Power Function is Strictly Increasing over Positive Reals/Natural Exponent

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:

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

$\blacksquare$