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