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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.

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

5 Answers 5

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

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  • $\begingroup$ You are welcome. I have added a proof that uses your very nice method. $\endgroup$ Mar 1, 2014 at 3:18
  • $\begingroup$ Saw it! Impressed!! $\endgroup$
    – Ishfaaq
    Mar 1, 2014 at 3:19
  • $\begingroup$ 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. $\endgroup$
    – Ishfaaq
    Mar 1, 2014 at 3:30
  • $\begingroup$ That is not what I wrote, it is an improvement (simplification) of what I wrote! $\endgroup$ Mar 1, 2014 at 3:33
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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), $\frac{1}{r^m}<x$.

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  • $\begingroup$ 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?? $\endgroup$
    – Ishfaaq
    Mar 1, 2014 at 2:35
  • $\begingroup$ 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.. $\endgroup$
    – Ishfaaq
    Mar 1, 2014 at 2:50
  • $\begingroup$ 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 $x$. $\endgroup$
    – cderwin
    Mar 1, 2014 at 2:51
  • $\begingroup$ 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. $\endgroup$
    – Ishfaaq
    Mar 1, 2014 at 2:55
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    $\begingroup$ 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. $\endgroup$
    – cderwin
    Mar 1, 2014 at 3:07
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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.

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Darrin
    Mar 1, 2014 at 3:40
  • $\begingroup$ To Rudin again yes? Page 10? $\endgroup$
    – Ishfaaq
    Mar 1, 2014 at 3:43
  • $\begingroup$ 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. $\endgroup$
    – Ishfaaq
    Mar 1, 2014 at 3:45
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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.

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