Power Function on Base Greater than One is Strictly Increasing/Integer

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Theorem

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

Let $f: \Z \to \R$ be the real-valued function defined as:

$\map f k = a^k$

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


Then $f$ is strictly decreasing.


Proof

Let $a > 1$.

By Power Function on Base Greater than One is Strictly Increasing: Positive Integer, the theorem is already proven for positive integers.

It remains to be proven over the strictly negative integers.


Let $i, j$ be integers such that $i < j < 0$.


From Order of Real Numbers is Dual of Order of their Negatives:

$0 < -j < -i$

So:

\(\ds a^{-j}\) \(<\) \(\ds a^{-i}\) Power Function on Base Greater than One is Strictly Increasing: Positive Integer
\(\ds \leadsto \ \ \) \(\ds \frac 1 {a^j}\) \(<\) \(\ds \frac 1 {a^i}\) Real Number to Negative Power: Positive Integer
\(\ds \leadsto \ \ \) \(\ds a^i\) \(<\) \(\ds a^j\) Ordering of Reciprocals

$\blacksquare$