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I am trying to figure out what I am suppose to prove for the following question on infinite sequences. The wording is confusing for me.

A real number $x$ is irrational, if we can find an ascending sequence of integers $(q_n)$, such that $xq_n$ is not an integer for any n, but if, when $p_n$ stands for the integer nearest to $xq_n$, $(xq_n - p_n)$ is a null sequence.

Is this an if and only if type question or am I asked to prove that $x$ is irrational or that $(xq_n - p_n)$ is a null sequence.

Thank you in advance

user10354138
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Seth
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2 Answers2

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There’s only one “if” in the statement of the proposition. You only need to show the implication in one direction.

Specifically, consider the part of the statement that comes after the word “if”. If all of that is true, then $x$ is irrational. That’s what you have to prove.

David K
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I don't know whether English is your native language, but I can certainly see why this could cause confusion for a non-native speaker, and probably for some native speakers as well, if they are unfamiliar with mathematical verbiage. Below is my attempt to make this as explicit and unambiguous as I can. Note that I've included a special case in the definition of $p_n,$ which won't matter in the long run (indeed, for each $n$ you can choose either the lesser or greater closest integer), but certainly in the short run, not including something like this creates some ambiguity.

Let $x$ be a real number and let $(q_n)$ be a strictly increasing sequence of integers. Also, for each positive integer $n,$ let $p_n$ be the integer closest to $xq_n$ (if there are two such closest integers, let $p_n$ be the greater of the two). Assume that both (1) and (2) hold: (1) for each positive integer $n,$ the real number $xq_n$ is not an integer; (2) $(xq_n - p_n) \rightarrow 0$ as $n \rightarrow \infty.$ Prove that $x$ is irrational.

Dave L. Renfro
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  • Thank you Dave. The question comes from Konrad Knopp's bigger book on infinite sequences. – Seth Jun 13 '19 at 20:00
  • I want to ask if you know if this question is somehow related to Dirichlet approximate theorem on rational approximation of irrationals. Also, in the literature, is there a name for this result? – Seth Jun 14 '19 at 08:47
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    @Seth Mai: Unless I'm overlooking something, this seems to be a peculiar way of stating a result that is an immediate consequence of a very elementary result, probably because this specific manner of stating the result is needed for something. Suppose $x$ were rational. Assume that $x = \frac{r}{s}$ where $r$ and $s > 0$ are integers having no common factors greater than $1.$ We are told that for each $n$ the number $\frac{r}{s} \cdot q_n$ is not an integer. Thus, $\frac{r}{s} \cdot q_n$ in lowest terms has a denominator that is greater than $1$ and less than or equal to $s.$ (continued) – Dave L. Renfro Jun 14 '19 at 09:38
  • Therefore, for each $n,$ the distance between $\frac{r}{s} \cdot q_n$ and any integer is greater than or equal to $\frac{1}{s},$ which contradicts the assumption that $(xq_n - p_n) \rightarrow 0.$ Here's what it seems to be mostly related to: Given a real number $y,$ let $| y |$ be the least distance between $y$ and an integer. Then one can show that $x$ is irrational if and only if the sequence $|x|,$ $|2x|,$ $|3x|,$ $\ldots$ is dense in the interval $[0,1].$ Your result seems to be the following version: (continued) – Dave L. Renfro Jun 14 '19 at 09:50
  • If the sequence $|x|,$ $|2x|,$ $|3x|,$ $\ldots$ has $0$ as a cluster point, then $x$ is irrational. Note that saying $|xq_n| \rightarrow 0$ implies $0$ is a cluster point of $|x|,$ $|2x|,$ $|3x|,$ $\ldots$ – Dave L. Renfro Jun 14 '19 at 09:51
  • In the last part of my 3-part comment, due to some late editing that I did to the wording, I left out "nonzero". The 2nd to last sentence there should be: "If the sequence of nonzero real numbers $|x|,$ $|2x|,$ $|3x|,$ $\ldots$ has $0$ as a cluster point, then $x$ is irrational." – Dave L. Renfro Jun 14 '19 at 10:29
  • @David L. Renfro, Thank you for the additional comments. The question came from the end of first chapter of Konrad Knopp's book Theory and Applications of infinite series. I saw another question somewhere on mathstackexchage which ask something similar to your comment. For me the trouble with this question is why the second condition would necessarily imply that x is irrational. It is just not intuitively clear. – Seth Jun 14 '19 at 15:11
  • @David L. Renfro, here is the question from mathstackexchange that is related to what you were saying.
    https://math.stackexchange.com/questions/903142/for-an-irrational-number-a-the-fractional-part-of-na-for-n-in-mathbb-n-is?rq=1     – Seth Jun 14 '19 at 15:35
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    It is just not intuitively clear --- The main idea is that, to use a specific case, any reduced fraction having a denominator of $7$ can't be closer than $\frac{1}{7}$ to any integer. Similarly, any reduced fraction having a denominator of $10$ can't be closer than $\frac{1}{10}$ to any integer. This is the idea behind my saying that $\frac{r}{s} \cdot q_n = \frac{r \cdot q_n}{s}$ can't be closer than $\frac{1}{s}$ to any integer, no matter what the integer $q_n$ is (unless $q_n$ is a multiple of $s,$ of course, but this is not allowed by assumption (1)). – Dave L. Renfro Jun 14 '19 at 15:54
  • @David L. Renfro. Does that mean because of the minimum distance of $1/s$ if we have a number x and the distance to its fractional part is greater than less than $1/s$, then x would have to be irrational. – Seth Jun 14 '19 at 16:05
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    Remember that the proof I gave (or sketched) begins by assuming that $x$ is rational. From this assumption and from (1), we find that the distance between each $xq_n$ and the set of integers is $\geq \frac{1}{s}.$ However, from (2) we find that the distance between $xq_n$ and the set of integers can be made arbitrarily small (and nonzero) --- in particular, there exists $n$ such that the distance between $xq_n$ and the set of integers is less than $\frac{1}{s},$ which contradicts what is asserted in the previous sentence. – Dave L. Renfro Jun 14 '19 at 18:19
  • Okay, thank you for clearing that up. – Seth Jun 14 '19 at 22:28