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Prove in any interval there exists a rational z

  1. Dec 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that in any interval there exists a rational z.


    2. Relevant equations



    3. The attempt at a solution
    My professor wrote this for me when trying to explain how to prove this:

    [itex]a \notin Q[/itex], [itex]\epsilon[/itex] rational

    [itex][r, s]\in a[/itex]

    [itex]l([r, s])<\frac{\epsilon}{2}[/itex]

    [itex]s-a<\frac{\epsilon}{2}[/itex]

    [itex]s-\frac{\epsilon}{2}<a[/itex]

    [itex]s-\frac{\epsilon}{2}+\epsilon<a+\epsilon[/itex]

    [itex]s<s+\epsilon<a+\epsilon[/itex]

    I don't see why you use [itex][r, s]\in a[/itex] or how my professor went from [itex]s-\frac{\epsilon}{2}+\epsilon<a+\epsilon[/itex] to [itex]s<s+\epsilon<a+\epsilon[/itex]

    I also don't know where to go from here to show what needs to be proved.
    I'd really appreciate any help.
     
    Last edited: Dec 17, 2009
  2. jcsd
  3. Dec 17, 2009 #2

    jgens

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    Here's how I would attack this problem: Consider the non-empty open interval [itex](a,b)\subset \mathbb{R}[/itex]. Clearly [itex](a,b)[/itex] consists of the set of all numbers [itex]x[/itex] such that [itex]a < x < b[/itex]. Choosing a natural number [itex]N > \frac{1}{b-a}[/itex], we can find an integer value [itex]k[/itex] such that [itex]Na < k < Nb[/itex] (this fact is trivial). Can you complete the proof for open intervals from here? Do you see how the case for closed intervals follows immediatly from this one?
     
    Last edited: Dec 17, 2009
  4. Dec 17, 2009 #3
    Thanks for the help, but I'd really like to understand the proof in the way my professor tried to show.
    If anyone else has ideas, I'd appreciate the help.
     
  5. Dec 17, 2009 #4

    jgens

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    I'm having a hard time reading your professor's "proof" especially since either his notation is incorrect/non-standard or you've copied his solution down incorrectly on this forum. For example, you say that [itex][r,s] \in a[/itex] when you probably mean [itex]a \in [r,s][/itex]. If you can clear up these notational errors, I (and others on the forum) can offer more help.
     
  6. Dec 17, 2009 #5
    This wasn't a full proof, just hints to help me through constructing the proof.
    My professor did write that [itex][r,s] \in a[/itex] which I thought was wrong as well. However, wouldn't saying that [itex]a \in [r,s][/itex] be assuming what is to be proved?
     
    Last edited: Dec 17, 2009
  7. Dec 17, 2009 #6

    jgens

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    Well [itex][r,s] \in a [/itex] makes no sense and saying [itex]a \in [r,s][/itex] is either assuming what's to be proven or assuming that there exists an irrational number in every interval, which while true, is probably not an appropriate assumption given the nature of the problem. I suppose you could probably split the problem up as follows:

    Let the open interval [itex](a,b)[/itex] be a non-empty subset of the real numbers. Since [itex](a,b) \neq \emptyset[/itex] we know that there is some real number [itex]x[/itex] such that [itex]x \in (a,b)[/itex]. Clearly [itex]x[/itex] must be either rational or irrational. We'll treat each case separately.
    1) Suppose [itex]x \in \mathbb{Q}[/itex], then there is a rational number in [itex](a,b)[/itex] which completes the proof for this case.
    2) Suppose [itex]x \notin \mathbb{Q}[/itex]. Let [itex]\varepsilon > ||(a,b)||[/itex] where [itex]||(a,b)||[/itex] denotes the length of [itex](a,b)[/itex]. Clearly [itex]x - a \leq b - a \leq \varepsilon[/itex] . . .

    And then you could try and carry on the proof from there, but I'm not entirely convinced that that's a fruitful approach. I guaruntee you that the method I posted earlier works (and it's nice and simple :wink:).
     
  8. Dec 18, 2009 #7
    I'll try to figure out how to prove this using your method, but I don't see where to go with your proof, or even why the things you've stated help to prove this.

    If anyone else has something to say about the proof my professor stated, I'd still appreciate hearing it.
     
  9. Dec 18, 2009 #8

    jgens

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    I'm sorry that I can't help you along with your professor's proof, but I can help you prove the desired result (at least that's helpful right?). So let's start at square one:

    Let's first consider the non-empty open interval [itex](a,b) \subset \mathbb{R}[/itex]. We know that [itex](a,b)[/itex] consists of the set of numbers [itex]x[/itex] such that [itex]a < x < b[/itex]. If we choose a natural number [itex]N > \frac{1}{b-a}[/itex] then we have that [itex]N(b-a) > 1[/itex]. It should be fairly obvious that there exists an integer [itex]k[/itex] such that [itex]Na < k < Nb[/itex] (if you don't find this obvious, you should definately prove this result). Dividing through by [itex]N[/itex] we find that [itex]a < \frac{k}{N} < b[/itex]. Thus, every open interval contains a rational number. Now, can you prove this result for the cases [itex][a,b)[/itex], [itex](a,b][/itex], and [itex][a,b][/itex]?

    And yes, I know this is essentially me reiterating exactly what I said earlier. Sometimes reading through things more than once can be very helpful.
     
  10. Dec 18, 2009 #9
    I just realized that I mistyped what I am trying to prove. It is supposed to be "prove that in every interval there exists an irrational z. That is why I was so confused with your proof. Thank you for all of your help, and I apologize for wasting your time. I will create a new thread about the proof I want help with.
     
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