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

## 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 $0 < r < 1$:

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

$\blacksquare$