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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,
[\[ x^{\frac{1}{n} } := \sup\{y \in \mathbb{R} : ? \] ]