Existence of Positive Root of Positive Real Number/Positive Exponent/Proof 2

Proof
We have that:
 * $0 < y_1 < y_2 \implies y_1^n < y_2^n$

so there exists at most one $y \in \R: y \ge 0$ such that $y^n = x$.

It remains to demonstrate that there exists at least such a $y$.

Let $S$ be the set consisting of the positive real numbers $t$ such that $t^n < x$.

Let $t = \dfrac x {1 - x}$.

Then $0 < t < 1$ and so:
 * $t^n < t < x$

demonstrating that $S \ne \O$.

Let $t_0 = 1 + x$.

Then $t > t_0$ implies $t^n \ge t > x$.

So $t \notin S$.

Hence $t_0$ is an upper bound of $S$.

By the Continuum Property, $S$ has a supremum.

Let $y = \sup S$.

$y_n < x$.

Choose $h \in \R$ such that $0 < h < 1$ and such that:
 * $h < \dfrac {x - y^n} {\paren {1 + y}^n - y^n}$

Let $\dbinom n m$ denote a binomial coefficient.

By the Binomial Theorem, the coefficient of $z^m$ in the expansion of $\paren {1 + z}^n$ is

We have:

Thus we have $y + h \in S$, which contradicts the fact that $y$ is an upper bound of $S$.

$y^n > x$.

Let $k \in \R$ be chosen such that:
 * $0 < k < 1$
 * $k < y$
 * $k < \dfrac {y^n - x} {\paren {1 + y}^n - y^n}$

Let $t \ge y - k$.

Then:

Thus we have $y - k$ is an upper bound of $S$

This contradicts the fact that $y$ is the supremum of $S$.

It has been shown that both $y^n < x$ and $y^n > x$ lead to a contradiction.

Hence it must be the case that $y^n = x$.

Hence the result.