I find it quite interesting that pi+e and pi*e are not proven to be irrational (although it's proven that at least one of them is irrational [1]).
It would be mind-blowing if either of them were rational numbers, yet it's very hard to prove either way.
I could believe pi*e rational but pi+e being rational would blow my mind.
I would be shocked if either of them were proven to be rational.
I'm kind of in the opposite camp. If Schanuel's conjecture is true, then e^iπ = 0 would be the only non-trivial relation between e, π, and i over the complex numbers. And the fact that we already found it seems unlikely.
you mean e^(i pi)=-1, which is known as Euler's identity and is a specific case of Euler's formula
e^(i theta) = cos theta + i sin theta
That formula gives infinitely many trivial relationships like this due to the symmetry of the unit circle
e^(i 2 pi) = 1
e^(3i/2pi)/i=1
e^(5i/2pi)/i=-1
e^(i 2n pi) = 1 for all n in Z ...
etc
Thanks for ringing some bells. It's been a long time since I used that equation.
Well, e^pi - pi = 20, is rational.
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).
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?
Nah. I'd read it as there being an expectation that the audience already knew the joke and they were playing the favorites.
Also that "black hat" character is consistently evil in every comic he's in.
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)
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.
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.
> 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.
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-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.
You didn't know that one because it's a lie. He's telling lies.
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`.
It is not exactly 20.
Wow, you just made my day with this! What a fantastic result! Beautiful.
Edit: looks like I swallowed the bait, hook like and sinker
I mean you just have to get to the point where all of the trailing decimal places (bits) form a repeating pattern with finite period. But since there are infinitely many such patterns it becomes extremely hard to rule out without some mechanism of proof.
Is the result of the addition or multiplication of an irrational number with any other real number not equal to it (and non-zero in the case of multiplication) always irrational? ex: pi + e, pi * e, but also sqrt(2) - 1 or sqrt(3) * 2.54 ?
No, sqrt(5)*sqrt(16*5)=20. More trivially, there's always a number y such that z = x*y for a given irrational x. You can give similar examples for all the other basic operations.
Take any irrational a where 1/a is also irrational. Then a * 1/a = 1.
Even moving from addition and multiplication to exponentials won’t save you: there are irrational numbers to irrational powers that are raational.
> Take any irrational a where 1/a is also irrational.
In other words: any irrational at all
The nonconstructive proof of that is simple and fun: either sqrt(2)^sqrt(2) or (sqrt(2)^sqrt(2))^sqrt(2) is just such an example.
Definitely not; consider the formula for calculating the log of any base given only the natural logarithm. That can result e.g. in two irrational numbers, the ratio of which are integers.
There is an important distinction to be made here. Examples in this thread show cases of irrational numbers multiplied by or added to other irrational numbers producing real numbers, but in the special case of a rational number added to or multiplied by an irrational number, the result is always irrational.
Otherwise, supposing for instance that (n/m)x is rational for integers n, m, both non-zero, and irrational x, we can express (n/m)x as a ratio of two integers p, q, q non-zero: (n/m)x = p/q if and only if x = (mp)/(qn). Since integers are closed under multiplication, x is rational, against supposition; thus by contradiction (n/m)x is irrational for any rational r = (n/m), with integers n, m both non-zero. Similarly for the case of addition.
pi and -pi are both irrational and their sum is zero.
irrational number + rational number = irrational number [1]
irrational number + irrational number could be rational or irrational.
5 - sqrt(2) is irrational
sqrt(2) is irrational
Add them up you get 5, which is rational
[1] If it were rational, you will be able to construct a rational representation of the irrational number using this equation.
To put the first equation more formally, we know that ℚ is closed under addition¹, so given k∊ℝ\ℚ, l∊ℚ then if k+l=m∊ℚ, then m-l=m+(-l)∊ℚ, but m-l=k which is not in ℚ so k+l∉ℚ.
⸻
1. For p,q∊ℚ, let p=a/b, q=c/d, a,b,c,d∊ℤ, then p+q=(ad+bc)/bd, but the products and sums of integers are integers, so p+q∊ℚ
x - (x - floor(x)) == x truncated to an integer
when x is an irrational number > 1:
"x - floor(x)" is just the fractional part of x, so it's an irrational number which is not equal to x.
Subtracting the fractional part from the original leaves only the integer part, which is obviously rational.
Out of curiosity, why would it be mind-blowing if either of them were a rational number?
If pi+e=a/b then you can write one as a/b minus the other
Which is pretty insane because these two numbers are not supposed to be related
> Which is pretty insane because these two numbers are not supposed to be related
Not really, there is Euler's identity: https://en.m.wikipedia.org/wiki/Euler%27s_identity
That one's "just" a special case of how complex numbers happen to work. I think the really cool relationship between e and pi is the fact that the Gaussian integral acts as a fixpoint/attractor when sampling and summing data from any distribution (this is the central limit theorem):
∫(−∞ to ∞) e^(-x²) dx = √π
I think the attractor property makes it a little more fundamental in some sense, whereas Euler's identity is "just" one special case of e^ix. The Gaussian is kind of the "lowest energy" or "highest entropy" state of randomness, which I think is really cool.
This is only true for distributions with finite variance (and the edge case of distribution with slowly growing infinite variance).
And for a given variance, gaussian distributions are exactly the maximal entropy distribution.
> because these two numbers are not supposed to be related
Says who? They’re not known to be related in that way, but it’s not like nature set out to prevent such a thing, or that large parts of mathematics would break down if it happened to be the case.
e and pi are highly related, both pop out of periodic phenomenon.
Same for pi^pi^pi^pi
> although it's proven that at least one of them is irrational
And not particular to e and pi. More generally, at least one of a+b and a*b must be irrational, if an and b are transcendental.