# 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:

$f \left({x}\right) = x^n$

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 Archimedean Principle, 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 $x \gt b$.

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:

$f' \left({p}\right) = \dfrac{x^n - b} {x - b}$
$f' \left({p}\right) = n p^{n - 1}$

Therefore $n p^{n - 1} \le 0$.

$p \gt 0 \implies n p^{n - 1} \gt 0$

But this is impossible because it was previously established that $p \ge b \gt 0$.

From this contradiction it follows that there can be no such $b$.

Hence the result.

$\blacksquare$