Irrational Integers :)
 

(n/(sqrt(n+1)-1))-sqrt(n+1)

Integers are equal to a subset of the rationals. For example an integer n is equal to n/1. :smile:

a=sqrt(3)-1=0.7321
2/a=2.7321

[QUOTE=a1call;458929](n/(sqrt(n+1)-1))-sqrt(n+1)[/QUOTE]

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 this as n/(sqrt(n+1)-1)+(n+1-sqrt(n+1))/(sqrt(n+1)-1) = (2n+1-sqrt(n+1))/sqrt(n+1)-1) which is rational any time n is one less than a perfect square.

While we're on the subject of irrationality: Show there is a rational number that is equal to an irrational number raised to the power of an irrational number. :geek:

[QUOTE=paulunderwood;458933]While we're on the subject of irrationality: Show there is a rational number that is equal to an irrational number raised to the power of an irrational number. :geek:[/QUOTE]

I know that one.
It's good looking.:smile:

[QUOTE=a1call;458934]I know that one.:smile:[/QUOTE]

Give the answer please with the use of spoiler tags. :boxer:

I dislike the spoiler tags.

[url]https://en.m.wikipedia.org/wiki/Euler%27s_identity[/url]

:no:

[url]https://en.wikipedia.org/wiki/Irrational_number[/url] says "In mathematics, the irrational numbers are all the [B]real[/B] numbers, which are not rational numbers, the latter being the numbers constructed from ratios (or fractions) of integers"
[SPOILER]
That page also gives the answer to the problem I set.[/SPOILER]

It is safe to assume that no one alive today is likely to come up with that identity without knowing it already.:smile:
Hence, no point in using the spoiler tags.

[QUOTE=paulunderwood;458938]:no:

[URL]https://en.wikipedia.org/wiki/Irrational_number[/URL] says "In mathematics, the irrational numbers are all the [B]real[/B] numbers, which are not rational numbers, the latter being the numbers constructed from ratios (or fractions) of integers"
[SPOILER]
That page also gives the answer to the problem I set.[/SPOILER][/QUOTE]
Doesn't
[IMG]https://wikimedia.org/api/rest_v1/media/math/render/svg/a7464809a40f9e486de3a454745f572fbf8bb256[/IMG]

satisfy the problem you set?

[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:

[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:

[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]

"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=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.

[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.
[$](\pi/(\sqrt{\pi+1}-1))-\sqrt{\pi+1} = 1[/$]