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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 \( E = \{y \in \mathbb{R} : y \ge 0 \text{ and } y^n < x \} \). We define the \( n^{th} \) root of \( x \), denoted as \( x^{\frac{1}{n}} \), to be \( x^{\frac{1}{n}} := \sup E \). Thus,
\[ x^{\frac{1}{n} } := \sup\{y \in \mathbb{R} : y \ge 0 \text{ and } y^n < x \} \]

The existence of the nth root, proposition
Let \( x \ge 0 \) be a positive real and let \( n \ge 1 \) be a positive integer. Then \( x^{\frac{1}{n}} \) is a real number (it exists).

An easy proof
Outline: we must show that the set \( \{y \in \mathbb{R} : y \ge 0 \text{ and } y^n < x \} \) has an upper bound; if it has an upper bound, it follows that it has a least upper bound and this least upper bound is the nth root of \( x \), by definition.

Let E be the set \( \{y \in \mathbb{R} : y \ge 0 \text{ and } y^n < x \} \). 0 is an element of E ( \( 0 \ge 0 \text{ and } 0^k = 0 \) for any natural number \( k \)). Thus E is non-empty. E also has an upper bound. To show this consider the two possible cases, \( x < 1 \) and \( x \ge 1 \). 

If \( x < 1 \), then 1 is an upper bound of E. If it isn't then \( 1 < y \) for some \( y \in E \). \( y > 1 \) implies that \( y^n > 1 \) so that \( y \) doesn't meet the criteria for being an element of E, a contradiction.

If \( x \ge 1 \), then \( x \) is an upper bound of E. If it isn't then \( y > x \) for some \( y \in E \). This implies that \( y^n > x \), so again, \( y \) cannot be an element of E, a contradiction.

So E must have an upper bound. As it is a non-empty set of reals with an upper bound, it must have a least upper bound. Thus \( x^{\frac{1}{n}} \) is a real number.