Existence of Positive Root of Positive Real Number

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Theorem

Let $x \in \R$ be a real number such that $x > 0$.

Let $n \in \Z$ be an integer such that $n \ne 0$.


Then there exists a $y \in \R: y \ge 0$ such that $y^n = x$.


Proof

Positive Exponent

Let $f$ be the real function defined on the unbounded closed interval $\left[{0 \,.\,.\, \to}\right)$ defined by $f \left({y}\right) = y^n$.

Consider first the case of $n > 0$.

By Strictly Positive Integer Power Function is Unbounded Above:

$\exists q \in \R_{>0}: f \paren q \ge x$

Since $x \ge 0$:

$f \paren 0 \le x$

By the Intermediate Value Theorem:

$\exists y \in \R: 0 \le y \le q, f \paren y = x$

Hence the result has been shown to hold for $n > 0$.

$\Box$


Negative Exponent

Let $m = -n$.

Then $m > 0$.

Let $g$ be the real function defined on $\hointr 0 \to$ defined by:

$\map g y = y^m$

Since $x \ge 0$:

$\dfrac 1 x \ge 0$

By Existence of Positive Root of Positive Real Number: Positive Exponent there is a $y > 0$ such that:

$\map g y = \dfrac 1 x$

It follows from the definition of power that:

\(\displaystyle \map f y\) \(=\) \(\displaystyle \dfrac 1 {\map g y}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {1 / x}\)
\(\displaystyle \) \(=\) \(\displaystyle x\)

$\blacksquare$


Also see