How to show $b^t < b^r$ for $b>1$, if $t < r$ and both are rationals, using only basic tools of analysis? Assume I have only read chapter 1 of Rudin up to exercise 6b). So we are still proving the properties of exponentiation and don't know how to fully manipulate exponents. Right now I am stuck showing the following (:
$$ \text{ if } t < r \implies b^t < b^r $$
where $b>1$ and $t,r \in \mathbb{Q}$
I know it's "obvious" because exponentiation is monotone (which can be easily shown since its derivative is positive, though we have not defined derivatives so that makes no sense as a justification just yet). Thus, I was wondering how one goes about proving the above statement.
What I attempted was to re-write exponents of rationals with the tools that we do know. We know $b^{t} = b^{t_1/t_2} = \left( b^{t_1} \right)^{\frac{1}{t_2}} = \left( b^{\frac{1}{t_2}} \right)^{t_1}$ and $b^{r} = b^{r_1/r_2} = \left( b^{r_1} \right)^{\frac{1}{r_2}} = \left( b^{\frac{1}{r_2}} \right)^{r_1}$ so I re-wrote it as follows:
$$ \left( b^{r_1} \right)^{\frac{1}{r_2}} \text{ vs } \left( b^{t_1} \right)^{\frac{1}{t_2}}$$
and raise both expression to the $t_2 r_2$ which yielded:
$$ b^{t_1 r_2} \leq b^{r_1 t_2}$$
since $t_1 r_2 \leq r_1 t_2$. This is obviously correct because both the exponents are just natural numbers and $b>1$ so nothing fishy is going on. However, its unclear how that implies what I want, specially because taking roots of both since (which is tempting to do just take $1/r_2$ root of both sides and then do that with $t_2$) and voila you would "get the answer". But taking roots of both sides and hoping the inequality keep holding is exactly what I am having trouble proving. Is there something I am missing? I am fine with the answer but I am also happy with hints (and maybe hide the answer?)? 

EDIT:
I just notice that:
$$t_1 r_2 \leq r_1 t_2$$
is not necessarily true. We know $r,t \in \mathbb{Q}$ but we don't know if the denominators or numerators are negative or not or if $r,t$ themselves are negative, so we don't know if there are any weird sign shifts when playing with the inequality.
 A: I will present a framework. Here, the "stronger theorem" is the theorem with "if" replaced by "if and only if"
1. Prove that the stronger theorem is true on natural numbers by induction.
This should be the most tedious part of the process, but I will leave that to you. Note that $b$ is still an arbitrary rational number.
2. Prove that the stronger theorem is true on integers.
This would require using the definition of negative power as the reciprocal of positive power, and then using some basic theorems (or axioms) of ordering. Note that $b$ is still an arbitrary rational number.
3. Prove that the theorem is true on rational numbers.
$$\begin{array}{rcll}
b^t &<& b^r \\
(b^t)^{t_2~r_2} &<& (b^r)^{t_2~r_2} & \text{by 2} \\
b^{t_1~r_2} &<& b^{t_2~r_1} \\
t_1 r_2 &<& t_2 r_1 & \text{by 2} \\
t &<& r
\end{array}$$
A: So you see that it suffices to show that  $1 < b \leq c$ and $n \in \mathbb{N}^*$ implies $b^{1/n} \leq c^{1/n}$. Assume to the contrary that $b^{1/n} > c^{1/n}$. Then raising both sides to the power of $n$ yields $b > c$, a contradiction$^*$.

$^*$ As a consequence of Proposition 1.18b) of Rudin, if $0 < x < y$, then $x^2 < xy$ (multiply both sides by $x$) and also $xy < y^2$ (multiply both sides by $y$). Therefore $x^2 < y^2$. An easy induction using this idea lets us conclude that $x^n < y^n$ for any $n \in \mathbb N ^*$ whenever $0 < x < y$.
A: Let $n \in \mathbb N$.  Then $b^n = b^n*1 < b^n*b = b^{n+1}$.  By transitivity of order and by induction, for $m,n \in \mathbb N$, $n < m \iff b^n < b^m$.
we can extend this to $n \in \mathbb Z$, as $b > 1 \implies \frac 1b < 1$ we have $b^n = b^n*1 < b^n*\frac 1b = b^{n-1}$ so by induction and transitivity of order, for $m,n \in \mathbb Z$, $n < m \iff b^n < b^m$.we can extend this to $n \in \mathbb Z$.
An aside:
For $b,c > 0$ and $ n \in \mathbb N$,  $b^n < c^n \iff b <c$ and in particular $b^n > 1 \iff b>1$.  This is easy to via induction as $b^n = b*.....*b < c*......*c = c^n \iff b < c$.
With that under our belts...
To extend to the rationals:
Remember $b^{\frac 1n}$ is the unique positive real $x$ so that $x^n = b$.  $x^n = b >  1$ so $x > 1$. 
In Exercise 6b) has you proved that for $r = \frac mn; m, n \in \mathbb Z; n > 0 $ defining $b^r = (b^{\frac 1n})^m$ is well defined.
Then $r= \frac mn < s = \frac pq \iff mq < pn \iff b^r= b^{\frac mn} = (b^{\frac 1 {nq}})^{mq} < (b^{\frac 1{nq}})^{pn}= b^{\frac pq} = b^s$
A: We consider:$$t<r \Rightarrow \exists x \in \mathbb{Q}_{>0} : t<r = t+x $$
Now the inequality to be proved becomes:
$$b^t<b^r\Rightarrow b^t<b^{r} = b^{t+x} \Rightarrow b^t<b^t\cdot b^x \Rightarrow 1<b^x.$$
Finally, raising both sides to the power $\frac1x$ (note that $x>0 \Rightarrow\frac1x>0)$ we get:
$$b>1.$$
A: The expression $b^{r} $ when $r$ is rational can be defined using algebraic processes and all its properties can be proved via algebra. The only part where we need analysis is the following theorem:

If $b>0$ and $n$ is a positive integer then there is a unique positive real number $a$ such that $a^{n} =b$. 

This theorem defines the symbol $b^{1/n}=a$ and if rational $r=m/n$ with $m, n\in\mathbb{Z}, n>0$ then we define $b^{r} =(b^{1/n})^{m}$. 
The question assumes $b>1$ and two rationals $r, t$ with $t<r$ and we are supposed to prove $b^{t} <b^{r} $. This involves only algebra. One approach (the simpler one) is to let $p=r-t>0$ and then show that $b^{p} >1$. We have $p=m/n$ with $m, n$ as positive integers. And we can see that $b^{m} \geq b >1$ and hence taking $n$-th root we get $b^{p} >1$ and multiplying by positive number $b^{t}$ we get $b^{r} >b^{t} $. 
The process of taking $n$-th roots preserves inequalities and technically we say that for any positive integer $n$ the function $f(x) =x^{1/n}$ is strictly increasing for $x>0$. The proof of this is dependent on the fact that $g(x) =x^{n} $ is strictly increasing for $x>0$. Moreover it is a very roundabout way to prove monotone nature of $f$ or $g$ via processes of calculus when the result is achieved via algebra.
The case for $g$ is simple. We have for positive $x, y$ $$\frac{x^{n}-y^{n}} {x-y} = x^{n-1}+x^{n-2}y+\dots+xy^{n-2}+y^{n-1}>0$$ so that $x^{n} - y^{n} $ has same sign as $x-y$. This proves that $g$ is strictly increasing in $(0,\infty)$.
Next we deduce the increasing nature of $f$ from  that of $g$ via contradiction. Let $0<x<y$ and assume that $f(x) \geq f(y) $ that is $x^{1/n}\geq y^{1/n}$ and raising to power $n$ we get (via the strictly increasing nature of $g$) $x\geq y$ contrary to $x<y$. Hence we arrive at a contradiction and we must have $f(x) <f(y) $ so that $f$ is strictly increasing in $(0,\infty)$.
Whatever has been proved above involves algebraic manipulation of inequalities and the use of words like "monotone" or "strictly increasing" does not in anyway indicate the use of processes of analysis. Also understand that the above results are typically studied when one starts to learn surds and rational exponents in high school. 
