Power of Strictly Positive Real Number is Strictly Positive/Positive Integer

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Theorem

Let $x \in \R_{>0}$ be a (strictly) positive real number.

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


Then:

$x^n > 0$

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


Proof

Proof by Mathematical Induction:

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$\forall x \in \R_{>0}: x^n > 0$


$\map P 0$ is true, as this just says:

\(\displaystyle x^0\) \(=\) \(\displaystyle 1\) Definition of Integer Power
\(\displaystyle \) \(>\) \(\displaystyle 0\)


Basis for the Induction

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

\(\displaystyle x^1\) \(=\) \(\displaystyle x\) Definition of Integer Power
\(\displaystyle \) \(>\) \(\displaystyle 0\)

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 0 $, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

$\forall x \in \R_{>0}: x^k > 0$


Then we need to show:

$\forall x \in \R_{>0}: x^{k + 1} > 0$


Inductive Step

This is our induction step:


\(\displaystyle x^k\) \(>\) \(\displaystyle 0\) Induction Hypothesis
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^{k + 1}\) \(>\) \(\displaystyle 0\) Multiply both sides by $x > 0$


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


Therefore:

$\forall n \in \Z_{\ge 0}: \forall x \in \R_{>0}: x^n > 0$

$\blacksquare$