Existence of Positive Root of Positive Real Number/Positive Exponent

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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 $n > 0$.


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


Proof 1

Let $f$ be the real function defined on the unbounded closed interval $\hointr 0 \to$ defined by $\map f y = y^n$.

Consider first the case of $n > 0$.

By Strictly Positive Integer Power Function is Unbounded Above:

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

Since $x \ge 0$:

$\map f 0 \le x$

By the Intermediate Value Theorem:

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

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

$\blacksquare$


Proof 2

Let $S$ be the set consisting of the positive real numbers $t$ such that $t^n < x$.

Let $t = \dfrac x {1 - x}$.

Then $0 < t < 1$ and so:

$t^n < t < x$

demonstrating that $S \ne \O$.

Let $t_0 = 1 + x$.

Then $t > t_0$ implies $t^n \ge t > x$.

So $t \notin S$.

Hence $t_0$ is an upper bound of $S$.

By the Continuum Property, $S$ has a supremum.


Let $y = \sup S$.

Aiming for a contradiction, suppose $y_n < x$.

Choose $h \in \R$ such that $0 < h < 1$ and such that:

$h < \dfrac {x - y^n} {\paren {1 + y}^n - y^n}$

Let $\dbinom n m$ denote a binomial coefficient.

By the Binomial Theorem, the coefficient of $z^m$ in the expansion of $\paren {1 + z}^n$ is

We have:

\(\displaystyle \paren {y + h}^n\) \(=\) \(\displaystyle y^n + \dbinom n 1 y^{n - 1} h + \dbinom n 2 y^{n - 2} h^2 + \dotsb + \dbinom n n h^n\) Binomial Theorem
\(\displaystyle \) \(\le\) \(\displaystyle y^n + h \paren {\dbinom n 1 y^{n - 1} + \dbinom n 2 y^{n - 2} + \dotsb + \dbinom n n}\) as $0 < h < 1$
\(\displaystyle \) \(=\) \(\displaystyle y^n + h \paren {\paren {1 + y}^n - y^n}\) Binomial Theorem
\(\displaystyle \) \(<\) \(\displaystyle y^n + \paren {x - y^n}\) Definition of $h$
\(\displaystyle \) \(=\) \(\displaystyle x\)

Thus we have $y + h \in S$, which contradicts the fact that $y$ is an upper bound of $S$.


Aiming for a contradiction, suppose $y^n > x$.

Let $k \in \R$ be chosen such that:

$0 < k < 1$
$k < y$
$k < \dfrac {y^n - x} {\paren {1 + y}^n - y^n}$

Let $t \ge y - k$.

Then:

\(\displaystyle t^n\) \(\ge\) \(\displaystyle \paren {y - k}^n\)
\(\displaystyle \) \(=\) \(\displaystyle y^n - \dbinom n 1 y^{n - 1} h + \dbinom n 2 y^{n - 2} h^2 - \dotsb + \paren {-1} \dbinom n n h^n\) Binomial Theorem
\(\displaystyle \) \(=\) \(\displaystyle y^n - k \paren {\dbinom n 1 y^{n - 1} + \dbinom n 2 y^{n - 2} k - \dotsb + \paren {-1} \dbinom n n k^{n - 1} }\)
\(\displaystyle \) \(\ge\) \(\displaystyle y^n - k \paren {\dbinom n 1 y^{n - 1} + \dbinom n 2 y^{n - 2} + \dotsb + \dbinom n n}\) as $0 < k < 1$
\(\displaystyle \) \(=\) \(\displaystyle y^n - k \paren {\paren {1 + y}^n - y^n}\) Binomial Theorem
\(\displaystyle \) \(>\) \(\displaystyle y^n - \paren {y^n - x}\) Definition of $k$
\(\displaystyle \) \(=\) \(\displaystyle x\)

Thus we have $y - k$ is an upper bound of $S$

This contradicts the fact that $y$ is the supremum of $S$.


It has been shown that both $y^n < x$ and $y^n > x$ lead to a contradiction.

Hence it must be the case that $y^n = x$.


Hence the result.

$\blacksquare$