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Simple demonstration with real, rational and integers

  1. Oct 6, 2017 #1
    1. The problem statement, all variables and given/known data
    Let ##\alpha \in \mathbb{R}## and ##n \in \mathbb{N}##. Show that exists a number ##m \in \mathbb{Z}## such that ##\alpha - \frac {m}{n} \leq \frac{1}{2n}## (1).


    3. The attempt at a solution
    If I take ##\alpha= [\alpha] +(\alpha)## with ##[\alpha]=m## (=the integer part) and ##(\alpha)=\frac{1}{2n}##(=the fractional part) I must have an equality in (1). Substituting, I obtain ##m(n-1)=0## and for ##n=1## I can always find a solution to the problem. Is this demonstration correct? How can I find a more general one?
     
  2. jcsd
  3. Oct 6, 2017 #2

    scottdave

    User Avatar
    Homework Helper
    Gold Member

    It looks to be it states: given any alpha & n, can you always find an m wich satisfies it. I don't think you choose alpha in advance.
     
  4. Oct 6, 2017 #3

    fresh_42

    Staff: Mentor

    I'm not sure what you've done there. The usual approach is to take the inequality ##\alpha -\frac{m}{n} \leq \frac{1}{2n}## and transform it until we get a condition on ##m##. That is ##m## on one side, the rest on the other. Now chose an ##m## which always satisfies this condition and write all steps from there on backwards to the condition you want to prove. Since ##\alpha## and ##n## are given, we only need a small enough or big enough ##m## to be chosen.

    Long explanation short: Don't bother about the choice of ##m## as long as you don't know what it has to provide.
     
  5. Oct 6, 2017 #4
    Thanks for the reply.
    So, rearranging I have ##m \geq \alpha n - \frac{1}{2}##.

    Now I have to choose m. I tried different values of ##\alpha ,n## and it seems that the right choice of m is ## m=\lfloor{\alpha \cdot n}\rfloor## (=integer part).
     
  6. Oct 6, 2017 #5

    fresh_42

    Staff: Mentor

    Don't bother ##\alpha## and ##n##. They are given and cannot be chosen. Only ##m## is up to your choice. And don't try to estimate as close as possible, there is no need to do so. It won't be rewarded in this example. The situation is a different one, if you try to estimate numerical calculations or the length of algorithms, but here it isn't needed. I took ##m=\lceil \alpha \cdot n \rceil## so I saved an additional thought on what happens at exactly a difference of ##0.5##. You could even chose ##m=10\cdot |\alpha n|##, it won't matter here.

    Anyway, it is correct. Now you have to write down ##m = \lceil \alpha n \rceil \geq \alpha n \geq \alpha n - \frac{1}{2} \Rightarrow \ldots ## and so on, until the last line is ##\alpha -\frac{m}{n} \leq \frac{1}{2n}\,##.
     
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