The End of an Era - That of The Natural Numbers
in [Math]
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From: Transfer Principle on 29 Apr 2010 15:20 On Apr 29, 8:14 am, A <anonymous.rubbert...(a)yahoo.com> wrote: > On Apr 29, 7:37 am, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> > wrote: > > This is a fine instance of using a very complicated construction to > > develop a form that is simpler than what you started with. Modulo > > three numbers are much simpler than the natural numbers. Why should > > you require so much top heavy math to construct them? Is that really > > the way that you think of such a limited construction? I'm sorry, but > > the mechanism of the modulo math is so simple and primitive that it > > does not deserve this treatment. If a gradeschool child can understand > > them should it take college mathematics to describe them? > [Y]ou contend that the integers > modulo n are "simpler" than the natural numbers. I think this isn't > true, if you want to have the addition and the multiplication on the > integers modulo n. How do you describe the multiplication on the > integers modulo n without referring to the multiplication on the > natural numbers? At first glance, one would think that _any_ finite object is, a priori, "simpler" than _any_ infinite object. (Actually, let's make that _hereditarily_ finite, in order to avoid questions such as "Is {omega} simpler than omega?") In particular, any object whose existence is provable in ZF-Infinity ought to be considered "simpler" than any object whose existence isn't provable in ZF without using the Axiom of Infinity. I've seen this trick mentioned in earlier threads -- what if we started with the axioms of PA, but then changed one of the axioms as follows. Instead of having zero not be a successor: An (~0=Sn) we allow zero to be a successor: En (0=Sn) The resulting theory has a model of every finite cardinality, and that would be the modulus. Then we try to define addition and multiplication recursively, as we normally do for PA: a+0 = a a+S(b) = S(a+b) a0 = 0 aS(b) = ab+b I assume the only problem here would be that there's no way to prove that these definitions are welldefined. Still, it's hard for me to believe that the only way to prove the existence of the following hereditarily finite set: {((0,0),0), ((0,1),0), ((0,2),0), ((1,0),0), ((1,1),1), ((1,2),2), ((2,0),0), ((2,1),2), ((2,2),1)} where (a,b) denotes the Kuratowski ordered pair and 0,1,2 are von Neumann ordinals (_not_ equivalence classes of naturals), would be first to use the Axiom of Infinity to construct the complete set of naturals (omega)! Recall that some posters are (ultra)finitists. In deference to the finitist, we ought to develop as much mathematics without the Axiom of Infinity before we invoke the axiom. To me, the Axiom of Infinity is second only to the Axiom of Choice as the axiom to avoid as long as possible. Indeed, some (ultra)finitists would like to work in mod 3, yet object to the existence of the set omega. I would find it extremely counterintuitive if it were really impossible to construct multiplication mod m in ZF-Infinity. (Notice that Golden is definitely _not_ a finitist, since he regularly refers to R and C in his posts. An example of a finitist who works in mod m is Doron Zeilberger. He once advocated replacing standard omega with the integers mod p, for some large prime p. Google "Doron Zeilberger" for more info on his views.)
From: MoeBlee on 29 Apr 2010 16:13 On Apr 29, 2:20 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > what if > we started with the axioms of PA, but then changed one of the > axioms as follows. Instead of having zero not be a successor: > > An (~0=Sn) > > we allow zero to be a successor: > > En (0=Sn) > > The resulting theory has a model of every finite cardinality, > and that would be the modulus. Then we try to define addition > and multiplication recursively, as we normally do for PA: > > a+0 = a > a+S(b) = S(a+b) > a0 = 0 > aS(b) = ab+b > > I assume the only problem here would be that there's no way to > prove that these definitions are welldefined. In PA, '+' and '*' are primitive, not defined. In PA we don't "define" addition and multplication, except in the sense that the axioms narrow down the structures for the language that are models of the theory. If those are axioms, then they don't have to meet the standards of eliminability and non-creativity of definitions. > Still, it's hard for me to believe that the only way to prove > the existence of the following hereditarily finite set: > > {((0,0),0), > ((0,1),0), > ((0,2),0), > ((1,0),0), > ((1,1),1), > ((1,2),2), > ((2,0),0), > ((2,1),2), > ((2,2),1)} > > where (a,b) denotes the Kuratowski ordered pair and 0,1,2 are > von Neumann ordinals (_not_ equivalence classes of naturals), > would be first to use the Axiom of Infinity to construct the > complete set of naturals (omega)! Of course we don't need the axiom of infinity to prove the existence of that set. All we need is are the pairing and union axioms (and extensionality for uniqueness). MoeBlee
From: A on 29 Apr 2010 16:37 On Apr 29, 3:20 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > On Apr 29, 8:14 am, A <anonymous.rubbert...(a)yahoo.com> wrote: > > > > > On Apr 29, 7:37 am, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> > > wrote: > > > This is a fine instance of using a very complicated construction to > > > develop a form that is simpler than what you started with. Modulo > > > three numbers are much simpler than the natural numbers. Why should > > > you require so much top heavy math to construct them? Is that really > > > the way that you think of such a limited construction? I'm sorry, but > > > the mechanism of the modulo math is so simple and primitive that it > > > does not deserve this treatment. If a gradeschool child can understand > > > them should it take college mathematics to describe them? > > [Y]ou contend that the integers > > modulo n are "simpler" than the natural numbers. I think this isn't > > true, if you want to have the addition and the multiplication on the > > integers modulo n. How do you describe the multiplication on the > > integers modulo n without referring to the multiplication on the > > natural numbers? > > At first glance, one would think that _any_ finite object is, > a priori, "simpler" than _any_ infinite object. (Actually, > let's make that _hereditarily_ finite, in order to avoid > questions such as "Is {omega} simpler than omega?") In > particular, any object whose existence is provable in > ZF-Infinity ought to be considered "simpler" than any object > whose existence isn't provable in ZF without using the Axiom > of Infinity. I don't think is true at all. Is the monster group "simpler" than the integers? Plenty of algebraic objects with finite underlying sets are much more complicated than certain algebraic objects with infinite underlying sets. > > I've seen this trick mentioned in earlier threads -- what if > we started with the axioms of PA, but then changed one of the > axioms as follows. Instead of having zero not be a successor: > > An (~0=Sn) > > we allow zero to be a successor: > > En (0=Sn) > > The resulting theory has a model of every finite cardinality, > and that would be the modulus. Then we try to define addition > and multiplication recursively, as we normally do for PA: > > a+0 = a > a+S(b) = S(a+b) > a0 = 0 > aS(b) = ab+b > > I assume the only problem here would be that there's no way to > prove that these definitions are welldefined. > > Still, it's hard for me to believe that the only way to prove > the existence of the following hereditarily finite set: > > {((0,0),0), > ((0,1),0), > ((0,2),0), > ((1,0),0), > ((1,1),1), > ((1,2),2), > ((2,0),0), > ((2,1),2), > ((2,2),1)} > > where (a,b) denotes the Kuratowski ordered pair and 0,1,2 are > von Neumann ordinals (_not_ equivalence classes of naturals), > would be first to use the Axiom of Infinity to construct the > complete set of naturals (omega)! > > Recall that some posters are (ultra)finitists. In deference > to the finitist, we ought to develop as much mathematics > without the Axiom of Infinity before we invoke the axiom. To > me, the Axiom of Infinity is second only to the Axiom of > Choice as the axiom to avoid as long as possible. Indeed, > some (ultra)finitists would like to work in mod 3, yet object > to the existence of the set omega. I would find it extremely > counterintuitive if it were really impossible to construct > multiplication mod m in ZF-Infinity. > > (Notice that Golden is definitely _not_ a finitist, since he > regularly refers to R and C in his posts. An example of a > finitist who works in mod m is Doron Zeilberger. He once > advocated replacing standard omega with the integers mod p, > for some large prime p. Google "Doron Zeilberger" for more > info on his views.)
From: William Hughes on 30 Apr 2010 08:27 On Apr 22, 12:50 am, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > It's widely believed our intuition of the natural numbers > has led to foundational understandings of mathematical > reasoning and not the least of which is the validity > of GIT proof. In this post, however, we'll demonstrate > that if there's an intuition of knowing the natural numbers, > there's also an intuition about not knowing them that > would invalidate GIT proof. > > Let F and F' be 2 formulas, we'll define the logical operator > xor as: > > F xor F' <-> (F \/ F') /\ ~(F /\ F') > > Let's now consider thew 2 formulas in L(PA): pGC ("pro GC") > and cGC("counter GC"): > > pGC <-> "There are infinitely many examples of GC" > cGC <-> "There are infinitely many counter examples of GC" > > Now, let's consider the following formula: > > (1) pGC xor cGC > > First observation: 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). What is this supposed to mean? As written it is false. (Suppose my intuition says that there are only a finite number of naturals). What if we replace GC with GC', "every even number is the sum of two odd numbers" or GC'' "every even number is the sum of a prime and a number with at most two prime factors" Whether or not (1) is decidable depends crucially on the status of GC. I suspect that rather that GC, you want some function of an integer, Und, such that there is no derivation of "There is no x such that Und(x) is true" However, you have to be a lot more precise about what you mean by "intuition about the naturals". -William Hughes
From: Nam Nguyen on 1 May 2010 02:01
William Hughes wrote: > On Apr 22, 12:50 am, Nam Nguyen <namducngu...(a)shaw.ca> wrote: >> It's widely believed our intuition of the natural numbers >> has led to foundational understandings of mathematical >> reasoning and not the least of which is the validity >> of GIT proof. In this post, however, we'll demonstrate >> that if there's an intuition of knowing the natural numbers, >> there's also an intuition about not knowing them that >> would invalidate GIT proof. >> >> Let F and F' be 2 formulas, we'll define the logical operator >> xor as: >> >> F xor F' <-> (F \/ F') /\ ~(F /\ F') >> >> Let's now consider thew 2 formulas in L(PA): pGC ("pro GC") >> and cGC("counter GC"): >> >> pGC <-> "There are infinitely many examples of GC" >> cGC <-> "There are infinitely many counter examples of GC" >> >> Now, let's consider the following formula: >> >> (1) pGC xor cGC >> >> First observation: 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). > > What is this supposed to mean? As written it is false. > (Suppose my intuition says that there are only a finite > number of naturals). I suppose you've entered a wrong auditorium listening to a wrong kind of presentation. The "standard theorists" and I are talking about the infinitely-many kind of natural numbers, NOT the finitely-many kind! (Not long ago AP was talking the 10**500 kind of naturals. I wouldn't think one would be interested in that talk though.) > What if we replace GC with GC', > "every even number is the sum of two odd numbers" > or GC'' > "every even number is the sum of a prime and a number with at > most two prime factors" If you replace something with something else, then you'd get something else. What's your point? > Whether or not (1) is decidable depends crucially on the > status of GC. So, if GC is false, would that "crucially" tell us if (1) decidable? > I suspect that rather that GC, > you want some function of > an integer, Und, such that there is no derivation of > "There is no x such that Und(x) is true" Why "rather than GC", while I'm talking about (1)? > However, you have to be a lot more precise about what > you mean by "intuition about the naturals". I have at least a couple of ways to be "more precise" on my "intuition about the naturals": a) it's a model of Q in which (1) is false, and b) it's a model of Q in which we can NOT assign a truth value for (1). Which one would you prefer? |