Prove if a < b, there is an irrational inbetween them

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    Irrational
  • #1
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Homework Statement



No giving up :biggrin:!

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

Homework Equations



A lemma and a theorem :

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

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

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.
 
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  • #2
The proofs is ok.

But, I really don't get this line:

Zondrina said:
##ℝ - \mathbb{Q} = \{ c \sqrt{d} \space | \space c, d \in ℝ\}##
 
  • #3
micromass said:
The proofs is ok.

But, I really don't get this line:

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

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

\setminus
 
  • #5
Don't forget that numbers like [itex]\pi[/itex] and e are also irrational (among many others).
 
  • #6
micromass said:
OK, but the right-hand side you wrote down really isn't equal to the set of irrationals...



\setminus

##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
Zondrina said:
##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.

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

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.
 
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