\( \newcommand{\matr}[1] {\mathbf{#1}} \newcommand{\vertbar} {\rule[-1ex]{0.5pt}{2.5ex}} \newcommand{\horzbar} {\rule[.5ex]{2.5ex}{0.5pt}} \)
deepdream of a sidewalk
Show Answer
\( \newcommand{\cat}[1] {\mathrm{#1}} \newcommand{\catobj}[1] {\operatorname{Obj}(\mathrm{#1})} \newcommand{\cathom}[1] {\operatorname{Hom}_{\cat{#1}}} \newcommand{\multiBetaReduction}[0] {\twoheadrightarrow_{\beta}} \newcommand{\betaReduction}[0] {\rightarrow_{\beta}} \newcommand{\betaEq}[0] {=_{\beta}} \newcommand{\string}[1] {\texttt{"}\mathtt{#1}\texttt{"}} \newcommand{\symbolq}[1] {\texttt{`}\mathtt{#1}\texttt{'}} \)
Math and science::Analysis::Tao::05. The real numbers

\( n^{th} \) root of a real

The definition of the nth root of a real utilizes the supremum, hence why the \( n^{th} \) doesn't exist for rationals.

Let \( 0 \ge x \) be a non-negative real. Let E be the set [\( E = \{ ? \} \)]. We define the \( n^{th} \) root of \( x \), denoted as \( x^{\frac{1}{n}} \), to be \( x^{\frac{1}{n}} := \sup E \). Thus,
[\[ x^{\frac{1}{n} } := \sup\{y \in \mathbb{R} : ? \] ]