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]





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.
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.


Gold Member
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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