Redpill me on this constant, Veeky Forums, because it seems to gain more and more popularity.
Why is it important?
Redpill me on this constant, Veeky Forums, because it seems to gain more and more popularity.
Why is it important?
well for one, of all thew uncountable number of irrational numbers, its one of the few that are in a countable subset (which almost none are.)
what
the rationality of gamma is still not known
fine, change 'irrational' to 'real' in if you want to be pedantic
all real numbers are in countable subsets...? just put that given number in a set
no, some real numbers cant be put in a countable set, there are some real numbers that cant be listed in any way.
what are you talking about? every real number can be put into a singleton set
>is calculated with the natural logarithm
>mathematicians still don't know if it's irrational
transcendental/irrationality questions are notoriously difficult
wat, no it cant.
Theres no way to pick out some of the real numbers from the real line to add to a singleton. You would have to define that number in some way, and if you can define it its countable.
>Theres no way to pick out some of the real numbers from the real line to add to a singleton.
why not? let x be a given real number and then {x} is such a singleton set
>and if you can define it its countable.
are you saying most real numbers aren't definable? what's wrong with their infinite decimal expansion?
[math]ln(e^2)[\math] is rational
Analysis fags love to define real numbers using convergent sums of rational numbers lol
Dont listen to that fag, to him, the square with side length of one doesnt have a diagonal.
Not a convincing argument, user.
You can only put natural numbers in the series which defines this constant.
It shows up a lot in multiplicative number theory, especially in asymptotic results.
>let x be a given real number
you need to be able to say what x is, and to do that it needs to be definable, and if its definable its countable. The best you can do about all the other real numbers as talk about a uncountable number of them between 2 definable numbers.
Thats a shitty argument since for example
[eqn]\frac{1}{1}\ln \left({\frac {2}{1}}\right) + \frac{1}{2}\ln \left({\frac {2^{2}}{1\cdot 3}}\right) + \frac{1}{3}\ln \left({\frac {2^{3}\cdot 4}{1\cdot 3^{3}}}\right) + \frac{1}{4}\ln \left({\frac {2^{4}\cdot 4^{4}}{1\cdot 3^{6}\cdot 5}}\right) \cdots + \frac{1}{n}\ln\left(\prod _{{k=0}}^{n}(k+1)^{{(-1)^{{k+1}}{n \choose k}}}\right)\cdots= \ln e = 1 [/eqn]
which is rational.
What you are trying to say is that the set of definable numbers is countable. You either improve your english or ,if you actually know english and are actually this dumb, learn how to talk precisely. The following things that you said are simply wrong:
>it is one of the few that are in a countable subset
All real numbers are in SOME countable subset. You need to specify if you have a specific countable subset in mind.
>Theres no way to pick out some of the real numbers from the real line to add to a singleton
Let S = {{x} : x is real}. By the axiom of choice I can pick out any {x}.
>you need to be able to say what x is, and to do that it needs to be definable, and if its definable its countable.
Numbers are not countable. Communicate clearly.
>mfw i dont understand you post :(
>axiom of choice
theres the main problem is seems, your working in some made up play framework while im discussing real math.
>le ebin all of contemporary mainstream math is wrong xDDD XD meme
Where the fuck did you get this from. Also, I'm pretty sure the nth term is wrong.
its the log of Guillera's product for e:
en.wikipedia.org
>Redpill me
gtfo pill-popping /pol/esmoker
is it just one person or multiple people who keep making these silly 'go back to /pol/' posts? it just seems like one butthurt leftist
funny how libtards will push fake politicized climate science but don't want people interested in politics posting on a science board for some reason
He's blaming /pol/ because they cut his welfare checks.