Prove in any interval there exists a rational z

In summary, to prove that in every interval there exists an irrational number, we can consider a non-empty open interval (a,b) and choose a natural number N greater than 1/(b-a). From this, we can find an integer k such that Na < k < Nb, and therefore, a < (k/N) < b. This shows that every open interval contains a rational number.
  • #1
cmajor47
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0

Homework Statement


Prove that in any interval there exists a rational z.


Homework Equations





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.
 
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  • #2
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?
 
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  • #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.
 
  • #4
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.
 
  • #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?
 
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  • #6
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:).
 
  • #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.
 
  • #8
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 definitely 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.
 
  • #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.
 

Related to Prove in any interval there exists a rational z

1. What does it mean to prove that there exists a rational number in any interval?

Proving that there exists a rational number in any interval means showing that there is at least one rational number between any two given numbers in the interval. This is known as the density of rational numbers.

2. How do you prove the density of rational numbers in an interval?

To prove the density of rational numbers in an interval, we use the Archimedean property, which states that for any two real numbers x and y, there exists a positive integer n such that nx > y. This allows us to find a rational number that falls between any two given real numbers.

3. Can you prove the density of rational numbers in any interval using the intermediate value theorem?

No, the intermediate value theorem only applies to continuous functions on a closed interval. It cannot be used to prove the density of rational numbers in an interval.

4. Does the density of rational numbers hold true for all intervals, including open, closed, and half-open intervals?

Yes, the density of rational numbers holds true for all intervals, regardless of whether they are open, closed, or half-open. This is because the Archimedean property applies to all real numbers and not just a specific type of interval.

5. How is the density of rational numbers related to the completeness of the real numbers?

The completeness of the real numbers means that every nonempty set of real numbers that is bounded above has a least upper bound. This property allows us to prove the density of rational numbers in any interval, as we can always find a rational number that is between any two given real numbers in the interval.

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