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[parent] if $a^n$ is irrational then ${a}$ is irrational (Theorem)
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\neq 0$ such that $ \displaystyle a=\frac{b}{c}$. Thus, $ \displaystyle a^n=\frac{b^n}{c^n}$, which is a rational number. $ \qedsymbol$

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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Cross-references: converse, rational number, rational, contrapositive, irrational, integer, real number
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This is version 10 of if $a^n$ is irrational then ${a}$ is irrational, born on 2004-04-18, modified 2008-04-30.
Object id is 5779, canonical name is IfAnIsIrrationalThenAIsIrrational.
Accessed 2693 times total.

Classification:
AMS MSC11J72 (Number theory :: Diophantine approximation, transcendental number theory :: Irrationality; linear independence over a field)
 11J82 (Number theory :: Diophantine approximation, transcendental number theory :: Measures of irrationality and of transcendence)

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