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

  • Thread starter STEMucator
  • Start date
  • Tags
    Irrational
In summary, the conversation discusses finding a proof for the existence of an irrational number between two rational numbers, and the set of irrationals. The proof involves using theorem 1.3 and the denseness of the set of rational numbers. However, there is no general way to write the elements of the irrationals, as most real and complex numbers are transcendental.
  • #1
STEMucator
Homework Helper
2,076
140

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.
 
Last edited:
Physics news on Phys.org
  • #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.
 

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

1. What does it mean to "prove" something in science?

In science, to "prove" something means to provide evidence or demonstrate through experimentation that a particular statement or hypothesis is true. The evidence must be replicable and supported by a logical and valid argument.

2. How can we prove that a < b, there is an irrational in between them?

To prove that a < b, there is an irrational in between them, we can use a proof by contradiction. We assume that there is no irrational number in between a and b, and then use logical reasoning and mathematical principles to show that this assumption leads to a contradiction. This contradiction proves that our initial assumption was false, and therefore, there must be an irrational number in between a and b.

3. Why is it important to prove the existence of an irrational number in between a and b?

Proving the existence of an irrational number in between a and b can provide a better understanding of the real numbers and their properties. It also helps to solidify our understanding of mathematical concepts and allows for the development of more complex mathematical theories and principles.

4. What is an irrational number?

An irrational number is a real number that cannot be expressed as a ratio of two integers. In other words, its decimal representation is non-terminating and non-repeating. Familiar examples of irrational numbers include pi (π) and the square root of 2 (√2).

5. Can you provide an example of an irrational number in between two given numbers?

Yes, for example, between 1 and 2, there is an irrational number √2 ≈ 1.414. This number is not a perfect square and cannot be expressed as a fraction, making it an irrational number. Therefore, it satisfies the condition of being an irrational number in between 1 and 2.

Back
Top