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Density of Rational Numbers


Date: 01/31/98 at 16:38:11
From: Keri McKenzie
Subject: Why the set of rational numbers is dense

Dear Dr. Math,

I am wondering what the following statement means:

"The set of rational numbers is dense." 

I know what rational numbers are thanks to my algebra textbook 
and your question sites. I have determined through brainstorming 
that rational numbers are dense because there are so many of them. 
In other words, they are densely crowded when defined as a set 
IE. Q={-4,0,2/3,4,5.56869}. Am I on the right track? 

Thanks for your help. I look forward to your input and I love your 
site!

Sincerely, 

Keri


Date: 01/31/98 at 19:26:52
From: Doctor Pete
Subject: Re: Why the set of rational numbers is dense

Hi,

The statement should read:

   "The set of rational numbers is dense in the real numbers."

(We will shortly see why we must say "...in the real numbers.")  

Now, let

    Z = the set of all integers
    Q = the set of all rational numbers
    R = the set of all real numbers
  R\Q = the set of all irrational numbers 
        (namely, those elements in R which are not in Q).

Then the meaning of "dense," in this case, is the following: Given any 
two distinct rational numbers p < q, there exists an irrational number 
x such that p < x < q; that is, given two distinct rational numbers we 
can always find an irrational number in between. And similarly, given 
any two distinct irrational numbers x < y, we can find a rational 
number q such that x < q < y.

So, we see why we must append the statement "...in the real numbers" 
because the property of "dense" of a set is dependent on the bigger 
set you contain the original set in. That is, in this case, Q is a 
proper subset of R, and to consider if Q is dense in R, we must 
examine elements which are in their difference R\Q.

A question:  Is Z (the set of integers) dense in Q?

The proof that Q is dense in R involves some basic analysis, and 
without knowing your background, I don't know how useful it would be 
for me to include it here. However, it is a somewhat "intuitively 
obvious" fact (but not so obvious once you see the proof). Density is 
a much stronger statement than just saying "there are a lot of them," 
because it is in effect saying, "there are a lot, *and* they're very, 
very closely packed together." Hence the choice of the word "dense" 
(as opposed to, say, "numerous").

-Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   


Date: 09/02/98 at 14:12:46
From: Joan Vilaseca
Subject: Cardinality of rationals and irrationals

In a section of the entry Density of Rational Numbers, you wrote the following:

   Given any two distinct rational numbers p < q, there exists an   
   irrational number x such that p < x < q; that is, given two 
   distinct rational numbers we can always find an irrational number 
   in between. And similarly, given any two distinct irrational 
   numbers x < y, we can find a rational number q such that x < q < y.

That is, there is no pair of irrationals without an irrational in 
between and also no pair of irrationals without a rational in between.

But then, this is a one-to-one relation (just map the first of every 
pair with the number in between.) And that means that cardinality of 
rationals and irrationals seems to be the same.

Where's the fault?

In geometry, that contradiction arises when we suppose a line to be 
equivalent to a set of points (0+0+0+....+0<>0).

Joan Vilaseca


Date: 09/02/98 at 16:24:35
From: Doctor Wilkinson
Subject: Re: Cardinality of rationals and irrationals

You have not defined a one-one relation. In the first place, every 
rational appears in infinitely many pairs, so you can't say for a 
given rational which irrational you're mapping to, and similarly for 
the irrational -> rational mapping. Secondly, you have not given any 
rule for choosing an irrational between two rationals or for choosing 
a rational between two irrationals. Finally, to get a one-one relation 
from these two mappings, even assuming they were defined, you would 
have to show that they were inverses of each other.

- Doctor Wilkinson, The Math Forum
  Check out our web site! http://mathforum.org/dr.math/   
    
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