Existence of Positive Root of Positive Real Number/Positive Exponent/Proof 2
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
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:
\(\ds \paren {y + h}^n\) | \(=\) | \(\ds y^n + \dbinom n 1 y^{n - 1} h + \dbinom n 2 y^{n - 2} h^2 + \dotsb + \dbinom n n h^n\) | Binomial Theorem | |||||||||||
\(\ds \) | \(\le\) | \(\ds y^n + h \paren {\dbinom n 1 y^{n - 1} + \dbinom n 2 y^{n - 2} + \dotsb + \dbinom n n}\) | as $0 < h < 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds y^n + h \paren {\paren {1 + y}^n - y^n}\) | Binomial Theorem | |||||||||||
\(\ds \) | \(<\) | \(\ds y^n + \paren {x - y^n}\) | Definition of $h$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x\) |
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:
\(\ds t^n\) | \(\ge\) | \(\ds \paren {y - k}^n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds y^n - \dbinom n 1 y^{n - 1} h + \dbinom n 2 y^{n - 2} h^2 - \dotsb + \paren {-1} \dbinom n n h^n\) | Binomial Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds y^n - k \paren {\dbinom n 1 y^{n - 1} + \dbinom n 2 y^{n - 2} k - \dotsb + \paren {-1} \dbinom n n k^{n - 1} }\) | ||||||||||||
\(\ds \) | \(\ge\) | \(\ds y^n - k \paren {\dbinom n 1 y^{n - 1} + \dbinom n 2 y^{n - 2} + \dotsb + \dbinom n n}\) | as $0 < k < 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds y^n - k \paren {\paren {1 + y}^n - y^n}\) | Binomial Theorem | |||||||||||
\(\ds \) | \(>\) | \(\ds y^n - \paren {y^n - x}\) | Definition of $k$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x\) |
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.
$\blacksquare$
Sources
- 1964: Walter Rudin: Principles of Mathematical Analysis (2nd ed.) ... (previous) ... (next): Chapter $1$: The Real and Complex Number Systems: Real Numbers: $1.37$. Theorem