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### Density of Rational Numbers

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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?

site!

Sincerely,

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

Hi,

"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/
```

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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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