Prove $b>1, r>0 \implies b^r > 1$ The proof is supposed to be extremely elementary (using on Rudin's Principles of Mathematical Analysis Chapter 1 material).
This actually is not the main problem, but I have simplified that problem down to this single lemma, but I cannot get it.
I have tried to use the Archimediean property, the fact that there is a real between any two integers, and several other assails, but to no avail.
The following is already avialable and can be used for the proof:
We have iterated multiplication for integer exponents. Rational exponents are given via the multiplicative property (which was proven) and the existence and uniqueness of positive roots of positive numbers. Then, for real $r$, $b^r = \operatorname{sup}(\{ b^t: t \in \mathbb {Q}, t \leq r \})$. We have the additive property for real exponents. We have the density of the rational in the reals, we have that every real is between an integer and its successor, and we have the Archimedean property.
Can you help me out?
 A: Hint: the function $x^y$ is increasing as a function of both $x$ and $y$. 
A: Let $b = 1 + x$, for $x > 0$.
Then you have $$b^r = (1+x)^r = 1 + S > 1,$$ where $$S = \sum C_n^k x^k >0.$$
A: Well you have then that if $r$>0 then there exists positive integers $p$ and $q$ such that $0<p/q<r$ and cosequently $b^{p/q} < b^r$ so it's enough to show that $b^{p/q}$ is greater than zero. But if $b > 1$ then $b^p > 1$ and if $x = b^{p/q}$ then $x^q = b^p$ (with x being positive by definition if $q$ is even), but being larger than $1$, $x$ could not be anything other than larger than one since otherwise $x^q$ would be less or equal to one.
This relies on the fact than $x^n$ will be larger than one if $x>1$, less if $x<1$ and $1$ if $x=1$ for positive integer $n$. This can be proven by induction as it's true for $n=1$ and if its true fore $n=j$ then $x^{j+1} = x^j x$ and $x^j$ and $x$ both being less than/equal to/greater than $1$ then so is their product.
A: I assume that you already have shown that
$$(b^t)^n=b^{tn}, t>0,n>0$$
at least for $t \in \mathbb{Q^+}$ and $n \in \mathbb{N}$.
Then you can prove the following monotonicity statements by contradiction.
Let $p,q \in \mathbb{N}$ and $b,c \in \mathbb{R^+}$:


*

*if $b>c$ then $b^\frac{p}{q}>c^\frac{p}{q}$.

*if $\frac{p}{q}<\frac{u}{v}$ and $b>1$ then $b^\frac{p}{q}<b^\frac{u}{v}$


Use this to show:
If $b>1$ and $r \in \mathbb{R}$ and $r>1$ then the set $\{b^t \mid t \in \mathbb{Q^+} \}$ is bounded above an so a supremum exists. This supremum is greater than $1$. So $b^r>1$.
A: You need to use the fact that
1) $b^{r} > 1$ if $b > 1, r > 0$ and $r$ is rational.
If $r$ is irrational and $r > 0$ then there is a rational $s$ with $0 < s < r$ and then we know that $b^{r} = \sup \{b^{t}\mid t \in \mathbb{Q}, t \leq r\}$ and since $s \in \mathbb{Q}, s < r$ it follows that $b^{r} \geq b^{s}$. Since $s > 0$ it follows that $b^{s} > 1$ and hence $b^{r} \geq b^{s} > 1$.
The proof of fact 1) mentioned in the beginning is easy. Let $r = m/n$ be a positive rational number so that $m, n$ are positive integers. Then we know that $$b^{m} - 1 = (b - 1)(b^{m - 1} + b^{m - 2} + \cdots + b + 1)$$ and since $b > 1$, it follows that $b^{m} > 1$. Therefore $b^{rn} > 1$. If $c = b^{r} \leq 1$ then as before we can say that $b^{m} = b^{rn} = c^{n} \leq 1$. Since $b^{m} > 1$ it thus follows that we can't have $b^{r} \leq 1$. Thus we have $b^{r} > 1$.
