Famous Theorems of Mathematics/Root of order n

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

Proof

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

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 $0<\varepsilon <1$ such that $(x+\varepsilon )^{n}<y$:

${\begin{aligned}(x+\varepsilon )^{n}&=\sum _{k\,=\,0}^{n}{\binom {n}{k}}x^{n-k}\varepsilon ^{k}\\&=x^{n}+\sum _{k\,=\,1}^{n}{\binom {n}{k}}x^{n-k}\varepsilon ^{k}\\&\leq x^{n}+\sum _{k\,=\,1}^{n}{\binom {n}{k}}x^{n-k}\varepsilon \qquad :\varepsilon ^{k}\leq \varepsilon \\&=x^{n}+\varepsilon \sum _{k\,=\,1}^{n}{\binom {n}{k}}x^{n-k}\\&=x^{n}+\varepsilon \left(\,\sum _{k\,=\,0}^{n}{\binom {n}{k}}x^{n-k}-x^{n}\right)\\&=x^{n}+\varepsilon {\bigl [}(x+1)^{n}-x^{n}{\bigr ]}<y\\\varepsilon &<\min \left\{1,{\frac {y-x^{n}}{(x+1)^{n}-x^{n}}}\right\}\end{aligned}}$

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

• Suppose that $x^{n}>y$.

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

${\begin{aligned}(x-\varepsilon )^{n}&=\sum _{k\,=\,0}^{n}{\binom {n}{k}}x^{n-k}(-\varepsilon )^{k}\\&=x^{n}+\sum _{k\,=\,1}^{n}{\binom {n}{k}}x^{n-k}(-\varepsilon )^{k}\\&\geq x^{n}+\sum _{k\,=\,1}^{n}{\binom {n}{k}}x^{n-k}(-\varepsilon )\qquad :(-\varepsilon )^{k}\geq -\varepsilon \\&=x^{n}-\varepsilon \sum _{k\,=\,1}^{n}{\binom {n}{k}}x^{n-k}\\&=x^{n}-\varepsilon \left(\,\sum _{k\,=\,0}^{n}{\binom {n}{k}}x^{n-k}-x^{n}\right)\\&=x^{n}-\varepsilon {\bigl [}(x+1)^{n}-x^{n}{\bigr ]}>y\\\varepsilon &<\min \left\{1,{\frac {x^{n}-y}{(x+1)^{n}-x^{n}}}\right\}\end{aligned}}$

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

Therefore $x^{n}=y$.

$\blacksquare$