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### Distances between Rational and Irrational Numbers

```Date: 01/13/2005 at 13:05:51
From: Herman
Subject: Math,  irrational and rational numbers

I know that if a rational number p/q is closer to an irrational number
x than it is from another rational number r/s (the fractions are
reduced so that the numerator and denominator do not have common
divisors) then q has to be larger than s, but I do not know how to
prove it.

That is, if |x - p/q| < |x - r/s| with x irrational and p/q, r/s
irreducible rationals then q > s.

I do not know where to start.  For example in pi=3.1415..., it is
clear than in the sequence 3/10, 31/100, 314/1000, the denominator is
getting large.  What bothers me is that the fractions are not
necessarily simplified, that is 314/1000 when simplified should be
157/500 and yes 500 > 100.  This involves things on number theory
(relative prime) and real analysis.

My problem arose when trying to find a function that is discontinuous
in all rationals and continuous in all irrationals.  I believe the
function is f(x) = 0 if x is irrational and f(x) = 1/q if x = p/q
reduced rational.  When proving continuity I need to be able to show
that for each epsilon > 0, there exists delta > 0 such that
|1/q| < epsilon always that |p/q - x| < delta.  If I show what I have
in my question, then I just need to find one (the first) p/q and after
that I do not have to do anything.

```

```
Date: 01/13/2005 at 14:19:26
From: Doctor Vogler
Subject: Re: Math,  irrational and rational numbers

Hi Herman,

Thanks for writing to Dr. Math.  There is a great deal of theory
involved in good rational approximations to irrational numbers.  What
you stated in that regard is not correct.  But you don't need anything
nearly as sophisticated to accomplish what you're trying to do.

I assume that you can show that your function is discontinuous at
rational points.  For that, let x = p/q be rational, choose a special
epsilon (anything smaller than 1/q will do) and show that any interval
around x contains points whose image is 0.

To show that your function is continuous at an irrational point x = r,
let epsilon > 0.  You want to show that 0 <= f(x) < epsilon near x =
r.  So find some q so that 1/q < epsilon.  Now show that there are
only finitely many rational points anywhere near x = r (say within 1
of r) whose denominators are no bigger than q.  Show that r is not one
of those points, and then let delta be small enough that the interval
(r-delta, r+delta) misses all of those points.

Can you fill in all of the details?

back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 01/13/2005 at 14:41:46
From: Herman
Subject: Thank you (Math,  irrational and rational numbers)

Doctor Vogler.

This is the first time I use "Ask Dr. Math".  I was expecting an
perfect, beautiful and clear!  I am so happy to see how this worked
out.  I was thinking on this problem for the last few days.

You said that what I stated in my question was not correct.  Could you

Thank you very, very much!

Herman

```

```
Date: 01/13/2005 at 15:02:23
From: Doctor Vogler
Subject: Re: Thank you (Math,  irrational and rational numbers)

Hi Herman,

You stated:

I know that if a rational number p/q is closer to an irrational number
x than it is from another rational number r/s (the fractions are
reduced so that the numerator and denominator do not have common
divisors) then q has to be larger than s.

This can be interpreted in a couple of different ways, but you
clarified when you also wrote

That is, if |x - p/q|<|x - r/s| with x irrational and p/q, r/s
irreducible rationals then q > s.

Here is a large collection of counterexamples:  Pick any x, and pick
any p/q.  In fact, pick any s > q as well, and I will find a number r
so that the (reduced) fraction r/s has

|x - p/q| < |x - r/s|.

Choose an integer

N > x + |x - p/q|

and then let

r/s = N + 1/s = (Ns + 1)/s,

which is to say, let

r = Ns + 1.

Then you have

|x - r/s| = r/s - x > N - x > |x - p/q|.

Anyway, the deep research I mentioned is called "Transcendental Number
Theory" and some of its first main theorems can be seen on the "more"

http://www.amazon.com/exec/obidos/tg/detail/-/052139791X/

which shows you a few pages of Alan Baker's book on the subject.

back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 01/13/2005 at 15:22:55
From: Herman
Subject: Thank you (Math,  irrational and rational numbers)

Great!  Thank you very much!

Herman
```
Associated Topics:
College Number Theory

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