alternative proof that $\sqrt{2}$ is irrationalAlternativeProofThatSqrt2IsIrrational2007-04-14 02:48:432007-08-23 19:22:46Proofsquare root of 20\usepackage{amssymb}
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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 not rational, 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 $n$.