# Homework Help: Prove in any interval there exists an irrational z

1. Dec 18, 2009

### cmajor47

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:

$a \notin Q$, $\epsilon$ rational

$[r, s]\in a$

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

$s-a<\frac{\epsilon}{2}$

$s-\frac{\epsilon}{2}<a$

$s-\frac{\epsilon}{2}+\epsilon<a+\epsilon$

$s<s+\epsilon<a+\epsilon$

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

I also don't know where to go from here to show what needs to be proved.
I'd really appreciate any help.

2. Dec 18, 2009

### ystael

As people were saying in the previous thread, the problem appears to be that much of what you have written down is nonsense ($$[r, s] \in a$$, 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 $$a$$, prove that there is an irrational number in every interval.

($$a$$ might be $$\sqrt{2}$$, using the standard proof.)

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

Does this make sense?

3. Dec 18, 2009

### cmajor47

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

4. Dec 18, 2009

### jgens

But, since $a \in \mathbb{Q}$ and $\epsilon \in \mathbb{Q}$ clearly $a + \epsilon \in \mathbb{Q}$. Thus $a + \epsilon$ cannot be irrational.

5. Dec 18, 2009

### cmajor47

Oops, meant to write $a\notin$ Q