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April 24[edit]

Completeness of axiomatic systems[edit]

Let A,B be two complete axiomatic systems, [having no finite model, and] phrased in a first order language, that denotes one operation only: It's a binary one, denoted: as "+" in the language of A, and as "×" in the language of B.

Let C be a third axiomatic system, that only contains the axioms of A and the axioms of B, along with the axiom of distributivity of "×" over "+". Is C necessarily ~~complete? Necessarily~~ incomplete? 185.46.76.1 (talk) 08:28, 24 April 2019 (UTC)[reply]

I suppose you mean a multiplication symbol '×', rather than a letter 'x'? If you edit in a desktop computer's browser, you can find the symbol in the 'Special characters'/'Symbols' above the edit field. If you use Windows, you can type the character directly from the numeric keypad with Alt+0215. I wouldn't be surprised, if it worked also in Linux. On mobile devices with Android, you will probably find the character in Symbols tab of the screen keyboard... --CiaPan (talk) 15:07, 24 April 2019 (UTC)[reply]

Thanks to your useful comment, I've just replaced "x" by "×". 185.46.78.10 (talk) 16:34, 24 April 2019 (UTC)[reply]

This sounds like homework: what have you tried? Hints: 1) think about Presburger arithmetic vs Peano arithmetic; 2) think about theories with finite models. 173.228.123.207 (talk) 05:32, 27 April 2019 (UTC)[reply]

No, it's not homework. Unfortunately, your hints didn't help me. I know that Presburger arithmetic (having one operation only) is complete, while Peano Arithmetic (having two operations controlled by the distributivity property) is not, so what? Anyway, my question is not limited to systems satisfied by finite models. 185.46.78.10 (talk) 21:44, 27 April 2019 (UTC)[reply]

Think of an axiomitized theory of a particular finite field (perhaps F₂). Write down just the axioms for its additive group as A, then for its multiplicative group as B, then put them together as C. Since the only model is finite, C has the properties you want. Now do something similar for Peano arithmetic: the additive fragment is Presburger arithmetic (which is complete), and I believe you can do something similar to get a complete theory for the multiplicative fragment. But when you put them together you get an incomplete theory. 173.228.123.207 (talk) 01:33, 28 April 2019 (UTC)[reply]

First, the additive fragment of Peano arithmetic is really complete, but the multiplicative fragment is not, so B can't be that fragment. Secondly, I don't assume that A,B have a finite model, nor do I assume they have any model at all - in which case C is undoubtedly complete. 185.46.78.10 (talk) 21:44, 27 April 2019 (UTC) 06:53, 28 April 2019 (UTC)[reply]

The multiplicative theory of the naturals is decidable, so let B be that theory. This demonstrates that C need not be complete.

As for the bit about finite fields, you've missed the point. It's an example showing that C need not be incomplete.--2404:2000:2000:5:0:0:0:C1 (talk) 23:23, 28 April 2019 (UTC)[reply]

Yes, sorry I was not more clear about the finite fields: since C's only model is finite, it is obviously decidable. I'll try to check whether any difficulties arise from combining the additive and multiplicative fragments of Peano arithmetic. Hmm. 173.228.123.207 (talk) 06:38, 29 April 2019 (UTC)[reply]

OP's comment. Thanks to your clarification about the finite models, I've just rephrased my original question. See above, under the title of this thread. 185.46.78.16 (talk) 07:51, 29 April 2019 (UTC)[reply]

Same answer as before, I think, except use the theory of real closed fields (which is decidable by a famous theorem of Tarski, see the article) instead of finite fields. Finite fields just made it extra clear. 173.228.123.207 (talk) 06:49, 30 April 2019 (UTC)[reply]

Way of obtaining Euler formula in logarithmic form by Roger Cotes[edit]

How has Roger Cotes arrived at the logarithmic form of Euler formula, as mentioned in the article, but lacking details?--109.166.136.255 (talk) 22:26, 24 April 2019 (UTC)[reply]

This may be useful. The full context (in Latin) can be found at [1]. Apparently Cotes didn't actually have a formula but wrote out the relationship in words, so it might be a bit of a stretch to say that Cotes anticipated Euler with the formula. Interpreting 300 year old math texts isn't easy though, and I don't claim to understand what Cotes wrote, but maybe that will give you a flavor of what Cotes was working on at the time. --RDBury (talk) 00:28, 25 April 2019 (UTC)[reply]

Actually footnote #5 in the Cotes article covers a lot of what I said above and has some additional explanation. The Stillwell book cited in the Euler formula article only states Cotes' original for its source and doesn't give an explanation as to how Cotes' prose is translated into the formula given; perhaps Stillwell is relying on an intermediate source for this. --RDBury (talk) 05:21, 25 April 2019 (UTC)[reply]