|
|
|
Viewing Version
3
of
'alternative proof that $\sqrt{2}$ is irrational'
|
[ view 'alternative proof that $\sqrt{2}$ is irrational'
|
back to history
]
| Title of object: |
alternative proof that $\sqrt{2}$ is irrational |
| Canonical Name: |
AlternativeProofThatSqrt2IsIrrational |
| Type: |
Proof |
| Created on: |
2007-04-14 02:48:43 |
| Modified on: |
2007-04-14 04:16:07 |
| Classification: |
msc:11J72, msc:11J82, msc:12E05, msc:13A05 |
Revision comment (for changes between this and next version):
| changed last occurrence of irrational to "not rational" |
Preamble:
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{psfrag}
\usepackage{graphicx}
\usepackage{amsthm}
\usepackage{xypic}
|
Content:
Following is a proof that $\sqrt{2}$ is irrational.
The polynomial $x^2-2$ is irreducible over $\mathbb{Z}$ by Eisenstein's criterion with $p=2$. Thus, $x^2-2$ is irreducible over $\mathbb{Q}$ by \PMlinkname{Gauss's lemma}{GausssLemmaII}. Therefore, $x^2-2$ does not have any roots in $\mathbb{Q}$. Since $\sqrt{2}$ is a root of $x^2-2$, it must be irrational.
This method generalizes to show that any number of the form $\sqrt[r]{n}$ is irrational, where $r \in \mathbb{Z}$ with $r>1$ and $n \in \mathbb{Z}$ such that there exists a prime $p$ dividing $n$ with $p^2$ not dividing $n$. |
|
|
|
|
|