Re: Archimedean Axiom and second order logic

From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 06/09/04 

Date: Wed, 09 Jun 2004 09:17:17 -0500

On Wed, 09 Jun 2004 14:04:11 GMT, "Edwin Clark" <eclark@math.usf.edu>
wrote:

>
>"A N Niel" <anniel@nym.alias.net.invalid> wrote in message
>news:080620040815232934%anniel@nym.alias.net.invalid...
>> Tarski showed that the first-order theory of the reals is decidable.
>> Goedel(?) showed that the first-order theory of the integers is not
>> decidable.
>> Therefore, any method of specifying the integers in the theory of the
>> reals is NOT first-order.
>>
>
>Well, my knowledge of such matters is limited. But perhaps you can tell me
>how one defines the reals in first order terms. Don't you need something
>equivalent to the least upper bound axiom (which is second order)?

Saying that the first-order theory of the reals is decidable does not
say anything about "defining the reals".

It actually doesn't say much at all until a few things are specified,
in particular what the language is. My quite possibly wrong
recollection is that we're talking about just the arithmetic
operators + - *, etc; which functions/operators we're including
is what I mean by specifying the language. The statement
"the first-order theory of the reals in the language [whatever]
is decidable" just means that there is an algorithm for determining
whether or not a sentence _in that language_ is true of the real
numbers.

>--Edwin Clark
>
>

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

David C. Ullrich

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