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'if $a^n$ is irrational then ${a}$ is irrational'
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| Title of object: |
if $a^n$ is irrational then ${a}$ is irrational |
| Canonical Name: |
IfAnIsIrrationalThenAIsIrrational |
| Type: |
Theorem |
| Created on: |
2004-04-18 15:40:32 |
| Modified on: |
2007-12-05 13:00:30 |
| Classification: |
msc:11J72, msc:11J82 |
Preamble:
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amsthm}
\newtheorem*{thm*}{Theorem} |
Content:
\begin{thm*}
If $a$ be a real number and $n$ is an integer such that $a^n$ is irrational, then $a$ is irrational.
\end{thm*}
\begin{proof}
We show this by way of contrapositive. In other words, we show that, if $a$ is rational, then $a^n$ is rational.
Let $a$ be rational. Then there exist integers $b$ and $c$ with $c\neq 0$ such that $\displaystyle a=\frac{b}{c}$. Thus, $\displaystyle a^n=\frac{b^n}{c^n}$, which is a rational number.
\end{proof}
Note that the converse is not true. For example, $\sqrt{2}$ is irrational and $\left(\sqrt{2}\right)^2=2$ is rational. |
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