Strictly Positive Integer Power Function is Unbounded Above
Theorem
Let $\R$ be the real numbers with the usual ordering.
Let $n \in \N_{>0}$.
Let $f: \R \to \R$ be defined by:
- $\map f x = x^n$
Then $f$ is unbounded above.
General Case
Let $\struct {R, +, \circ, \le}$ be a totally ordered ring with unity.
Suppose that $R$ has no upper bound.
Let $n \in \N_{>0}$.
Let $f: R \to R$ be defined by:
- $\map f x = \circ^n x$
Then the image of $f$ is unbounded above in $R$.
Proof
If $n = 1$, then $f$ is the identity function.
By the Axiom of Archimedes, the real numbers are unbounded above.
Thus by definition of the identity function: $f$ is unbounded above.
$\Box$
Let $n \ge 2$.
Aiming for a contradiction, suppose that $f$ is bounded above by $b \in \R$.
Without loss of generality suppose that $b > 0$.
Then by the definition of an upper bound:
- $\forall x \in \R: x^n \le b$
Let $x > b$.
Then:
- $\dfrac {x^n - b} {x - b} \le 0$
By the Mean Value Theorem, there exists a point $p$ between $b$ and $x$ such that:
- $\map {f'} p = \dfrac {x^n - b} {x - b}$
- $\map {f'} p = n p^{n - 1}$
Therefore $n p^{n - 1} \le 0$.
By:
and:
- Strictly Positive Real Numbers are Closed under Multiplication it follows that:
- $p > 0 \implies n p^{n - 1} > 0$
But this is impossible because it was previously established that $p \ge b > 0$.
From this contradiction it follows that there can be no such $b$.
Hence the result.
$\blacksquare$