Strictly Positive Integer Power Function is Unbounded Above

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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:

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


By Derivative of Power:

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


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

By Power of Strictly Positive Real Number is Positive and Positive Real Numbers Closed under Multiplication it follows that:

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


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.

$\blacksquare$


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