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Proof: Rational q and Irrational p

Date: 03/03/2003 at 21:36:26
From: Brandon
Subject: Topology

Prove that if a and b are rationals, then there is a rational q and an 
irrational p so that both p and q are between a and b.


Prove that if a and b are irrationals, then there is a rational q and 
an irrational p so that q and p are both between a and b.

I was thinking it might have to do with taking an average or possibly 
taking the rational and putting it over something irrational [i.e. 
sqrt(2)]. I'm really not sure.

Thank you!

Date: 03/04/2003 at 02:24:34
From: Doctor Jacques
Subject: Re: Topology

Hi Brandon,

First problem

Let a and b be rational numbers, with a < b.

To find a rational q between a and b, your idea of taking an average 
is correct: it should be easy to prove it.

We want also to find an irrational p between a and b.

You can certainly find an irrational number between 0 and 1 (for 
example, 1/sqrt(2) is such a number) - call that number z. For general 
a and b, consider the function:

   f(x) = (b-a)x + a

The point is that f maps 0 to a and 1 to b, and any number between 0 
and 1 to a number between a and b. In addition, as the coefficients 
are rationals, f maps rationals to rationals and irrationals to 

Specifically, if we consider the number 1/sqrt(2) (an irrational 
between 0 and 1), its image will be:

  a + (b-a)*(1/sqrt(2))

and that number is irrational and between a and b. I used the letter 
z because there is nothing special with 1/sqrt(2) - any irrational 
between 0 and 1 would do.

Second problem

We assume now that a and b are irrational numbers, with a < b.

The idea is as follows:

Consider the set of numbers:

  A = {n/r | n in Z}

where r is an integer greater than 2/(b-a). Note that all elements of 
A are rational.

On the number line, these numbers correspond to points spaced (1/r) 
apart. Note that

  (1/r) < (b-a)/2.

The length of the segment (a,b) is (b-a), and the spacing of the 
numbers of A is less than half that interval. It is intuitively 
obvious that there will be one point of A in the interval (a,b).

To prove it rigorously, define n as the least integer greater than 

We have:

  n-1 <= ra < n


 (n-1)/r <= a < n/r

Can you prove that (n/r) < b, using the fact that

  1/r < (b-a)/2 ?

This will give you the rational q you are looking for.

To find an irrational between a and b, consider (a+q)/2. That number 
is certainly between a and b; can it be rational ?

Does this help?  Write back if you'd like to talk about this 
some more, or if you have any other questions.

- Doctor Jacques, The Math Forum 

Date: 03/05/2003 at 11:37:49
From: Brandon
Subject: Topology

Thank you for what you have given me so far. But I still have a 
question on how to prove n/r < b using 1/r < (b-a)/2. 

Thank you for your help.

Date: 03/06/2003 at 02:10:06
From: Doctor Jacques
Subject: Re: Topology

Hi Brandon,

Concerning the second part, we have:

  (n-1)/r <= a
  1/r     < (b-a)/2

If we add these inequalities, we get:

  n/r < a + (b-a)/2 = (a+b)/2 < (b+b)/2 = b

since a < b. As we also know that n/r > a and n and r are integers,
n/r is a rational between a and b.

Does this make sense ?

- Doctor Jacques, The Math Forum 

Date: 03/06/2003 at 12:57:09
From: Brandon
Subject: Thank you (Topology)

Yes, that makes perfect sense. Thank you Dr. Jacques.

Associated Topics:
High School Sets

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