Famous Theorems of Mathematics/Root of order n

For all $$y>0$$ and for all $$n\in\N$$ there exists $$x\in\R$$ such that $$x^n=y$$.

Proof
Let us define a set $$A=\Big\{r\in\R:r\ge0,r^ny+1$$ we get $$r^n>r>y$$).

Therefore, by the completeness axiom of the real numbers it has a supremum $$x$$. We shall show that $$x^n=y$$.
 * Suppose that $$x^n<y$$.
 * It is sufficient to find $$0<\varepsilon<1$$ such that $$(x+\varepsilon)^n<y$$:
 * $$\begin{align}(x+\varepsilon)^n&=\sum_{k\,=\,0}^n\binom{n}{k}x^{n-k}\varepsilon^k\\&=x^n+\sum_{k\,=\,1}^n\binom{n}{k}x^{n-k}\varepsilon^k\\&\le x^n+\sum_{k\,=\,1}^n\binom{n}{k}x^{n-k}\varepsilon\qquad:\varepsilon^k\le\varepsilon\\&=x^n+\varepsilon\sum_{k\,=\,1}^n\binom{n}{k}x^{n-k}\\&=x^n+\varepsilon\left(\,\sum_{k\,=\,0}^n\binom{n}{k}x^{n-k}-x^n\right)\\&=x^n+\varepsilon\bigl[(x+1)^n-x^n\bigr]y$$.
 * As before, it is sufficient to find $$0<\varepsilon<1$$ such that $$(x-\varepsilon)^n>y$$:
 * $$\begin{align}(x-\varepsilon)^n&=\sum_{k\,=\,0}^n\binom{n}{k}x^{n-k}(-\varepsilon)^k\\&=x^n+\sum_{k\,=\,1}^n\binom{n}{k}x^{n-k}(-\varepsilon)^k\\&\ge x^n+\sum_{k\,=\,1}^n\binom{n}{k}x^{n-k}(-\varepsilon)\qquad:(-\varepsilon)^k\ge-\varepsilon\\&=x^n-\varepsilon\sum_{k\,=\,1}^n\binom{n}{k}x^{n-k}\\&=x^n-\varepsilon\left(\,\sum_{k\,=\,0}^n\binom{n}{k}x^{n-k}-x^n\right)\\&=x^n-\varepsilon\bigl[(x+1)^n-x^n\bigr]>y\\\varepsilon&<\min\left\{1,\frac{x^n-y}{(x+1)^n-x^n}\right\}\end{align}$$
 * hence $$x-\varepsilon\notin A$$, but $$x-\varepsilon<x$$ and so $$x-\varepsilon\in A$$. A contradiction.

Therefore $$x^n=y$$.

$$\blacksquare$$