Re: A basic question on real numbers

On 12 Jun 2006 03:55:13 -0700, aone1504@xxxxxxxxx wrote:

I have the following basic question:-

Irrational numbers can be obtained as the limit of cauchy sequence of
rational numbers.

Is there a proof that a convergent infinite series having irrational
terms converges to a number which can always be obtained as the (above)
limit of cauchy sequence of rationls.

I would like to see the proof for this statement.

Well, in fact _any_ real number is the limit of a
cauchy sequence of rationals - this has nothing to
do with the fact that the number is a "coonvergent infinite
series having irrational terms".

How you prove this depends on where you're starting. I don't
know what the context is: (i) if this is a situation where
you have some well-defined set of axioms that the real numebers
satisfy then tell us what your axioms are and someone will
show you the proof from those axioms. (ii) if on the other
hand this is all very informal, we haven't said yet
exactly what the real numebers are, then consider the
following:

Probably you believe this:

(*) If x is any real number then there exists an integer
n such that

n <= x < n + 1.

We call this integer the floor of x, written [x].

If you believe (*) then it's easy to prove the following
theorem:

Thm. If x is a real number then there exists a sequence
of rationals that converges to x.

Proof: First note that for any x we have

[x] <= x < [x] + 1,

which implies that

0 <= x - [x] < 1,

which implies that

(*) |x - [x]| < 1.


Now suppose that x is a real number, and for
n = 1, 2, ... define

r_n = [n*x]/n.

Then r_n is certainly rational, since [n*x] is
an integer. And

|x - r_n| = |(n*x - [n*x]) / n|
= |n*x - [n*x]| / n
< 1/n,

by (*). This shows that r_n -> x as n -> infinity.

QED.

Thanks.


************************

David C. Ullrich
.

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