| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Hello all,
-2
As many people know how the prove 2 is irrational.
-n
But, is it exist a general method to prove m is irrational/
-4 -7
rational ? (for example, 3 , 4 , ...)
Thanks and Rgds,
-- Joseph
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1608.1 | An amendment ? | HGOVC::EDDIELEUNG | Eddie Leung @HGO | Mon May 11 1992 03:37 | 11 |
(1/2) -(1/2)
You must be thinking of 2 or 2 and in general
(1/n)
m m,n being positive integers.
Eddie.
PS : A mathematician is someone who writes A, pronounces B but thinks
C while actually the right answer is D.
| |||||
| 1608.2 | SSAG::LARY | Laughter & hope & a sock in the eye | Mon May 11 1992 04:43 | 16 | |
Its a fairly simple extension of a proof known to the ancient Greeks to prove that if n and m are integers then n^(1/m) is either an integer or irrational; i.e. you can't have 1/m n = p/q, q>1, p and q relatively prime If you could, then you could raise both sides of that equation to the m-th power to get n = p^m / q^m or n * q^m = p^m Since p and q have no common factors, p^m must divide n; i.e. n = k*p^m, but then k * p^m * q^m = p^m, or k * q^m = 1, which means that k = q = 1 and that contradicts the premise q>1. | |||||