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

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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.

$\blacksquare$

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