# 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:

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

## 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 $a > 1$:

\(\displaystyle r\) | \(<\) | \(\displaystyle 1\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle r - 1\) | \(<\) | \(\displaystyle 0\) | Subtract $1$ from both sides | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle a^{r - 1}\) | \(<\) | \(\displaystyle a^0\) | Power Function on Base Greater than One is Strictly Increasing: Rational Number | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle a^{r - 1}\) | \(<\) | \(\displaystyle 1\) | Definition of Integer Power | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle -1\) | \(<\) | \(\displaystyle -a^{r - 1}\) | Order of Real Numbers is Dual of Order of their Negatives | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 0\) | \(<\) | \(\displaystyle 1 - a^{r - 1}\) | adding $1$ to both sides | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 0\) | \(<\) | \(\displaystyle r \left({1 - a^{r - 1} }\right)\) | multiplying both sides by $r > 0$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 0\) | \(<\) | \(\displaystyle D_a \map {g_r} a\) | 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:

\(\displaystyle 0\) | \(<\) | \(\displaystyle r\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle a^0\) | \(<\) | \(\displaystyle a^r\) | Power Function on Base Greater than One is Strictly Increasing: Rational Number | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 1\) | \(<\) | \(\displaystyle a^r\) | Definition of Integer Power |

So, for $0 < r < 1$:

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

$\blacksquare$