The End of an Era - That of The Natural Numbers
in [Math]
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From: Nam Nguyen on 1 May 2010 13:20 William Hughes wrote: > On May 1, 1:14 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: >> William Hughes wrote: >>> On May 1, 12:55 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: >>>> William Hughes wrote: >>>>> On May 1, 12:11 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: >>>> William Hughes wrote: >>>>>>> However, I cannot see why you should think the >>>>>>> fact that you have "any intuition about the >>>>>>> naturals" means that cGC is undecidable. >>>>>> My memory might be bad, but where did I say that? >>>>> You said >>>>> "if we have any intuition about the naturals >>>>> then we'd also have the intuition that >>>>> we can't know the arithmetic >>>>> truth or falsehood of (1)" >>>>> Now apply the fact that (1) is decidable iff cGC >>>>> is decidable >>>> So, for example, suppose cGC is provable, how would you demonstrate >>>> (1) is decidable? >>> pGC is provably true. >> First, what would your definition of a formula being decidable be? (Note: you >> were using the term "decidable"). > > A formula P is decidable, iff P is provable or ~P > is provable. > > >> Secondly, can you prove pGC is _true_ in the >> naturals? > > Yes, it follows almost trivially from the fact that there > are infinitely many primes. > > >>> So if cGC is provably true then (1) >>> is provably flase and if cGC is provably false then (1) is >>> provably true. >> >
From: Nam Nguyen on 1 May 2010 13:21 William Hughes wrote: > On May 1, 1:14 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: >> William Hughes wrote: >>> On May 1, 12:55 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: >>>> William Hughes wrote: >>>>> On May 1, 12:11 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: >>>> William Hughes wrote: >>>>>>> However, I cannot see why you should think the >>>>>>> fact that you have "any intuition about the >>>>>>> naturals" means that cGC is undecidable. >>>>>> My memory might be bad, but where did I say that? >>>>> You said >>>>> "if we have any intuition about the naturals >>>>> then we'd also have the intuition that >>>>> we can't know the arithmetic >>>>> truth or falsehood of (1)" >>>>> Now apply the fact that (1) is decidable iff cGC >>>>> is decidable >>>> So, for example, suppose cGC is provable, how would you demonstrate >>>> (1) is decidable? >>> pGC is provably true. >> First, what would your definition of a formula being decidable be? (Note: you >> were using the term "decidable"). > > A formula P is decidable, iff P is provable or ~P > is provable. So, why did you mention the word "true" here? Is a formula being provable equivalent to it being true? >> Secondly, can you prove pGC is _true_ in the >> naturals? > > Yes, it follows almost trivially from the fact that there > are infinitely many primes. The definition of pGC was given before, in a high level as: pGC <-> "There are infinitely many examples of GC" From "there are infinitely many primes", how do you conclude pGC is true?
From: MoeBlee on 1 May 2010 13:35 On Apr 30, 6:24 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > set theory also gives a > definition of the multiplication of finite ordinals in terms > of addition In set theory we could also define multiplication of finite ordinals by Cartesian product. > although finite induction/recursion in ZF, > ironically, requires the Axiom of Infinity) I don't know what you mean by 'finite induction/recursion', but in set theory we don't need the axiom of infinity to prove that we can do induction on finite sets. Moreover, even without the axiom of infinity, we can define terms for operations that turn out to work how we would like them for finite ordinals. What we need the axiom of infinity for is to prove that, for such operations, there is a SET (i.e., the function itself) that exists that is the operation. >. In ZF, note that > _cardinal_ multiplication (Cartesian products, as A is using) > is not the same as _ordinal_ multiplication (recursive > addition, as Golden is using). I don't know what Golden is using, but just to be clear, for natural numbers, cardinal multiplication is the same as ordinal multiplication. And for finite sets, multiplication could be defined recursively or by Cartesian products. > Regardless of whether one is using cardinal or ordinal > multiplication, if one uses Cartesian products or disjoint > unions or finite ordinal arithmetic, is one actually working > with "the natural numbers" (omega)? Your question is not clear. Without the axiom of infinity, we can still quantify over all natural numbers; it's just that without the axiom of infinity we don't have a SET of all the natural numbers. MoeBlee
From: William Hughes on 1 May 2010 13:42 On May 1, 2:21 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: <snip> > The definition of pGC was given before, in a high level as: > > pGC <-> "There are infinitely many examples of GC" > > From "there are infinitely many primes", how do you conclude > pGC is true? Clearly. I do not understand what is meant by "There are infinitely many examples of GC" I take it to mean there are infinitely many even integers greater than 4 that are the sum of two primes. This follows immediately from the fact that there are an infinite number of primes (just add 3 to every prime, if the prime is not 2 you get an x for which GC(x) is true). What do you mean by the statement? - William Hughes
From: Nam Nguyen on 1 May 2010 14:15
William Hughes wrote: > On May 1, 2:21 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > <snip> > >> The definition of pGC was given before, in a high level as: >> >> pGC <-> "There are infinitely many examples of GC" >> >> From "there are infinitely many primes", how do you conclude >> pGC is true? > > Clearly. I do not understand what is meant by > > "There are infinitely many examples of GC" > > I take it to mean there are infinitely many even > integers greater than 4 that are the sum of two primes. > This follows immediately from the fact that there are > an infinite number of primes (just add 3 to every prime, > if the prime is not 2 you get an x for which GC(x) is true). Ah, yes. It's trivial indeed as you said. (As long long as we don't in meta level conclude from "there are infinitely many primes" as a 1st order theorem to "pGC is true" as a [meta] true statement. We were talking about "decidable" which is of syntactical notion. I forgot that I had asked you about being "_true_ in the naturals". My bad!). OK then, let's get back to my question before this issue: NN asked: > So, for example, suppose cGC is provable, how would you demonstrate > (1) is decidable? to which WH responded: > pGC is provably true. My question is based entirely on syntactical notion, free of notion of truth of the naturals. Why does your answer have the word "true"? Iow, why does your answer have something to do with my question? |