5 ms·

Well, e^pi - pi = 20, is rational.

hollerith 2y ago
It is not exactly 20.
toth 2y ago
Very nice, didn't know about that one! In a similar vein, Ramanujan famously proved that e^(sqrt(67) pi) is an integer. And obviously exp(i pi) is an integer as well, but that's less fun. (Note: only one of the above claims is correct)
- mvdtnz 2y ago
  You didn't know that one because it's a lie. He's telling lies.
  - toth 2y ago
    Charitably it was a joke, as was my quip about `e^(sqrt(67) pi)`. It is a funnier joke without a disclaimer at the end, but unlike GP I couldn't bring myself to leave one out and potentially mislead some people... What I meant was that I didn't know that `e^pi - pi` is another transcendental expression that is very close an integer. You might think this is just an uninteresting coincidence but there's some interesting mathematics around such "almost integers". Wikipedia has a quick overview [1]. I didn't realize it before, but they have GP's example and also the awesome `e + pi + e pi + e^pi + pi^e ~= 60`. [1] https://en.wikipedia.org/wiki/Almost_integer https://en.wikipedia.org/wiki/Almost_integer
- isaacfrond 2y ago
  The number you are looking for is e^(sqrt(163) pi). According to Wikipedia: In a 1975 April Fool article in Scientific American magazine,[8] "Mathematical Games" columnist Martin Gardner made the hoax claim that the number was in fact an integer, and that the Indian mathematical genius Srinivasa Ramanujan had predicted it – hence its name. It is not an integer of course.
  - toth 2y ago
    Actually `e^(sqrt(n) pi)` is very close to being an integer for a couple of different `n`s, including 67 and 163. For 163 it's much closer to an integer, but for 67 you get something you can easily check in double precision floats is close to an integer, so I thought it worked better as a joke answer :) FYI, the reason you get these almost integers is related to the `n`s being Heegner numbers, see https://en.wikipedia.org/wiki/Heegner_number https://en.wikipedia.org/wiki/Heegner_number.
  - Someone 2y ago
    > The number you are looking for is e^(sqrt(163) pi) […] It is not an integer of course. Of course? I’m not aware that we have some theorem other than “we computed it to lots of decimals, and it isn’t an integer” from which that follows.
    - less_less 2y ago
      It's not really "of course", and I don't think we have such a theorem in general. But in this case, I believe the fact that it's not an integer follows from the same theorem that says it's very close to an integer. See eg https://math.stackexchange.com/questions/4544/why-is-e-pi-sqrt163-almost-an-integer https://math.stackexchange.com/questions/4544/why-is-e-pi-sq... Basically e^(sqrt(163)*pi) is the leading term in a Laurent series for an integer, and the other (non-integer) terms are really small but not zero.
nimih 2y ago
Do you have a citation for the rationality of e^pi - pi? I couldn't find anything alluding to anything close to that after some cursory googling, and, indeed, the OEIS sequence of the value's decimal expansion[1] doesn't have notes or references to such a fact (which you'd perhaps expect for a rational number, as it would eventually be repeating). [1] https://oeis.org/A018938 https://oeis.org/A018938
- chowells 2y ago
  https://xkcd.com/217/ https://xkcd.com/217/
  - nimih 2y ago
    Is the joke here that if you lie to people (on the Internet or otherwise), they’ll take it at face value for a little bit and then decide you’re either a moron or an asshole once they realize their mistake?
    - chowells 2y ago
      Nah. I'd read it as there being an expectation that the audience already knew the joke and they were playing the favorites.
      - Izkata 2y ago
        Also that "black hat" character is consistently evil in every comic he's in.
c0redump 2y ago
Wow, you just made my day with this! What a fantastic result! Beautiful. Edit: looks like I swallowed the bait, hook like and sinker

It is not exactly 20.

toth 2y ago

Very nice, didn't know about that one! In a similar vein, Ramanujan famously proved that e^(sqrt(67) pi) is an integer. And obviously exp(i pi) is an integer as well, but that's less fun. (Note: only one of the above claims is correct)

mvdtnz 2y ago
You didn't know that one because it's a lie. He's telling lies.
- toth 2y ago
  Charitably it was a joke, as was my quip about `e^(sqrt(67) pi)`. It is a funnier joke without a disclaimer at the end, but unlike GP I couldn't bring myself to leave one out and potentially mislead some people... What I meant was that I didn't know that `e^pi - pi` is another transcendental expression that is very close an integer. You might think this is just an uninteresting coincidence but there's some interesting mathematics around such "almost integers". Wikipedia has a quick overview [1]. I didn't realize it before, but they have GP's example and also the awesome `e + pi + e pi + e^pi + pi^e ~= 60`. [1] https://en.wikipedia.org/wiki/Almost_integer https://en.wikipedia.org/wiki/Almost_integer
isaacfrond 2y ago
The number you are looking for is e^(sqrt(163) pi). According to Wikipedia: In a 1975 April Fool article in Scientific American magazine,[8] "Mathematical Games" columnist Martin Gardner made the hoax claim that the number was in fact an integer, and that the Indian mathematical genius Srinivasa Ramanujan had predicted it – hence its name. It is not an integer of course.
- toth 2y ago
  Actually `e^(sqrt(n) pi)` is very close to being an integer for a couple of different `n`s, including 67 and 163. For 163 it's much closer to an integer, but for 67 you get something you can easily check in double precision floats is close to an integer, so I thought it worked better as a joke answer :) FYI, the reason you get these almost integers is related to the `n`s being Heegner numbers, see https://en.wikipedia.org/wiki/Heegner_number https://en.wikipedia.org/wiki/Heegner_number.
- Someone 2y ago
  > The number you are looking for is e^(sqrt(163) pi) […] It is not an integer of course. Of course? I’m not aware that we have some theorem other than “we computed it to lots of decimals, and it isn’t an integer” from which that follows.
  - less_less 2y ago
    It's not really "of course", and I don't think we have such a theorem in general. But in this case, I believe the fact that it's not an integer follows from the same theorem that says it's very close to an integer. See eg https://math.stackexchange.com/questions/4544/why-is-e-pi-sqrt163-almost-an-integer https://math.stackexchange.com/questions/4544/why-is-e-pi-sq... Basically e^(sqrt(163)*pi) is the leading term in a Laurent series for an integer, and the other (non-integer) terms are really small but not zero.

nimih 2y ago

Do you have a citation for the rationality of e^pi - pi? I couldn't find anything alluding to anything close to that after some cursory googling, and, indeed, the OEIS sequence of the value's decimal expansion[1] doesn't have notes or references to such a fact (which you'd perhaps expect for a rational number, as it would eventually be repeating). [1] https://oeis.org/A018938 https://oeis.org/A018938

chowells 2y ago
https://xkcd.com/217/ https://xkcd.com/217/
- nimih 2y ago
  Is the joke here that if you lie to people (on the Internet or otherwise), they’ll take it at face value for a little bit and then decide you’re either a moron or an asshole once they realize their mistake?
  - chowells 2y ago
    Nah. I'd read it as there being an expectation that the audience already knew the joke and they were playing the favorites.
    - Izkata 2y ago
      Also that "black hat" character is consistently evil in every comic he's in.

c0redump 2y ago

Wow, you just made my day with this! What a fantastic result! Beautiful. Edit: looks like I swallowed the bait, hook like and sinker