IfAnIsIrrationalThenAIsIrrational 2004-04-18 15:40:32 2008-04-30 17:19:31 Theorem irrational \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amsthm} \newtheorem*{thm*}{Theorem} \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.