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Given a set $ \{ \sqrt{3}, 2\sqrt{3}, 3\sqrt{3},...\}$, prove that some of the elements have fractional part less than 0.01 when written in decimal form.

Here is my attempt so far:

Divide the range $[0, 1.0]$ in blocks of size 0.01. So there would be 100 such blocks. Now there blocks would act as our pigeon holes. Fractional part of subsequent set elements would act as our pigeons. Now either of two cases can happen:

  1. Fractional part of a set element lies in block $[0, 0.01]$. In this case we are done.
  2. 2 set elements lie in same block. Let us say $n\sqrt{3}$ and $m\sqrt{3}$ where $n > m$. If mantissa of $n\sqrt(3)$ is greater than fractional part of $(n-m)\sqrt{3}$ would lie between $[0, 0.01]$. In this case we are done. I don't know what to do when fractional part of of $n\sqrt{3}$ is smaller.

Any suggestions or alternate solutions?

VividD
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gibraltar
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    Here is an ugly way. Suppose the mantissa $(n-m)\sqrt{3}$ is near $1$ but a little below, like $0.996\dots\approx 1-0.004$. Consider $k(n-m)\sqrt{3}$. Each time we push up $k$ by $1$, the mantissa goes down by $0.004$, so for some $k$ it falls in $(0,0.01)$. – André Nicolas Nov 09 '13 at 09:06
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    @AndréNicolas Your solution suggests that limit 0.01 can brought down to any arbitrary decimal below 1.0. Maybe we can try proving the harder version :) – gibraltar Nov 09 '13 at 09:16
  • In general, if $\alpha$ is any irrational number, then the fractional parts of $n\alpha$ (the mantissas) are dense in the interval $[0,1)$. For any $x$ in that interval, and any positive $\epsilon$, there is an $n$ such that the mantissa is within $\epsilon$ of $x$. Same proof. – André Nicolas Nov 09 '13 at 09:21
  • Isn't the mantissa of any nonzero number by definition confined to the range either $(0.1,1]$ or $[1,10)$ (depending on conventions used)? – Hagen von Eitzen Nov 09 '13 at 09:28
  • @HagenvonEitzen Here mantissa refers to fractional part. I actually don't what is the correct terminology. – gibraltar Nov 09 '13 at 09:31
  • @gibraltar: Fractional part is the usual term. Here is a proof of the general result that André mentioned. – Brian M. Scott Nov 09 '13 at 09:32
  • Oh, I see. It is $\log_{10}x -\lfloor \log_{10} x\rfloor$ (and linguitically a false friend) – Hagen von Eitzen Nov 09 '13 at 09:34
  • @Hagen: Mantissa has the same meaning in English; gibraltar was simply using the wrong term. – Brian M. Scott Nov 09 '13 at 09:35
  • @BrianM.Scott Statement corrected. – gibraltar Nov 09 '13 at 09:37
  • In fact your second case always occurs, no need for case 1 – Ewan Delanoy Nov 09 '13 at 09:38
  • @gibraltar: I saw; that was quick. Thanks! – Brian M. Scott Nov 09 '13 at 09:38
  • Historical note: we use the term, "Dirichlet's pigeonhole principle" because Dirichlet used it in his proof of the density of the fractional part of $n\alpha$ for irrational $\alpha$. – Gerry Myerson Nov 09 '13 at 10:54
  • @GerryMyerson Proof of Dirichlet's pigeonhole principle assumes $n\alpha$ series contains elements with $n < 0$ too. But this not true in my case. $n > 0$ in the series I have mentioned so same proof can't be borrowed. Correct me if I am wrong. – gibraltar Nov 09 '13 at 12:32
  • Yes, you are wrong. – Gerry Myerson Nov 09 '13 at 22:44

1 Answers1

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Based on original question and comments I will try to put together a complete proof:

Lema: Given a set $ \{ \sqrt{3}, 2\sqrt{3}, 3\sqrt{3},..., 101\sqrt{3}\}$, at least one of elements will have fractional part less than 0.01 or greater than 0.99.

Divide the range $[0, 1]$ in 100 cells of size 0.01. Assign all elements of the set to cells based on their fractional part. Since there are 101 elements in the given sequence, at least one cell will contain 2 elements. Lets say they are $p\sqrt{3}$ and $q\sqrt{3}$. Let also say that fractional part of $p\sqrt{3}$ is greater than fractional part of $q\sqrt{3}$. (Two fractional parts can not be equal, otherwise $\sqrt{3}$ would be rational number.)

If $p>q$, than $(p-q)\sqrt{3}$ will have fractional part less than 0.01, and will be one of elements of the initial set. And the element is found.

If $p<q$, than $(p-q)\sqrt{3}$ will have fractional part less than 0.01, but it will be negative number (lets say -2.998, -22.993, or similar), and $(q-p)\sqrt{3}$ will have fractional part greater than 0.99, and will be one of elements of the initial set. Then a procedure that Andre described is applied. And the element is again found.

VividD
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