 666,262,452,970,848,505 decimal places we'd have a 75% chance of knowing it isn't an integer? Meanwhile to prove it isn't rational... – fleablood Feb 11 '16 at 02:28
  • @fleablood: If you could compute the number to that many decimal places you'd have an 80-90% chance of showing it's not an integer. I'm not sure how much error the exponentiation would introduce, so maybe you'd need a few more to be safe. But you're already asking for hundreds of millions of billions of digits, so what's a few -- or a few trillion -- between friends? – Charles Feb 11 '16 at 04:19
  • But basically, that explains how it can not be known. Exponenientiation grows .... well, exponentially. – fleablood Feb 11 '16 at 09:44
  • @fleablood In this case, it's more to do with tetration. – Ben J Mar 21 '19 at 17:33
  • As of 2021, the computational space hurdle doesn't seem to be all that much - you can buy a 16 TB external hard drive for $329, so that's about $5.5 mil for 270k TB - well within the realm of possibility for some large organization that wants to generate publicity by computing this numerically. Are there any additional hurdles in computing this number, e.g. with the time-complexity? – Mike Battaglia Feb 27 '21 at 20:08
  • @MikeBattaglia You should probably ask that as a separate question. But the world record for computing $\pi$ found 100 trillion decimal places with about $30,000 of compute resources. You’d need 10,000 times more resources at a minimum. But even with the money it’s not likely you could efficiently address all that memory, and that would lead to a time and money slowdown of at least a factor of 10. So billions of dollars, best case, which isn’t chump change like $5 mill. – Charles Jun 21 '22 at 01:58
1

it is not difficult. The only problem seems to be is that it is to large number to present here all $10^{18}$ digits and the first digit after delimiter

Michael Hardy
  • 1
user56396
  • 49
  • 7
    It may be really difficult if it IS an integer – TROLLHUNTER Jul 01 '14 at 17:23
  • 2
    If you did get the first 1000 digits after the decimal point and they were all $0$, then the question of whether it's an integer would be undecided: maybe the 1001th digit would be $1$. ${}\qquad{}$ – Michael Hardy Jul 01 '14 at 17:35
  • 2
    agree with Michael Hardy, but don't believe that even first 2 are 0. The chances are 1 to 100, I would bet on such condition :) – user56396 Jul 01 '14 at 17:47