My question is: is there a sensible way to define a measure how 'interesting' is a given theorem (or theory)? — Mephist
we can use Shannon information theory to compute the surprisal, i.e. the information content — alcontali
So how could that be applied in such an abstract 'possibility space', so to speak? — Wayfarer
I don’t know but i think work on measuring abstract human concepts like how interesting, original, artistic, aesthetic and so forth things are should be useful in the work on developing artificial intelligence — Frotunes
But I also suspect it too will suffer from what psychology and ethics suffer from, which is that are rather vague, and what humanities suffers from, which is that they’re rather generalising and socio-political, or what formal logic suffers from, which is that it’s rather dismissive of plain common sense (which, though sometimes prone to irrationality, is often very useful) — Frotunes
But I agree with you that some kind predictability measure (such as the one defined by Shannon information theory) should have a role. — Mephist
In fact, I believe that even if you took a very talented mathematician that has no knowledge at all about a new theory, and you presented him with the purely formal expression of a theorem, he wouldn't be able to say if that is something interesting or absolutely boring. — Mephist
o, from this point of view, the representation of a theory in Category theory as a functor from the logical proposition to the the model should be the object that is taken into account to calculate Shannon's probability. — Mephist
And that, of course, would be essential to build some kind of A.I. software that "understands" mathematics. — Mephist
Well, what I meant to ask was is if could exist (or if somebody invented) any "automatic" and "objective" way to recognize meaningful mathematical theorems (or theories) when they are expressed in a formal language.
Probably the answer is NO. But if there is no such thing as an objective "value" of a mathematical sentence, how can mathematicians be able to recognize an "interesting" new theory when they see one? — Mephist
So, could a computer program recognize this theorem as surprising (or improbable)? — Mephist
A model could be of course a computer-vision software plus the ability to paint triangles and observe them — Mephist
In other words, both terms of the logical equivalence have a very small complexity (or information measure) if interpreted taking as model an axiomatization of real numbers. — Mephist
Then, if we give a computer a "physical" model of geometry (for example a world made of pixels), we could recognize, experimenting with it, that every length can be measured, and every measure is in fact a fraction. So, he could recognize as "surprising" (or statistically very improbable) a theorem that says that there is a measure that is not a fraction. — Mephist
and "pretend" that it had a solution by adding a new symbol "i" to the number system, then you discover that now all polynomials have a solution. — Mephist
Would a computer be able to "discover" complex numbers? I don't think so — Mephist
But I think that a compute could be able to RECOGNIZE that complex numbers are an interesting concept by using a measure of how "improbable" are theorems when interpreted on the right models — Mephist
It is exactly this kind of visual puzzles that Immanuel Kant rejected in his Critique of Pure Reason. Classical Euclidean geometry is not pure reason, because it requires solving visual puzzles. Immanuel Kant was adamant: Pure reason is language only. It may only make use of symbol manipulation. — alcontali
Agreed. Pythogoras theorem is only surprising in a visual puzzling environment such as classical Greek geometry. — alcontali
Experimental testing is forbidden in math. Therefore, this approach by number sampling is very, very un-mathematical. In fact, you do not prove anything by sampling lots of numbers. The proof must be the result of judicious symbol manipulation instead. — alcontali
For polynomials with coefficients in the rationals, you can indeed still reduce an otherwise irreducible polynomial by adding the appropriate field extension. Still, even adding i does not guarantee a closed-form solution for the roots (constructed using only supported field operators: + - * /), because we are not sure that there is a tower of radical field extensions available to achieve that. Even though adding i guarantees a solution, it may not spare you from having to approximate. — alcontali
Indeed, no. Gödel's incompleteness precludes that. It is generally not possible to discover new theorems by enumerating the domain of theorems and then verifying if they are provable in the theory. — alcontali
In my impression, the surprising element is that you do not need to extend it further to guarantee a solution for all polynomials with rational coefficients, but not necessarily in closed form. — alcontali
I think that the practice of using field extensions more or less guaranteed that i would be discovered ... — alcontali
From a formal point of view, it's a logical equivalence between orthogonality of two vectors and an equation between real numbers (vectors' lengths).
Should an equivalence of this kind be surprising? (or improbable)? The answer is NO: in fact, whatever angle you take between two vectors, there will be an induced equation between vectors' lengths, and vice-versa. — Mephist
I am very curious about this. Doesn't the vector's length follow from the Pythagorean theorem? If so, then it seems it isn't a simple logical equivalence, since the equation between real numbers (a vector's length) depends on the application of the theorem. Or is there an independent way of measuring a vector's length? — Kornelius
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