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across
 5 years ago
I am having trouble understanding the theorem below, so I will post it here as a reminder to myself to review it as often as I can. Also, I may get lucky enough to have someone clarify the questions that I will most likely post here as I read through it time and time again.
across
 5 years ago
I am having trouble understanding the theorem below, so I will post it here as a reminder to myself to review it as often as I can. Also, I may get lucky enough to have someone clarify the questions that I will most likely post here as I read through it time and time again.

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across
 5 years ago
Best ResponseYou've already chosen the best response.3Dirichlet's Approximation Theorem "If \(\alpha\) is a real number and \(n\) is a positive integer, then there exist integers \(a\) and \(b\) with \(1\leq a\leq n\) such that \(a\alphab<1/n\)." Proof. Consider the \(n+1\) numbers \(0,\{\alpha\},\{2\alpha\}\), ..., \(\{n\alpha\}\). These \(n+1\) numbers are the fractional parts of the numbers \(j\alpha\), \(j=0,1\), ..., \(n\), so that \(0\leq\{j\alpha\}<1\) for \(j=0,1\), ..., \(n\). Each of these \(n+1\) numbers lies in one of the \(n\) disjoint intervals \(0\leq x<1/n\), \(1\leq x<2/n\), ..., \((j1)\leq x<j/n\), ..., \((n1)\leq x<1\). Because there are \(n+1\) number under consideration, but only \(n\) intervals, the pigeonhole principle tells us that at least two of these numbers lie in the same interval. Because each of these intervals has length \(1/n\) and does not include its right endpoint, we know that the distance between two numbers that lie in the same interval is less than \(1/n\). It follows that there exist integers \(j\) and \(k\) with \(0\leq j < k\leq n\) such that \(\{k\alpha\}\{j\alpha\}<1/n\). Now let \(a=kj\) and \(b=[k\alpha][j\alpha]\). Because \(0\leq j < k\leq n\), we see that \(1\leq a\leq n\). Moreover,\[a\alphab=(kj)\alpha([k\alpha][j\alpha])\]\[=(k\alpha[k\alpha])(j\alpha[j\alpha])\]\[=\{k\alpha\}\{j\alpha\}<1/n.\]Consequently, we have found integers \(a\) and \(b\) with \(1\leq a\leq n\) and \(a\alphab<1/n\), as desired. \(\blacksquare\)

asnaseer
 5 years ago
Best ResponseYou've already chosen the best response.0Your question prompted me to search this topic and try and learn about it. During my google quest, I came across the original German paper on this topic and thought it might interest you as I /think/ you are a German speaker. http://bibliothek.bbaw.de/bibliothekdigital/digitalequellen/schriften/anzeige?band=07abh/1837&seite:int=00000286

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sabes algo de eso jose?
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