Famous Theorems of Mathematics/Root of order n

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

Proof

Let us define a set $A=\{r\in ℝ:r\ge 0,r^n<y\}$.

This set is non-empty (for $0\in A$) and has an upper-bound $y+1$ (for all $r>y+1$ 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 $0<\epsilon <1$ such that $(x+\epsilon {)}^n<y$:

$\begin{array}{cc}(x+\epsilon {)}^n& =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})x^{n-k}{\epsilon }^k\\ & =x^n+\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})x^{n-k}{\epsilon }^k\\ & \le x^n+\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})x^{n-k}\epsilon :{\epsilon }^k\le \epsilon \\ & =x^n+\epsilon \sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})x^{n-k}\\ & =x^n+\epsilon (\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})x^{n-k}-x^n)\\ & =x^n+\epsilon [(x+1{)}^n-x^n]<y\\ \epsilon & <\min \{1,\frac{y-x^n}{(x+1{)}^n-x^n}\}\end{array}$

hence $x+\epsilon \in A$, but $x<x+\epsilon$ and so $x+\epsilon \notin A$. A contradiction.

• Suppose that $x^n>y$.

As before, it is sufficient to find $0<\epsilon <1$ such that $(x-\epsilon {)}^n>y$:

$\begin{array}{cc}(x-\epsilon {)}^n& =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})x^{n-k}(-\epsilon {)}^k\\ & =x^n+\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})x^{n-k}(-\epsilon {)}^k\\ & \ge x^n+\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})x^{n-k}(-\epsilon ):(-\epsilon {)}^k\ge -\epsilon \\ & =x^n-\epsilon \sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})x^{n-k}\\ & =x^n-\epsilon (\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})x^{n-k}-x^n)\\ & =x^n-\epsilon [(x+1{)}^n-x^n]>y\\ \epsilon & <\min \{1,\frac{x^n-y}{(x+1{)}^n-x^n}\}\end{array}$

hence $x-\epsilon \notin A$, but $x-\epsilon <x$ and so $x-\epsilon \in A$. A contradiction.

Therefore $x^n=y$.

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