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 |
$\blacksquare$