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 $x^{n} = y$ .

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

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 $x^{n} = y$ .

It is sufficient to find

0 < ε < 1

such that

(x + ε)^{n} < y

:

\begin{matrix} (x + ε)^{n} & = \sum_{k = 0}^{n} (\binom{n}{k}) x^{n - k} ε^{k} \\ = x^{n} + \sum_{k = 1}^{n} (\binom{n}{k}) x^{n - k} ε^{k} \\ \leq x^{n} + \sum_{k = 1}^{n} (\binom{n}{k}) x^{n - k} ε : ε^{k} \leq ε \\ = x^{n} + ε \sum_{k = 1}^{n} (\binom{n}{k}) x^{n - k} \\ = x^{n} + ε (\sum_{k = 0}^{n} (\binom{n}{k}) x^{n - k} - x^{n}) \\ = x^{n} + ε [(x + 1)^{n} - x^{n}] < y \\ ε & < \min {1, \frac{y - x^{n}}{(x + 1)^{n} - x^{n}}} \end{matrix}

hence

x + ε \in A

, but

x < x + ε

and so

x + ε \notin A

. A contradiction.

As before, it is sufficient to find

0 < ε < 1

such that

(x - ε)^{n} > y

:

\begin{matrix} (x - ε)^{n} & = \sum_{k = 0}^{n} (\binom{n}{k}) x^{n - k} (- ε)^{k} \\ = x^{n} + \sum_{k = 1}^{n} (\binom{n}{k}) x^{n - k} (- ε)^{k} \\ \geq x^{n} + \sum_{k = 1}^{n} (\binom{n}{k}) x^{n - k} (- ε) : (- ε)^{k} \geq - ε \\ = x^{n} - ε \sum_{k = 1}^{n} (\binom{n}{k}) x^{n - k} \\ = x^{n} - ε (\sum_{k = 0}^{n} (\binom{n}{k}) x^{n - k} - x^{n}) \\ = x^{n} - ε [(x + 1)^{n} - x^{n}] > y \\ ε & < \min {1, \frac{x^{n} - y}{(x + 1)^{n} - x^{n}}} \end{matrix}

hence

x - ε \notin A

, but

x - ε < x

and so

x - ε \in A

. A contradiction.

Therefore $x^{n} = y$ .

$◼$

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