AnotherKevin comments on Gödel and Bayes: quick question - Less Wrong

1 Post author: hairyfigment 14 April 2011 06:12AM

You are viewing a comment permalink. View the original post to see all comments and the full post content.

Comments (36)

You are viewing a single comment's thread. Show more comments above.

Comment author: Sniffnoy 14 April 2011 06:46:31AM *  12 points [-]

The key is not just addition and multiplication, the key is addition and multiplication of whole numbers. The first-order theory of the real numbers, by contrast - with addition and multiplication - is decidable. And probability deals with real numbers, not integers.

On top of that you have the issue that the "axioms" of probability are not axioms in the sense of an axiomatic system, they're just "axioms" in the sense of "requirements".

In short this whole post seems badly misinformed.

Comment author: AnotherKevin 15 April 2011 09:42:37PM *  0 points [-]

Either I am confused or this discussion is confused.

N(X) iff (X=0) || ((X > 0) && N(X-1)) iff X is natural or 0
Z(X) iff ( (X >= 0) -> N(X) ) && ( (X < 0) -> N(0 - X) ) iff X is an integer

equivalently

X is a natural number

X is an integer

I'm also under the impression that the algebraic numbers are countable, dense in R, and that

Edit: note to all, mixing latex and plain text on a line looks messy. Further edited for formatting due to lack of preview.

Comment author: Sniffnoy 15 April 2011 10:54:42PM 1 point [-]

If you're attempting to define N as a first order predicate, that doesn't work; you've defined it in terms of itself. You can't directly define predicates recursively; predicates must be finite. If you want to do get a "recursive" predicate you have to do quite a bit more work than that, and in particular you need tools not available in the first order theory of the reals (with addition and multiplication, as usual).

Your definition of Z has additional minor problems; you mean and, not implies. (X>=0 => N(X)) is automatically satisfied for any X<0.

Your last statement is correct (if a bit less general than it could be :) ), though your notation is a bit strange. (Again, assuming + and * as usual.)

Might I ask what the relevance of all this is?

Comment author: AnotherKevin 16 April 2011 12:48:35AM 0 points [-]

Z is defined correctly. When X >= 0 the formula becomes N(X) AND TRUE when X < 0 the formula becomes TRUE AND N(0-X).

Otherwise I was confused. I was trying to define N implicitly which I should have recognized as invalid. Explaining what I was trying to say at the end would be pointless given that I didn't say it and it's also wrong =P. Mea culpa

Comment author: Sniffnoy 16 April 2011 08:25:43AM *  1 point [-]

Oh, you said "and"; my apologies, I implicitly read an "or" there!