Re: Rational numbers

From: Bob Silverman (anonymous_at_mathforum.org)
Date: 07/28/04 

Date: Wed, 28 Jul 2004 21:07:46 +0000 (UTC)

On 28 Jul 2004, Paul A. Tanner III wrote:
>
<snip>

>Sorry, but Bob's question is not the rhetorical question I pose and
>answer for students. The rhetorical question I pose and answer is not
>why irrationals exist, but, if you wish more precise language, what is
>it about irrationals in decimal form (or in expanded digital form in
>other bases) that makes it impossible to translate them into a (field)
>defined ratio of integers.

The question, posed this way, is nonsense. When you ask about
"irrationals in decimal form", you are mixing PROPERTIES of the
numbers with REPRESENTATION. Irrationals can be represented
many ways. Some of these representations are finite, some are not.

Perhaps what you mean to ask is

Why can't numbers whose decimal expansion is infinite, but not
 periodic be represented as the ratio of two integers?

Is this what you have in mind?

And the answer is quite simple. Numbers that are rational have
decimal expansions that are either finite or are periodic.

Therefore, numbers whose decimal expansion is infinite, but not
periodic are not rational. We call these "irrational"

Why is this so hard???

Furthermore, there ARE fields in which irrationals CAN be represented
as the ratio of integers in that field. But these fields are 'larger'
that the rationals in the sense that they are countably infinite, but
contain the rationals as a proper subset. Consider, for example,
the field Q(sqrt(2)). In this field, sqrt(2) CAN be represented as
the ratio of elements of the ring of integers of that field.

You keep asking the same question in different ways. You seem to be
asking in disguised ways: why are the irrationals not part of the
field of rationals. The field of rationals forms a set. Simply put:
irrationals are not IN that set.

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