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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.