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Topic: Calculus without irrational numbers
Replies: 2   Last Post: Aug 11, 2013 2:09 PM

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 gnasher729 Posts: 419 Registered: 10/7/06
Re: Calculus without irrational numbers
Posted: Aug 10, 2013 9:18 PM

On Sunday, August 11, 2013 12:30:35 AM UTC+1, G Patel wrote:
> It is possible to do differentiation over the rationals and (definite) integration, but the FTOC would not be true because f'=g' => f = g + C requires mean value theorem, which requires extreme value theorem to prove it, which requires the underlying field to be complete (i.e., supremums to exist in the proof of extreme value theorem).
>
> So Calculus would exist, but FTOC would not as "antiderivative" would not be well defined.

Wrong. Definite integrals wouldn't exist in general.

With rational numbers, we would only have functions defined over rational numbers, and all function values would also be rational.

Integral (from 1 to 2) (1/x) dx in the real numbers is ln 2. With rational numbers only, it doesn't exist.

Date Subject Author
8/10/13 gaya.patel@gmail.com
8/10/13 gnasher729
8/11/13 David C. Ullrich