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-   -   Irrational Integers :) (https://www.mersenneforum.org/showthread.php?t=22301)

paulunderwood 2017-05-13 21:46

[QUOTE=a1call;458942]Doesn't
[IMG]https://wikimedia.org/api/rest_v1/media/math/render/svg/a7464809a40f9e486de3a454745f572fbf8bb256[/IMG]

satisfy the problem you set?[/QUOTE]

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

a1call 2017-05-13 22:12

[QUOTE=paulunderwood;458946]No, because an irrational is a real number by definition. At least, I do not think Euler's Identity is a solution. :smile:[/QUOTE]

I see. Thank you for the correction.

I like the "non-[URL="https://en.wikipedia.org/wiki/Constructive_proof"]constructive proof[/URL]" method.:smile:

science_man_88 2017-05-13 22:14

[QUOTE=paulunderwood;458946]No, because an irrational is a real number by definition. At least, I do not think Euler's Identity is a solution. :smile:[/QUOTE]

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

paulunderwood 2017-05-13 22:23

"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 [url]https://en.wikipedia.org/wiki/Transcendental_number[/url].

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

science_man_88 2017-05-13 22:48

[QUOTE=paulunderwood;458951]"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 [url]https://en.wikipedia.org/wiki/Transcendental_number[/url].

Therefore not all transcendental numbers are irrational -- only the real ones.[/QUOTE]

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 2017-05-13 22:50

[QUOTE=science_man_88;458932]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 ... [/QUOTE]
If you plug any n (rational or irrational) except 0, you will get 1.
You can plug n = [FONT=Fixedsys][SIZE=3][FONT=&quot]π[/FONT][/SIZE][/FONT], -- it will still be 1.
[$](\pi/(\sqrt{\pi+1}-1))-\sqrt{\pi+1} = 1[/$]


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