Irrational Integers :)

[paulunderwood]

Originally Posted by a1call

Doesn't
https://wikimedia.org/api/rest_v1/me...5f572fbf8bb256

satisfy the problem you set?

No, because an irrational is a real number by definition. At least, I do not think Euler's Identity is a solution.

[a1call]

Originally Posted by paulunderwood

No, because an irrational is a real number by definition. At least, I do not think Euler's Identity is a solution.

I see. Thank you for the correction.

I like the "non-constructive proof" method.

[science_man_88]

Originally Posted by paulunderwood

No, because an irrational is a real number by definition. At least, I do not think Euler's Identity is a solution.

it leads to one as it shows e^(i*pi)=-1 which comes from e^ trancendental =
rational

[paulunderwood]

"In mathematics, a transcendental number is a real or complex number that is not algebraic—that is, it is not a root of a non-zero polynomial equation with integer (or, equivalently, rational) coefficients." says https://en.wikipedia.org/wiki/Transcendental_number.

Therefore not all transcendental numbers are irrational -- only the real ones.

[science_man_88]

Originally Posted by paulunderwood

"In mathematics, a transcendental number is a real or complex number that is not algebraic—that is, it is not a root of a non-zero polynomial equation with integer (or, equivalently, rational) coefficients." says https://en.wikipedia.org/wiki/Transcendental_number.

Therefore not all transcendental numbers are irrational -- only the real ones.

okay so what polynomial is i*pi the root of ? if none it is transcendental I think you'll find the article is in error in that the real numbers don't even encompass the algebraic numbers i is algebraic it's the root of x^2+1=0 so I wouldn't believe everything you read on wikipedia.

[Batalov]

Originally Posted by science_man_88

if you plug in n=3 you get 1 as your answer 3/(sqrt(4)-1) - 2/1 = 3/1-2/1=1/1 = 1 in fact you can restate ...

If you plug any n (rational or irrational) except 0, you will get 1.
You can plug n = π, -- it will still be 1.
\((\pi/(\sqrt{\pi+1}-1))-\sqrt{\pi+1} = 1\)