# 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.

• Since $r>1$ so the sequence $(r^n)$ diverges and so given a $x>0$ there is some $m\in \mathbb{N}$ such that $x<r^m$. Mar 1, 2014 at 3:16
• @Jose Antonio: Thanks for the input. But I haven't gotten to covering sequences yet. Mar 1, 2014 at 3:17

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 $\frac{2r}{1+r}\gt 1$.

• You are welcome. I have added a proof that uses your very nice method. Mar 1, 2014 at 3:18
• Saw it! Impressed!! Mar 1, 2014 at 3:19
• Although there is a slight confusion with the terms I believe. I think this is what you suggest (and it is simplistically brilliant btw). $r \gt 1 \implies \frac c r \lt c \implies \exists r^n \in A$ such that $\frac c r \lt r^n \implies c \lt r^{n + 1}$ leading to a contradiction. Mar 1, 2014 at 3:30
• That is not what I wrote, it is an improvement (simplification) of what I wrote! Mar 1, 2014 at 3:33

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), $\frac{1}{r^m}<x$.

• Couple of issues: $(1):$ The problem is equivalent to showing the existence of $m \in \Bbb N$ such that $r^m$ is greater than $x$... $(2):$ Then we lose a considerable amount of generality assuming $x \lt 1$. But you are partly right as in we would be done if we can pick $m$ such that $m \lt \log_{r}{x}$. But the existence of such a natural number is not guaranteed is it?? Mar 1, 2014 at 2:35
• What we can do instead may be is to split two cases $x \le r$ and $x \gt r$. The first part is trivial. For the second we can pick $m$ such that $m \gt \log_{r}{x}$ which is always guaranteed. So thanks for suggesting the method. But I still think there is a better solution. Splitting cases isn't nice. There should be a sure-fire way to show this $\forall x$. Thanks loads though.. Mar 1, 2014 at 2:50
• As for (1): that's not what you have in the title; I am simply changing $\frac{1}{r^m}=r^{-m}=r^n$ and I'm just letting $n$ be negative. As for (2) We don't lose any generality; if $m=1$ and $x\geq 1$, $\frac{1}{r^1}$ is already less than $x$. Mar 1, 2014 at 2:51
• I think I've just followed your method in the above comment without making the substitution $r = \frac 1 {r'}$. So thanks. But like I said there should be a way based simply on the completeness property of $\Bbb R$. To be honest the Analysis book I'm reading has not even defined logarithms formally yet. It comes in the section on sequences I believe. Mar 1, 2014 at 2:55
• Well if you want something more based on completeness, $\inf\left\{\frac{1}{r^n}|n\in\mathbb{N}\right\}=0$, and if you can prove this , you should be good. You can try to do this by saying there is rational $r=\frac{p}{q}$ between $0$ and $x$ for any $x$ (density of $\mathbb{Q}\subset\mathbb{R}$), and then there is some $m$ such that $r^m>q$, implying that $\frac{1}{r^m}<\frac{1}{q}<x$ and this should be good. Mar 1, 2014 at 3:07

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.

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.

• NOTE: There may need to be some justification for the statement that there is an $s>0$ with $r^s=n$. I do not know your set of assumptions in your study's development to this point. Mar 1, 2014 at 3:40
• To Rudin again yes? Page 10? Mar 1, 2014 at 3:43
• Or maybe not?? That is just the existence of the root. Anyway I got my hands on a lovely solution above. But thanks for the input anyway. Nice to show it from a direct implication instead of a contradiction. Mar 1, 2014 at 3:45

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.