Power of Strictly Positive Real Number is Strictly Positive/Positive Integer
Jump to navigation
Jump to search
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:
\(\ds x^0\) | \(=\) | \(\ds 1\) | Definition of Integer Power | |||||||||||
\(\ds \) | \(>\) | \(\ds 0\) |
Basis for the Induction
$\map P 1$ true, as this just says:
\(\ds x^1\) | \(=\) | \(\ds x\) | Definition of Integer Power | |||||||||||
\(\ds \) | \(>\) | \(\ds 0\) |
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:
\(\ds x^k\) | \(>\) | \(\ds 0\) | Induction Hypothesis | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^{k + 1}\) | \(>\) | \(\ds 0\) | 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:
- $\forall n \in \Z_{\ge 0}: \forall x \in \R_{>0}: x^n > 0$
$\blacksquare$