Prove in any interval there exists an irrational z

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1. The problem statement, all variables and given/known data
Prove that in any interval there exists an irrational 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.
 
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As people were saying in the previous thread, the problem appears to be that much of what you have written down is nonsense ([tex][r, s] \in a[/tex], which is backward, and the dubious inequality deduction you mention).

What might be going on here is an attempt to make the following argument:
Given that there is a rational number in every interval, and the existence of one single irrational number [tex]a[/tex], prove that there is an irrational number in every interval.

([tex]a[/tex] might be [tex]\sqrt{2}[/tex], using the standard proof.)

Here is how this argument would go. Let [tex](r,s)[/tex] be the interval in which we wish to find an irrational number. We try to do this by adding a rational number [tex]q[/tex] to [tex]a[/tex], because the sum of a rational and an irrational is irrational. So we need to find [tex]q[/tex] so that [tex]r < q + a < s[/tex]. Clearly it suffices to have [tex]r - a < q < s - a[/tex], i.e., choose [tex]q[/tex] to be any rational number in [tex](r - a, s - a)[/tex]. We can do this by hypothesis, so the proof is done.

Does this make sense?
 
I think so. Is this proof correct:
Let [itex]a\in[/itex] Q, [itex]\epsilon[/itex] rational, where [itex]\epsilon\in[r-a, s-a][/itex].
Then [itex]r-a<\epsilon<s-a[/itex], so [itex]r<\epsilon+a<s[/itex]
So [itex]\epsilon+a\in[r, s][/itex] and [itex]\epsilon+a[/itex] is irrational.
Therefore in any interval, there exists an irrational number.
 

jgens

Gold Member
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But, since [itex]a \in \mathbb{Q}[/itex] and [itex]\epsilon \in \mathbb{Q}[/itex] clearly [itex]a + \epsilon \in \mathbb{Q}[/itex]. Thus [itex]a + \epsilon[/itex] cannot be irrational.
 
Oops, meant to write [itex]a\notin[/itex] Q
 

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