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.

$\blacksquare$