Power Function on Base greater than One tends to One as Power tends to Zero/Rational Number/Lemma
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Theorem
Let $a \in \R$ be a real number such that $a > 1$.
Let $r \in \Q_{> 0}$ be a strictly positive rational number such that $r < 1$.
Then:
- $1 < a^r < 1 + a r$
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Proof
Define a real function $g_r: \R_{> 0} \to \R$ as:
- $\map {g_r} a = 1 + a r - a^r$
Then differentiating with respect to $a$ gives:
- $D_a \map {g_r} a = r \paren {1 - a^{r - 1} }$
We show now that the derivative of $g_r$ is positive for all $a > 1$:
\(\ds r\) | \(<\) | \(\ds 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds r - 1\) | \(<\) | \(\ds 0\) | Subtract $1$ from both sides | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a^{r - 1}\) | \(<\) | \(\ds a^0\) | Power Function on Base Greater than One is Strictly Increasing: Rational Number | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a^{r - 1}\) | \(<\) | \(\ds 1\) | Definition of Integer Power | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds -1\) | \(<\) | \(\ds -a^{r - 1}\) | Order of Real Numbers is Dual of Order of their Negatives | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\ds 1 - a^{r - 1}\) | adding $1$ to both sides | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\ds r \left({1 - a^{r - 1} }\right)\) | multiplying both sides by $r > 0$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\ds D_a \map {g_r} a\) | from the formula found for $D_a \map {g_r} a$ above |
So $D_a \map {g_r} a$ is positive for all $a > 1$.
Whence, by Derivative of Monotone Function, $g_r$ is increasing for all $a > 1$.
Now:
- $\map {g_r} 1 = r > 0$
So $\map {g_r} a$ is positive for all $a > 1$.
That is:
- $0 < 1 + a r - a^r$
Adding $a^r$ to both sides of the above yields:
- $a^r < 1 + a r$
Finally:
\(\ds 0\) | \(<\) | \(\ds r\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds a^0\) | \(<\) | \(\ds a^r\) | Power Function on Base Greater than One is Strictly Increasing: Rational Number | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 1\) | \(<\) | \(\ds a^r\) | Definition of Integer Power |
So, for $0 < r < 1$:
- $1 < a^r < 1 + a r$
$\blacksquare$