Talk:Richard's paradox

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[edit] Pre-major-revision comments

A google search seems to prefer the name "Richard's paradox". --Brion 23:49 Oct 1, 2002 (UTC)

I don't believe that anyone not previously familiar with this paradox would be able to follow this article. I sure can't. For example, this

One example sentence would be "That positive real number whose square is two."

The sentence quoted is not a sentence. Ortolan88 Probably the one who wrote above isn't a native English speaker (neither am I, anyway) and got confused between sentence and phrase.

[edit] Post-major-revision comments

The paradox is implying that the set of these phrases (let's call it P) is countable. If it is, then the "paradox" is quite obvious, as R is an uncountable set. If P isn't countable, then it makes no sense to say "the nth phrase" because doing so requires a mapping between N and P. And it is by no means explained why couldn't P be uncountable.

I`m absolutely not mathematical, and maybe I`m just being dumb, but I don`t understand this:-

"Thus, if a number is Richardian, then the definition corresponding to that number is a property that the number itself has. (More formally, "x is Richardian" is equivalent to "x does not have the property designated by the defining expression with which x is correlated in the serially ordered set of definitions".)"

Seems to be self-contradictory.

You're right. Thanks for the catch. Typos usually aren't that bad, but they tend to be really nasty when you forget to use the word "not". Heh. Eric Herboso 01:05, 23 Jan 2005 (UTC)
Also, the reason why the set P is definitely countable is because every element in P is a string of characters, and we order the set by putting each element with a lower amount of characters before any element with a higher amount; and for any two elements with the same number of characters, we order according to standard alphabetical order (including extra characters in the order as well). If this isn't sufficiently clear in the article, feel free to more specifically address this issue there. Hope this helps. Eric Herboso 01:13, 23 Jan 2005 (UTC)
That's true if we explicitly require to use finitely long phrases. If we use infinitely long phrases, we can express any real number, for example the one for pi begins "the real number whose floor is three, whose first decimal digit is one, whose second decimal digit is four..." And in the article, nowhere is it stated that definition must be finite. --Army1987 22:06, 9 August 2005 (UTC)

[edit] POSSIBLE COPYRIGHT VIOLATION (sorry about the caps, but it's important)

A substantial portion of this article appears to be taken from the book Gödel's Proof by Nagel & Newman (link goes to Amazon). While there aren't any direct quotes, phrases such as "an essential but tacit assumption" appear in both, and are unlikely to have been developed independently. I don't think it quite warrants blanking the article, but a serious rewrite and/or citations are needed. I don't have time to work on it, though. Benandorsqueaks 23:55, 27 March 2006 (UTC)

I was the one who wrote that substantial portion of the article. I did use Nagel & Newman's book as a reference when rewriting this article, so this might explain my particular choice of phrases being similar to Nagel and Newman. Indeed, the flow of the argument is also taken from Nagel and Newman, since I felt their order of presentation was very good for their discussion of Richard's paradox. But I do not feel that there has been a copyright violation here, as the source material was merely used as a reference when rewriting the article due to its ideas on how to present Richard's paradox most efficiently and readably. If such a use of reference were considered a copy vio, then when writing an article on calculus, you wouldn't be able to reference a calculus book's presentation flow, or when writing an article on history, you wouldn't be able to make the same points that some recent book you'd read had talked about. Because I merely took the ideas from Nagel and Newman and used them to write the article, rather than quote from the text, I feel that this does not constitute a copyright violation. Of course, I am no lawyer, so if someone who knows more on this disagrees with me, then they should say so on my talk page, and I will rewrite the appropriate sections. — Eric Herboso 20:42, 3 April 2006 (UTC)

[edit] Disputed

Hang on... in the article it says "Explaining away Richard's Paradox is as easy as being careful to distinguish between statements within arithmetic (which make no reference to any system of notation) and statements about some system of notation in which arithmetic is codified." Half the trick of Godels Incompleteness Theorem is the invoking of Godel numbering of notation symbols so that mathematical notation becomes equivalent to arithmetic. I get the feeling that Richard's Paradox is not supported by expert logic mathematicians. User:rem120 5th April, 2005

Gödel's proof is a formalization of Richard's paradox, in which metamathematics (about statements in the formal system) is represented by arithmetic (on the Gödel numbers of those statements). Gödel proves that this is valid, and that the arithmetical statements are provable within the formal system. The key is that we're representing metamathematics, not discussing it directly. Benandorsqueaks 20:26, 2 April 2006 (UTC)

So are you saying that the article is incorrect in stating that Richard's paradox is fallacious? Remy B 05:46, 3 April 2006 (UTC)
Richard's paradox is fallacious, but this is because of how it is set up. The initial conditions specified that we number statements concerning (A): the arithmetical properties of integers. Then we set up the property of Richardian concerning (B): statements concerning the arithmetical properties of integers. The important point is that these two things are different; the list we set up in the beginning is a list of statements concerning (A), while the property of being Richardian does not concern (A); it just concerns (B). Therefore, the property of being Richardian was not enumerated in that list.
The reason why it is fallacious is because we set up our initial list such that the property of being Richardian is not in the list. But a very similar argument was taken by Gödel where he took a system that could codify self-referential statements. The Richardian paradox is fallacious, but Gödel's proof is not. — Eric Herboso 20:57, 3 April 2006 (UTC)
In the article it says that you could consider "the first natural number" and "not divisible by any integer other than 1 and itself" as arithmetical properties. Why cant you consider "is Richardian" as one too? What is it about being Richardian that isnt an arithmetical property? Remy B 09:27, 4 April 2006 (UTC)
The statements of arithmetical properties apply to numbers and you can see them as functions, like e.g. f(x) where f is the property and x is some number or set of numbers. Now the property of "being Richardian" is not a property of a number or set of numbers but of a pair (x,y) where x is a statement about numbers and y is the number assigned to it. Hence, to be RIchardian is a property of the type f(x,y). It is hence not on the same level as the other definitions and statemnts, because it predicates someting about them. When trying to apply it to itself, you get a paradox not unlike the Liar's paradox. Therefore we have to distinguish mathematical statements from statements about mathematical statements, just as we have to distinguish sentences and assertions about the truth-value of sentences. See also Alfred Tarski and the distinction between language and metalanguage. Cat 12:09, 27 April 2006 (UTC)
Forgive my persistence, but I'm still not convinced from that line of reasoning. Why cant f(x) be a function that returns 0 or 1 (or whatever results are desired) whether the number x is Richardian or not? That is formulated in the same way as, say, how an f(x) might be a function that returns 0 or 1 (or whatever) whether the number x is "the first natural number", or perhaps "not divisible by any integer other than 1 and itself". I dont see why it has to be a double-parameter function when you are dealing only with the Richardian property. (Just to clarify, I do actually believe that the logic is false, purely because prominent mathematicians state that it is false, and I believe their abilities far above mine. I just dont see the truth in the reasoning given in the article, and on this talk page, and that makes me think there is something I am missing). Remy B 12:49, 27 April 2006 (UTC)
Why cant f(x) be a function that returns 0 or 1 (or whatever results are desired) whether the number x is Richardian or not? Carefully consider the input neede for such a function. It would need 1) the statement or definition about numbers as input and 2) the number assigned to that very statement in order to decide whether it is Richardian or not. Consider the other statements and definitions: "1 is the first natural number", "a number divisible by 2 is even", independently from the number assigned to them they are applicable to numbers. The property of being RIchardian depends on the the pair of number and statement. Look at the definition itself: "x is Richardian" is equivalent to "x does not have the property designated by the defining expression with which x is correlated in the serially ordered set of definitions" Contrast this with "x is even" is equivalent to "x is divisible by two". In the latter statement no reference is being made to the "serially ordered set of definitions" or the numbers correlated to them. That is why we would get a paradox if we would add "Richardian" to the set of definitions. The fact that we would get a paradox indicates that "being Richardian" is a different kind of property that being even or odd. Cat 13:16, 27 April 2006 (UTC)
I'm not sure why it is a paradox to include a statement which contains "serially ordered set of definitons" to the set of definitions. If it is well-defined, it has a definition, which means it has a unique number which can be included in the order (eg. taking the Godel number of its algorithm). Also, why do you have to input the statement "Is Richardian" (or its equivalent) to the function? You do not need to pass the statement "is the first natural number" to a function f(x) which determines whether or not x is the first natural number. Remy B 13:50, 27 April 2006 (UTC)
It generates a paradox because it predicates something about itself and then you get an infinite loop at each stage of which the truth value flips. If you consider "being Richardian" a statement of the form f(x), then when evaluating the property of "being Richardian" of itself as part of a list of statements, you effectively plug it into itself thus f(f(x)). Thereby, if it applies, it renders itself untrue, but at the next iteration f(f(f(x))) it would be true again, in virtue of the fact of being untrue of itself at the earlier stage ... and so on ... this is a paradox. We can "solve" the paradox by looking at the nature of the statement being made. A statement like "S is P" (Subject is Predicate, like "snow is white") assigns a property to an object, e.g. whiteness to snow, which is similar to f(x) -> White(Snow) and FirstNN(1). A statement like the property of being Richardian is different. Why? This property is formulated in a conditional way: IF the statement does not apply to its own number THEN the number is Richardian. This is not anymore an "S is P" case, but rather an "IF S is not P THEN S is R". Now the paradox occurs when we take "being Richardian" itself as a possible candidate for "being Richardian". So in the statement "IF S is not P THEN S is R", where R=being Richardian, we plug in "being Richardian" also as P, hence getting the form "IF S is not R THEN S is R" which is a paradox. Seeing as we have obtained an untruth, we have to revise our premises: Clearly there is something fishy going on with "being Richardian". The property R should not be applicable to itself, like a function for one kind of object is not applicable to objects of other domains. So if R can be applied to a structure like "S is P" by saying "If S is not P THEN S is R", we must reformulate this into some less vulnerable form. For instance we can say that "R" is not a property of S, but of statements like "S is P", and it is true exactly when "S is P" is not true (when the statement does not apply to its own assigned number). Then R is not a property of an object, but of a statement, and hence it is a higher order function: R(f(x)), which cannot be applied to itself, but depends on the outcome of the function f(x). Then you also see again that we would get an infinite loop alternating between true and false if we would apply it to itself, as the truth value would come to depend solely from R applied to itself and not from R applied to f(x). R applied to itself does not have a definite value, but f(x) does in every case, and hence R(f(x)) does too. The error lies in considering R a statement at the same level of f(x), while it is not. Cat 08:20, 28 April 2006 (UTC)

I don't know, I think Remy is right. We can number the one-place predicates of Peano arithmetic, using arithmetic. The idea that the system Gödel took is somehow different from this system doesn't seem right. All he needed was addition, multiplication, and first-order logic. The article mentions "not divisible except by 1 and itself", which seems to imply that the system does have all this. Therefore no matter what convention we use, "is Richardian" is, I'm thinking, also an arithmetical property. -Dan 14:22, 24 May 2006 (UTC)

I suppose the point is that this is being done in natural language, not in a formal system, which is not arithmetical. My mistake. Followup question: Any recursive predicate can be expressed in arithmetic. Why doesn't this prove people are not algorithms? -Dan 15:07, 24 May 2006 (UTC)
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