Proof that if $a \gt 1$ and $r \gt s$, then $a^{r} \gt a^{s}$ I'm trying to prove the sentence above but I'm not getting results, I'm learning precalculus yet. Could someone help me find the starting point to prove the sentence above?
 A: This may be difficult, since exponentials for non-integer exponents are defined in a rather complicated way. But, you could argue that if we let $t=r-s$, we get:
\begin{align*}
a^r = a^{s+t} = a^{s} \cdot a^t
\end{align*}
Since $a>1$ and $t>0$, we have that $a^t > 1$ (can you convince yourself of this?). Therefore $a^r > a^s$ (in particular since $a^r > 0$).
A: You can write $$a^r = a^s a^{r-s}$$
Because $a>1$, $a^{r-s}$ is greater than 1 when the exponent is greater than 0, and given to us is that $r>s$ so $r-s>0$. This gives that $a^r$ is $a^s$ times some number greater than one. Therefore, $a^r > a^s$
A: Step 1: If $a > 0$ then $a^s > 0$ for all $s$.
Step 2:  If $a > 1$ then if $s > 0$ then $a^s> 1$ and if $s = 0$ then $a^s = 1$ and if $s < 0$ then $0 < a^s < 1$.
Step 3:  If $r > s$ 
The $r - s > 0$ so $a^{r-s} > 1$
But $a^{r-s} = \frac {a^s}{a^r}$ so $\frac {a^s}{a^r} > 1$.
And as $a^s > 0$ we get $a^r > a^s$.
....
Proving 1) and 2) for integers is easy.  Proving 1) and 2) for rationals is clear if you raise everything to a common multiple of the denominators of $r$ and $s$.  Prove 1) and 2) for reals depends on how real exponents were defined.  But if they were defined as limits then the result follows by taking limits.
A: There exist rational numbers satisfying 
$$\frac{q_{1}}{p_{1}}> r> \frac{q_2}{p_2} > s > \frac{q_3}{p_3}.$$
(Assume $p_1, p_2, p_3$ are positive integers.)
Because 
$$\frac{q_1 p_2}{p_1 p_2} =\frac{q_{1}}{p_{1}}> \frac{q_2}{p_2} = \frac{p_1 q_2}{p_1 p_2},$$
and because $p_1 p_2$ is positive integer, we know that $q_1 p_2 > p_1 q_2$. 
$a>1$. Hence, 
$$\cdots>a^2>a^1 = a > a^0=1 > a^{-1} = \frac{1}{a} > a^{-2} = \frac{1}{a^2} > \cdots.$$
It means that if $m>n$ are integers, then $a^m > a^n$. Thus, 
$$a^{q_1 p_2} > a^{p_1 q_2} > 0.$$
If $x>y>0$ and $t$ is a positive integer, then $x^{\frac{1}{t}} > y^{\frac{1}{t}}>0$. Thus, 
$$a^{\frac{q_{1}}{p_{1}}} = a^{\frac{q_1 p_2}{p_1 p_2}} > a^{\frac{p_1 q_2}{p_1 p_2}}
= a^{\frac{q_2}{p_2}}. $$
And (by omitting rigorous explanation,) 
$$ a^{\frac{q_{1}}{p_{1}}} > a^r > a^{\frac{q_2}{p_2}}. $$
Similarly, we can show that 
$$a^{\frac{q_2}{p_2}} > a^s > a^{\frac{q_3}{p_3}}. $$
Thus, $a^r > a^s$. 
