Power Function on Base greater than One tends to One as Power tends to Zero/Rational Number/Lemma

From ProofWiki
Jump to navigation Jump to search

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:

$1 < a^r < 1 + a r$




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 $0 < r < 1$:

$1 < a^r < 1 + a r$

$\blacksquare$