# Homework Help: Prove if a < b, there is an irrational inbetween them

1. Apr 30, 2013

### Zondrina

1. The problem statement, all variables and given/known data

No giving up !

The question : http://gyazo.com/08a3726f30e4fb34901dece9755216f3

2. Relevant equations

A lemma and a theorem :

http://gyazo.com/f3b61a9368cca5a7ed78a928a162427f
http://gyazo.com/ca912b6fa01ea6c163c951e03571cecf

The fact $\sqrt{2}$ and $2^{-1/2}$ are irrational.

3. The attempt at a solution

Suppose 0 < b - a. We must show that $\exists x \in ℝ - \mathbb{Q} \space | \space a < x < b$

Since 0 < b - a, we know we can apply theorem 1.3 to find $r \in \mathbb{Q} \space | \frac{a}{\sqrt{2}} < r < \frac{b}{\sqrt{2}}$ because of the denseness of $\mathbb{Q}$.

We know : $ℝ \setminus \mathbb{Q}$ is the set of irrationals.

Also, since $r \in \mathbb{Q}$, we can take $r = \frac{p}{q}$ for some $p, q \in \mathbb{Z}$

This yields $a < \frac{p \sqrt{2}}{q} < b$.

Therefore, because $\frac{p \sqrt{2}}{q} \in ℝ - \mathbb{Q}$ the claim is proven true.

EDIT : Fixed a small error.

Last edited: Apr 30, 2013
2. Apr 30, 2013

### micromass

The proofs is ok.

But, I really don't get this line:

3. Apr 30, 2013

### Zondrina

I was just trying to highlight that it was the set of irrationals.

Do you know how to write backslashes in latex? I was trying to write R\Q, but it treats '\' like an escape character.

4. Apr 30, 2013

### micromass

OK, but the right-hand side you wrote down really isn't equal to the set of irrationals...

\setminus

5. Apr 30, 2013

### LeonhardEuler

Don't forget that numbers like $\pi$ and e are also irrational (among many others).

6. Apr 30, 2013

### Zondrina

$ℝ \setminus \mathbb{Q}$ Yay :)!

Hmm would general elements look like $c + d \sqrt{e}$ instead then? Trying to get a grasp on what the set elements look like.

7. Apr 30, 2013

### micromass

There is no real good description of general elements of the irrational numbers. Elements like $c+d\sqrt{e}$ are still very special.

The reality is that the set of irrationals is huge. Most elements of the irrational numbers can't be explicitely described.

8. Apr 30, 2013

### VantagePoint72

There is no general way of writing the elements of the irrationals in the way you are trying to do. The transcendental numbers are a subset of the irrationals and they cannot, by definition, be expressed as the roots of polynomials with rational coefficients. We only know the general form of a few families of transcendental numbers. However, we know that almost all real (and complex) numbers are transcendental, so you've only captured a very tiny subset with your definition.