Exercise 1.18

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Prove that $\sqrt[n]{m}$ is irrational if $m$ is not the $n$-th power of an integer.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
Here $m \in \N$.

Suppose that $r = \sqrt[n]{m} \in \Q$. As $r\geq 0$, $r = a/b, a\geq 0, b >0, a\wedge b = 1$. Moreover $r^n = m$, thus $a^n = m b^n$.

For every prime $p$, $n\  \mathrm{ord}_p(a) = \mathrm{ord}_p(m)  + n\  \mathrm{ord}_p(b)$, so $n \mid \mathrm{ord}_p(m) $.

From Ex. 1.15, we conclude that $m$ is a $n$-th power.

Conclusion : if $m \geq 0$ is not the $n$-th power of an integer, $\sqrt[n]{m}$ is irrational.  
\end{proof}

Answers