Power Function is Strictly Increasing over Positive Reals/Natural Exponent

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Theorem

Let $n \in \Z_{>0}$ be a strictly positive integer.

Let $f: \R_{>0} \to \R$ be the real function defined as:

$\map f x = x^n$

where $x^n$ denotes $x$ to the power of $n$.


Then $f$ is strictly increasing.


Proof

Proof by induction on $n$:

Let $x, y \in \R_{>0}$ be strictly positive real numbers.

For all $n \in \Z_{>0}$, let $P \left({n}\right)$ be the proposition:

$x < y \implies x^n < y^n$


Basis for the Induction

$\map P 1$ is true, as this just says:

\(\ds x^1\) \(=\) \(\ds x\) Definition of Integer Power
\(\ds \) \(<\) \(\ds y\) by hypothesis
\(\ds \) \(=\) \(\ds y^1\) Definition of Integer Power


This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

$x < y \implies x^k < y^k$


Then we need to show:

$x < y \implies x^{k + 1} < y^{k + 1}$


Induction Step

This is our induction step:

First:

\(\ds x\) \(<\) \(\ds y\)
\(\ds \leadsto \ \ \) \(\ds x^k\) \(<\) \(\ds y^k\) Induction Hypothesis
\(\ds \leadsto \ \ \) \(\ds x^{k+1}\) \(<\) \(\ds x \times y^k\) Multiply both sides by $x > 0$
\(\ds \) \(<\) \(\ds y \times y^k\) Multiply both sides of $x < y$ by $y^k > 0$
\(\ds \) \(=\) \(\ds y^{k+1}\) Definition of integer power


So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \Z_{>0}: x < y \implies x^n < y^n$

$\blacksquare$

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