# 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:

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

$\blacksquare$