Given any real numbers $x \gt 0$ and $r \gt 1$ there exists $m \in \Bbb N$ such that $\frac 1 {r^m} \lt x$ I know it can be proven that given any real number $x \gt 0$ there exists  $m \in  \Bbb N$ such that $\frac 1 {2^m} \lt x$. I tried to generalise it but am surprisingly not getting anywhere. I couldn't find a similar question on the site and if this is in fact a duplicate I apologise and please point me towards the original.
So this is the problem: "Given any real numbers $x \gt 0$ and $r \gt 1$ there exists  $m \in  \Bbb N$ such that $\frac 1 {r^m} \lt x$"
Here's what I've tried so far. I have assumed the negation which leads to $r^n \le \frac 1 x \forall n \in \Bbb N$. Then $\sup A = \{ r^n \ | \ n \in \Bbb N \} = c$ exists. I can prove that $r^n \gt r^{n - 1} \forall n \in \Bbb N$. So I can get a contradiction if I can say that $c$ is also of the form $r^k$ for some natural number $k$. To this end I assumed $c \not \in A$ but am not getting anywhere from there. Maybe there is a counter-example??
Would be grateful for any help. Thanks in advance. 
 A: Let $r=1+t$, where $t\gt 0$. Note that by the Binomial Theorem, or more simply by the Bernoulli Inequality, we have $(1+t)^n \ge 1+tn\gt tn$ if $n\ge 1$. 
Thus to make $\frac{1}{r^n}\gt x$, it is enough to pick $n \gt \frac{x}{t}$. There is such an integer, by the Archimedean property of the reals.  
Remark: Your approach will work nicely. By your choice of $c$ (which is clearly $\gt 0$) there is a positive integer $n$ such that $(1+r)^n \ge \frac{c}{(1+r)/2}$. Then $(1+r)^{n+1}\ge c\frac{2r}{1+r}$. But it is easy to verify that $\frac{2r}{1+r}\gt 1$. 
A: So this is what I'd do:
First of all, this is equivalent to saying there exists some integer $m$ such that $r^m<x$ for $r>1, x>0$.  Without loss of generality, let $x<1$.  Then note that $m=\log_r{r^m}<\log_r{x}$.  Therefore, if we pick any $m<\log_{r}{x}$ (which is obviously possible), $\frac{1}{r^m}<x$.
A: I think this is similar to what you thought. 
Suppose $r>1$ and let $s= \sup\{r^n: n\in \mathbb{N}\}$. We claim that $s= \infty$. Suppose for sake of  contradiction that $s=c$ for some $c\in \mathbb{R}^{>0}$. Then $s/r<s$ so there is some $n$ such that $s/r<r^n$. Then $s<r^{n+1}$ contrary to the choice of $s$. Then the set is not bounded and given a $x>0$ there is a $m\in \mathbb{N}$ such that $x<r^m$ since otherwise the set is bounded and the least upper bound must be finite.
A: By the Archimedean Property (see Rudin, "Principles of Mathematical Analysis," 3rd edition page 9) there exists an $n \in \mathbb{N}$ such that $\frac{1}{r}<nx$. Therefore $\frac{1}{nr}<x$. Since $r>1,$ there is some $s>0$ such that $r^s =n$. Let $m-1$ be the smallest nonnegative integer bigger than $s$. Then we have $r^{m-1}>r^s$ and so $$\frac{1}{r^m}=\frac{1}{r^{m-1}r}\leq \frac{1}{r^sr}=\frac{1}{nr} < x$$ as desired. 
A: Let's continue your approach which is right in spirit. Assume that for all $n \in \mathbb{N}$ we have $\dfrac{1}{r^{n}} \geq x$ and then the set $A = \{x_{n} = r^{n}: n \in \mathbb{N}\}$ is bounded above by $1/x$ and thus $c = \sup A$ exists. Next you point out the fact that since $r > 1$ we have $r^{n} > r^{n - 1}$ so that the sequence $x_{n} = r^{n}$ is strictly increasing. We claim that $\lim_{n \to \infty}x_{n} = c = \sup A$.
Clearly from the definition of supremum we can see that for any $\epsilon > 0$ there is some $m \in \mathbb{N}$ such that $c - \epsilon < x_{m}$. Since $x_{n}$ is increasing it follows that for $n > m$ we have $c - \epsilon < x_{n} \leq c < c + \epsilon$. It follows that $\lim_{n \to \infty}x_{n} = c$. Now we know that $x_{n + 1} = rx_{n}$ and taking limits we get $c = rc$. This is a contradiction as $ c \geq x_{1} = r > 1$. It follows that our initial assumption is wrong.
