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I've wondered this for a while. Most people these days believe ZFC is consistent on intuitive and other grounds, so let's assume that it really is consistent. What reason do we have to believe that it is arithmetically sound? Could ZFC be consistent but false in the sense that it proves arithmetic statements that are false about the standard integers? Nik Weaver wrote an article suggesting this is actually the case, that ZFC has no models whose ω is the standard integers and that it is -unsound, or anyway suggesting that we don't have much evidence to the contrary. But, I get the impression that his writings are considered a bit out of the mainstream. If we assume PA is sound, then since ZFC proves PA, any arithmetically false theorem of ZFC would have to be independent of PA, already putting such theorems in a pretty rarefied realm. Are there some numerical experiments that might give evidence one way or the other? 67.122.209.126 ( talk) 00:40, 1 June 2009 (UTC)
About Sigma-1, I was stating Weaver's conjecture that ZFC is Sigma-1 unsound, which is of course an even stronger statement than ZFC being unsound at some higher quantifier depth. It turns out this topic (of ZFC unsoundness) was discussed extensively on the FOM mailing list earlier this month: [1], see many posts from Weaver and others on "arithmetical soundness of ZFC" in that thread. I just came across this while googling.
Of course ZFC in the abstract proves many arithmetical statements that PA does not. By "rarefied", I mean there aren't many such theorems that actually turn up in mathematics as practiced by humans, e.g. Kruskal's theorem. All the examples I can think of are provable in second-order arithmetic, but I'm no expert. I do know about Harvey Friedman's artificial examples that use ZFC, large cardinals, etc. 67.122.209.126 ( talk) 14:15, 1 June 2009 (UTC)
Just to check I understand: when Nik Weaver claims that ZFC is likely -unsound, he means that probably either asserts the existence of functions that are not functions over the naturals in the standard model, or denies the existence of such functions that are? And, since his complaint is that ZFC overgenerates, presumable that ZFC asserts the existence of such phantom functions? My gut feeling is that overgeneration here would lead to inconsistency. — Charles Stewart (talk) 08:37, 4 June 2009 (UTC)
This whole business, I'm really struggling to understand it, and this seems to be the right place to ask about the thing lol.
When I read the derivation given here I have the impression of having a really good card trick done on me, that it's impressive and you can't figure out why or how it does what it does, but you don't really believe that it's doing what it appears to be. My confusion is that deriving it requires measuring angles in radians, and that seems to be the only reason for bringing π into it in the first place. Wouldn't it work equally well as if you were using degrees? Or any positive number in fact, depending on what you were measuring angles in?
Am I massively understanding the ideas here lol, all I have is AS level Maths & Further at the moment and my understanding is rather limited heh, I was just astonished by the beauty of this thing and I was wondering how it worked in greater detail. —Preceding unsigned comment added by Dan Hartas ( talk • contribs) 20:47, 1 June 2009 (UTC)
This is an excellent question whose answer is worthy of its own Wikipedia page. If we set a = eπ/180, then a180i = −1. In a similar way, you can find a number b such that b15i = −1, or put any other number (except 0) in the role in which we see 15 in this example. If the number you want in that role is π, then what you need in the role of a or b is e. Recall something about exponential functions and about e. If we let ƒ(x) = cx, then ƒ is an exponential function, satisfying the important law
Now suppose we let
Using the two trigonometric identities
and the definition of multiplication of complex numbers
we can see that
and this is that same "important identity" that ƒ satisfies. In that way, cis is seen to behave like an exponential function, as if we had this identitiy:
But what number would be the base, d? Now remember something about exponential functions: if g(x) = cx, then the derivative is
and that is easy to derive from the definition of the derivative. The "constant" is equal to 1 precisely if the base c is equal to e, and one can say that's why e is so "natural". In other words, when x = 0, then g(x) is changing 1 times as fast as x is changing, precisely if the base c is equal to e. Now suppose x = 0, so that g(x) = 1, and x is changing at a rate of i. That means as x passes the number 0 while moving "upward", along the "imaginary" axis, g(x) passes the number 1 while moving to the "upward", in the "imaginary" direction, at the same rate. Now the point(cos(x), sin(x)) moves along the circle at the same rate as the rate at which x moves, precisely when x is in radians—that's how radians are defined. (So the answer to the question posed earlier, "what is d if cis(x) = d x, is d = e i.)
Bottom line: measuring the angles in radians is the same as letting e be the base of the exponential function. Michael Hardy ( talk) 02:32, 2 June 2009 (UTC)
Mathematics desk | ||
---|---|---|
< May 31 | << May | June | Jul >> | June 2 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
I've wondered this for a while. Most people these days believe ZFC is consistent on intuitive and other grounds, so let's assume that it really is consistent. What reason do we have to believe that it is arithmetically sound? Could ZFC be consistent but false in the sense that it proves arithmetic statements that are false about the standard integers? Nik Weaver wrote an article suggesting this is actually the case, that ZFC has no models whose ω is the standard integers and that it is -unsound, or anyway suggesting that we don't have much evidence to the contrary. But, I get the impression that his writings are considered a bit out of the mainstream. If we assume PA is sound, then since ZFC proves PA, any arithmetically false theorem of ZFC would have to be independent of PA, already putting such theorems in a pretty rarefied realm. Are there some numerical experiments that might give evidence one way or the other? 67.122.209.126 ( talk) 00:40, 1 June 2009 (UTC)
About Sigma-1, I was stating Weaver's conjecture that ZFC is Sigma-1 unsound, which is of course an even stronger statement than ZFC being unsound at some higher quantifier depth. It turns out this topic (of ZFC unsoundness) was discussed extensively on the FOM mailing list earlier this month: [1], see many posts from Weaver and others on "arithmetical soundness of ZFC" in that thread. I just came across this while googling.
Of course ZFC in the abstract proves many arithmetical statements that PA does not. By "rarefied", I mean there aren't many such theorems that actually turn up in mathematics as practiced by humans, e.g. Kruskal's theorem. All the examples I can think of are provable in second-order arithmetic, but I'm no expert. I do know about Harvey Friedman's artificial examples that use ZFC, large cardinals, etc. 67.122.209.126 ( talk) 14:15, 1 June 2009 (UTC)
Just to check I understand: when Nik Weaver claims that ZFC is likely -unsound, he means that probably either asserts the existence of functions that are not functions over the naturals in the standard model, or denies the existence of such functions that are? And, since his complaint is that ZFC overgenerates, presumable that ZFC asserts the existence of such phantom functions? My gut feeling is that overgeneration here would lead to inconsistency. — Charles Stewart (talk) 08:37, 4 June 2009 (UTC)
This whole business, I'm really struggling to understand it, and this seems to be the right place to ask about the thing lol.
When I read the derivation given here I have the impression of having a really good card trick done on me, that it's impressive and you can't figure out why or how it does what it does, but you don't really believe that it's doing what it appears to be. My confusion is that deriving it requires measuring angles in radians, and that seems to be the only reason for bringing π into it in the first place. Wouldn't it work equally well as if you were using degrees? Or any positive number in fact, depending on what you were measuring angles in?
Am I massively understanding the ideas here lol, all I have is AS level Maths & Further at the moment and my understanding is rather limited heh, I was just astonished by the beauty of this thing and I was wondering how it worked in greater detail. —Preceding unsigned comment added by Dan Hartas ( talk • contribs) 20:47, 1 June 2009 (UTC)
This is an excellent question whose answer is worthy of its own Wikipedia page. If we set a = eπ/180, then a180i = −1. In a similar way, you can find a number b such that b15i = −1, or put any other number (except 0) in the role in which we see 15 in this example. If the number you want in that role is π, then what you need in the role of a or b is e. Recall something about exponential functions and about e. If we let ƒ(x) = cx, then ƒ is an exponential function, satisfying the important law
Now suppose we let
Using the two trigonometric identities
and the definition of multiplication of complex numbers
we can see that
and this is that same "important identity" that ƒ satisfies. In that way, cis is seen to behave like an exponential function, as if we had this identitiy:
But what number would be the base, d? Now remember something about exponential functions: if g(x) = cx, then the derivative is
and that is easy to derive from the definition of the derivative. The "constant" is equal to 1 precisely if the base c is equal to e, and one can say that's why e is so "natural". In other words, when x = 0, then g(x) is changing 1 times as fast as x is changing, precisely if the base c is equal to e. Now suppose x = 0, so that g(x) = 1, and x is changing at a rate of i. That means as x passes the number 0 while moving "upward", along the "imaginary" axis, g(x) passes the number 1 while moving to the "upward", in the "imaginary" direction, at the same rate. Now the point(cos(x), sin(x)) moves along the circle at the same rate as the rate at which x moves, precisely when x is in radians—that's how radians are defined. (So the answer to the question posed earlier, "what is d if cis(x) = d x, is d = e i.)
Bottom line: measuring the angles in radians is the same as letting e be the base of the exponential function. Michael Hardy ( talk) 02:32, 2 June 2009 (UTC)