# Nested Rational Interval - Prove it doesn't contail any Rational Point

• Apr 14th 2009, 06:08 AM
aman_cc
Nested Rational Interval - Prove it doesn't contail any Rational Point
This is rather elementary but I'm struggling with a formal prove:

Consider this closed nested interval with Rational end points, the usual definition are:

Closed nested interval [an,bn] where an,bn belong to Q
sqr(an) < 2 < sqr(bn)
and [an+1,bn+1] is subset of [an,bn]

Prove the there is no x (in Q) which belongs to this nested interval.

Thanks

PS: Also if someone can tell me if there is a way use mathematical symbols and notations in your post
• Apr 14th 2009, 12:11 PM
aleph1
Entering mathematical symbols
Quote:

Originally Posted by aman_cc
PS: Also if someone can tell me if there is a way use mathematical symbols and notations in your post

This should get you started:

http://www.mathhelpforum.com/math-he...-mathtype.html
• Apr 21st 2009, 09:56 PM
Quote:

Originally Posted by aleph1
This should get you started:

Infact if you want to get started you should read the first sticky thread's
attachment. Latex Help

The whole subsection in the above link is dedicated for that purpose.You can ask any question about it.
• May 4th 2009, 11:09 AM
aleph1
Quote:

Originally Posted by aman_cc
This is rather elementary but I'm struggling with a formal prove:

Consider this closed nested interval with Rational end points, the usual definition are:

Closed nested interval [an,bn] where an,bn belong to Q
sqr(an) < 2 < sqr(bn)
and [an+1,bn+1] is subset of [an,bn]

Prove the there is no x (in Q) which belongs to this nested interval

It took me a while to find a formal answer to this simple problem.

From the nested interval theorem, the intersection of nested closed intervals is a singleton set, or another closed interval. Continue selecting nested closed intervals until the intersection is a singleton set. This smallest intersection interval consists of three points, the rational closing points and an interior singleton point. The rationals are not continuous. The singleton point cannot be rational.

(response inspired by posting http://www.mathhelpforum.com/math-he...30-post10.html)
• May 5th 2009, 02:47 AM
aman_cc
Thanks, but by assuming Rationals are not continous aren't we somewhere using what we have to prove. I feel so.
Thanks
• May 5th 2009, 06:47 AM
aleph1
Quote:

Originally Posted by aman_cc
Thanks, but by assuming Rationals are not continous aren't we somewhere using what we have to prove. I feel so.
Thanks

I am not sure what proofs and definitions you have at you disposal to work on this.

Here is a discussion of the Dirichlet function that shows that it is possible to find an irrational number as close as desired to a rational number. Possibly the article's provided δ ε approach will show you how to prove what you need.
• May 5th 2009, 06:49 AM
HallsofIvy
You don't need to use "rationals aren't continuous". (In fact, only functions are continuous. What is meant is that "no non-singleton set in the set of rationals is connected".)

As aleph1 said, the intersection of any such collection of intervals consists of a single point. It should be obvious from $\displaystyle \sqrt{a_n}< \sqrt{2}<\sqrt{b_n}$ that $\displaystyle \sqrt{2}$ is in everyone of those intervals and so in the intersection. Since there is only one point in the intersection, that point is $\displaystyle \sqrt{2}$ which is not rational.