# Power Function on Base between Zero and One is Strictly Decreasing/Positive Integer

## Theorem

Let $a \in \R$ be a real number such that $0 < a < 1$.

Let $f: \Z_{\ge 0} \to \R$ be the real-valued function defined as:

- $\map f n = a^n$

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

Then $f$ is strictly decreasing.

## Proof

Proof by induction on $n$:

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

- $a^{n + 1} < a^n$

$\map P 0$ is the case:

\(\ds a^1\) | \(=\) | \(\ds a\) | Definition of Integer Power | |||||||||||

\(\ds \) | \(<\) | \(\ds 1\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds a^0\) | Definition of Integer Power |

### Basis for the Induction

$\map P 1$ is true, since:

\(\ds a\) | \(<\) | \(\ds 1\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds a \times a\) | \(<\) | \(\ds 1 \times a\) | Real Number Ordering is Compatible with Multiplication | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds a^2\) | \(<\) | \(\ds a^1\) | 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:

- $a^{k + 1} < a^k$

Then we need to show:

- $a^{k + 2} < a^{k + 1}$

### Induction Step

This is our induction step:

\(\ds a^{k + 1}\) | \(<\) | \(\ds a^k\) | Induction Hypothesis | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds a \times a^{k + 1}\) | \(<\) | \(\ds a \times a^k\) | Real Number Ordering is Compatible with Multiplication | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds a^{k + 2}\) | \(<\) | \(\ds a^{k + 1}\) | Definition of Integer Power |

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

Therefore:

- $\forall n \in \Z_{\ge 0}: a^{n + 1} < a^n$

Hence the result, by definition of strictly decreasing.

$\blacksquare$