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visits | member for | 2 years, 11 months |
seen | 17 hours ago | |
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Sep 30 |
comment |
Why have mathematicians used differential equations to model nature instead of difference equations
"(That highest number plus one is zero.)" This has always struck me as a curious view. What is gained by supposing the highest number plus one is zero, rather than supposing the highest number plus one does not exist? Indeed, what is gained by supposing there is a highest number, rather than just being agnostic about the matter (and just not assuming the axiom "every number has a successor")? |
Mar 15 |
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Is there a theory of abuse of notation?
Is the poster looking for sufficient or necessary conditions? |
Jan 18 |
awarded | Yearling |
Nov 2 |
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Illustrating Edward Nelson's Worldview with Nonstandard Models of Arithmetic
@EN. Many thanks for your comments. |
Nov 1 |
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Illustrating Edward Nelson's Worldview with Nonstandard Models of Arithmetic
Also, if I may, one other line of question (I think this interests many many more people than myself, so it is why I am presuming to take up your time). Is the notion of "actual number" different from the notion of "natural number"? If the answer is yes, does induction hold for the actual numbers? If the answer is no, why the use of "actual" instead of "natural"? |
Nov 1 |
comment |
Illustrating Edward Nelson's Worldview with Nonstandard Models of Arithmetic
@EN. Why not? For example it would take only around 13 tries to bifurcate 5000 successfully. First question: do you see no reason to believe that ψ(80^i) is a theorem of F if i = 2500? If you answer yes, second question has i = 1250; and if you answer no, second question has i = 3750. etc. Is it a question of vagueness, like 5000 hairs is a lot of hair, but you would say there is no least i such that i hairs is a lot of hairs because "lot" is vague? |
Nov 1 |
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Illustrating Edward Nelson's Worldview with Nonstandard Models of Arithmetic
Is there a least i such that you see no reason to believe that ψ(80^i) is a theorem of F? |
Oct 21 |
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Decidability of equality of elementary expressions
Is it possible to give an answer to the easier question where i is replaced by 1, i.e. one considers as expressions: 1; exp(x); ln(x); (x⋅y) ? |
Sep 30 |
comment |
Has anyone pursued Frege's idea of numbers as second-order concepts?
Suppose there are ten objects, so you can prove the second-order numbers up to 10 exist, are different, and occur in a successoring chain. Let P be the third-order object containing the primes less than 10. Then the claim that there are four of them is the claim there exists a second order set A and a third-order relationship R such that R is 1-1 onto from P to A, where #A = 4. (Perhaps I'm getting mixed up in the orders here, but I hope that works.) |
Sep 24 |
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Has anyone pursued Frege's idea of numbers as second-order concepts?
In arithmetic without the successor axiom, you could say it in this fashion. If S(S(S(S(S(S(S(S(S(S(0))...) exists, then the number of primes less than it is 4. |
Sep 22 |
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Has anyone pursued Frege's idea of numbers as second-order concepts?
@Keshav. I think you can define third-order sequential operator S, third-order natural number predicate N, and second-order 0. Then you're right that you're not going to be able to prove every number has a successor, because you have no first-order objects. But with the definition of N you get induction, and the successor-existence theorem is not essential to the development of arithmetic; you get "arithmetic without the successor axiom." I haven't worked it through, so the buyer should beware, but I don't see any problems. |
Sep 16 |
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Has anyone pursued Frege's idea of numbers as second-order concepts?
If your logic is limited to first- and second-order concepts, then I don't see how you can define anything equivalent to N (the predicate satisfied by the natural numbers) if numbers are second-order, so you won't have induction. If you allow third-order concepts (and have comprehension to state their existence), then I imagine the usual Frege Arithmetic approach would go through, where second-order concepts in FA are replaced by third-order concepts in your logic. |
Jul 30 |
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What proportion of chess positions that one can set up on the board, using a legal collection of pieces, can actually arise in a legal chess game?
Perhaps an easier question first: what proportion of positions are legal when all the original pieces are on the board? |
Jun 2 |
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One can earn nothing on the Brownian motion, true ?
Well, that was meant to be a comment. Can't see how to change it, though. |
Jun 2 |
answered | One can earn nothing on the Brownian motion, true ? |
May 30 |
comment |
Deduction theorem
@Francois. I wasn't aware I was loosening the rules. Looking at Mendelson, he defines a formal axiomatic theory for the propositional calculus with three axioms. Keep only the first of the three, which is A => (B => A). Then A => A isn't provable, but at least according to Mendelson's definition, it is a formal axiomatic theory (just not an interesting one). Point taken that there are some systems where not even A follows from A. |
May 29 |
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Deduction theorem
Perhaps one should include "interesting" in front of "axiomatic system"? Even in an empty axiomatic system, A always follows from A, but in an empty axiomatic system, one cannot prove anything, much less A => A. By considering any set of axioms which do not allow the proof of A => A, the deduction theorem would still evidently not hold. |
May 19 |
answered | Reference request: Minimal Axiomatizations of PA over (+,x,<=). |
May 18 |
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Reference request: Minimal Axiomatizations of PA over (+,x,<=).
Are you looking for something like: 1/ Induction: From phi(0) & (n)(phi(n) => phi(n+1)) infer (n)phi(n) 2/ x ≤ y iff (there exists z)(z + x = y) 3/ x + 0 = 0 4/ x + (y + 1) = (x + y) + 1 5/ x * 0 = 0 6/ x * (y + 1) = (x * y) + x 7/ There is no ≤ maximal element I think that works... |
Mar 1 |
answered | Is there any straightforward way to substitute for Gaussian/Brownian assumptions in financial mathematics? |