# if ${a}^{n}$ is irrational then $a$ is irrational

###### Theorem.

If $a$ be a real number and $n$ is an integer such that ${a}^{n}$ is irrational, then $a$ is irrational.

###### 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 $c\ne 0$ such that $a={\displaystyle \frac{b}{c}}$. Thus, ${a}^{n}={\displaystyle \frac{{b}^{n}}{{c}^{n}}}$, which is a rational number^{}.
∎

Note that the converse^{} is not true. For example, $\sqrt{2}$ is irrational and ${\left(\sqrt{2}\right)}^{2}=2$ is rational.

Title | if ${a}^{n}$ is irrational then $a$ is irrational |
---|---|

Canonical name | IfAnIsIrrationalThenaIsIrrational |

Date of creation | 2013-03-22 14:18:50 |

Last modified on | 2013-03-22 14:18:50 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 13 |

Author | Wkbj79 (1863) |

Entry type | Theorem |

Classification | msc 11J82 |

Classification | msc 11J72 |