'When Are Relations Neither True Nor False?
in [Math]
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From: Tim Golden BandTech.com on 30 Mar 2010 09:06 On Mar 27, 11:25 am, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > Tim Golden BandTech.com wrote: > > On Mar 26, 5:55 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > >> Nam Nguyen wrote: > >>> Alan Smaill wrote: > >>>> Nam Nguyen <namducngu...(a)shaw.ca> writes: > >>>>>> Seriously, if you could demonstrate a truly absolute abstract truth > >>>>>> in mathematical reasoning, I'd leave the forum never coming back. > >>>>> If you can't (general "you") then I'm sorry: my duty to the Zen council, > >>>>> so to speak, is to see to it that "absolute" truths such as G(PA) is a > >>>>> thing of the past, if not of oblivion. > >>>> one day you will realise that your duty to the Zen council > >>>> is to overcome your feeling of duty to what is purely subjective ... > >>> I'm sure your belief in the "absolute" truth of G(PA) is subjective, which > >>> you'd need to overcome - someday. Each of us (including Godel) coming to > >>> mathematics and reasoning has our own subjective "baggage". > >>> Is it FOL, or FOL=, that you've alluded to? For example. > >> Note how much this physical reality has influenced and shaped our > >> mathematics and mathematical reasonings. Euclidean postulates had their > >> root in our once perception of space. From P(a) we infer Ex[P(x)] > >> wouldn't be an inference if the our physical reality didn't support > >> such at least in some way. And uncertainty in physics is a form > >> relativity. > > >> The point is relativity runs deep in reality and you're not fighting > >> with a lone person: you're fighting against your own limitation! > > >> Any rate, enough talk. Do you have even a single absolute truth you > >> could show me so that I'd realize I've been wrong all along? Let's > >> begin with the natural numbers: which formula in the language of > >> arithmetic could _you_ demonstrate as absolutely true? > > > There is a fairly straightforward construction that can yield both > > boolean logic and continuous higher forms, and even a lower form that > > I will call universal. > > > Constrain the real numbers to those values whose magnitude is unity. > > We see two options > > +1, -1 . > > It's relative as to how many real numbers one could "constrain". So > "constraint" is a relative notion, not an absolute one. > > In any rate, in all the below (including the URL) I still couldn't > see an absolute truth. Could you state such truth here? By accepting the generalization of sign the existence of dimension follows directly. That is the most absolute truth that I've come up with. The logical concept of true/false is a binary logic. It is discrete. One could treat the signs which are already discrete as the basis for boolean logic, but I find the spherical construction that I gave more interesting. From that breed we see the binary option's next stage is on a continuum rather than a discrete triplet. The spherical logic is likewise interesting to physical concepts, where working in a 4D space which is constrained to unity magnitude elements we see our ordinary 3D world locally and we see an upper limit on universe size. This spherical paradigm allows the definition of translation as rotation, whereas the Euclidean version will have us always defining rotation in terms of translation. This is an important feature but I admit it is askance to the OP's problem. I do find a continuum logic acceptable. In philosophical problems when one feels uncertain of a conclusion then we should not assume that the individual's interpretation is flawed. Qualities of many sorts can fall on a continuum, and the dimensionality of the problem leaves us with the obvious option of constraining some of those qualities in order to make a true statement. Each different wording of a statement makes subtle changes in sensibility and to even admit that there is a best wording is to accept that the lesser wordings were on a continuum of truth. To accept that in the future a finer wording might exist leaves open all interpretations to superior replacements. When we cast our individual belief into the socially accepted version we should only do so partially, if we hope to find superior replacements. Nam, you are too cryptic. You have already presupposed some conclusion that you have not stated clearly. I suppose that the only statement you will find acceptable as absolutely true is: true is true; but then there will be no productive work to do. If you refute the existence of absolute truth then I believe you are in support of a truth continuum. - Tim > > > > > Using polysign numbers extend this system to P3. > > (http://bandtechnology.com/PolySigned) > > One might initially consider there to be a three verticed logic here, > > but in the general form we see that the unity values now form a > > continuous circle. > > This is a nice exercise in the continuous/discrete paradigms of > > throught. In one dimension we see a discrete type, not unlike charge. > > In two dimensions we see that the same procedure yields a continuum of > > values, though there are arguably those three unique positions > > -1, +1, *1 . > > > Inspecting the product logic back in P2 (the boolean or constrained > > real number case) > > - + = - > > + - = - > > + + = + > > - - = + > > and likewise in the three signed case (overlooking the above > > redundancy) > > - - = + > > - + = * > > - * = - > > + + = - > > + * = + > > * * = * > > > Does a false false yield a true? The english language discourages the > > usage of double negatives, yet their use does exist within in it with > > such phrases as > > 'I am not an atheist.' > > Back in ordinary logic it is no problem to see that the math holds up > > in P2 so that > > Not(Not(A)) = A. > > The meaning of false and true cannot be reused in P3 and it is a nice > > human puzzle to consider that we and our dualistic thought patterns > > have artificially limited us. The P3 language is not sensible to the > > human mind, yet it may be entirely accurate. > > > Treading on P1 is difficult for most, but there we see just one > > instance within this logical paradigm > > -1 . > > Thus the polysign allow for a universal but fairly inanimate form at > > the bottom of the hierarchy > > universality > > duality > > triality (not to be confused with Clifford form) > > ... > > By leaving the Euclidean and working the sphere these forms exist > > naturally. > > > - Tim
From: Jesse F. Hughes on 30 Mar 2010 09:07 Newberry <newberryxy(a)gmail.com> writes: > On Mar 28, 9:01 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> Newberry <newberr...(a)gmail.com> writes: >> > On Mar 28, 5:54 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> >> Newberry <newberr...(a)gmail.com> writes: >> >> >> > But I can. In a system with gaps Tarski's theorem does not apply. We >> >> > can then simply equate truth with provability. >> >> >> Your second sentence does not follow. You have to show that you have >> >> a logic in which provability turns out to be equivalent to truth. >> >> Tarski's theorem may not preclude this possibility, but it doesn't >> >> follow that you can then "simply equate truth with provability." >> >> > Did I say it follows? I meant that it is possible. In classical logic >> > withuot gaps it is impossible. Why did you not interpret what I said >> > this way? >> >> "We can then simply equate truth with provability." > > It does automatically folow but we can nevertheless do that. You have to *show* that this can be done in your system. And, indeed, the word "equate" is still misleading, since it suggests that define true to mean "provable". That can certainly be done. I can say that, hereafter, when I say that a statement of PA is true, I mean that there is a proof of P in PA. Of course, such semantic play is unsatisfactory. -- "After years of arguing I realize that your intellects are too limited to fully grasp my work. [...] Still, no matter how child-like your minds are, [...] since you have language, [...] there's a chance that I'll be able to find something that your minds can handle." --JSH
From: Jesse F. Hughes on 30 Mar 2010 09:04 Newberry <newberryxy(a)gmail.com> writes: > Indeed. But if we leave out all the vacuous sentences we can still do > all the useful arithmetic as we know it. Although all the people on > this board believe that such sentences are true nobody argued that > they were useful. Aatu even said that they did not belong in ordinary > mathematical reasoning. Furthermore there is a reason to think that > they are neither true nor false. I cannot think of any good reason for > claiming that 1 + 1 = 2 is not true. You seem to have misrepresented Aatu's claims. Moreover, you're just wrong. I've argued repeatedly that some sentences of the form ~(Ex)(P & Q) occur in ordinary mathematical reasoning (and hence are useful), even when (Ex)P is false. An example occurred in sci.math recently. Simon C. Roberts gave a purported proof of FLT[1], by arguing: ~(En)(Ea,b,c)(a^n + b^n = c^n & a, b, c are pairwise coprime). Let's denote a^n + b^n = c^n by P(a,b,c,n) and a, b, c are pairwise coprime by Q(a,b,c), so that Simon's argument attempts to show that ~(En)(Ea,b,c)( P(a,b,c,n) & Q(a,b,c) ). (1) Of course, I am *not* claiming that he proved what he claims. That's beside my point. A poster named bill replied that (1) is not Fermat's last theorem[2], which has the form ~(En)(Ea,b,c) P(a,b,c,n). (2) Arturo responded[3] by proving (En)(Ea,b,c) P(a,b,c,n) -> (En)(Ea,b,c)( P(a,b,c,n) & Q(a,b,c) ). (3) Hence, a proof of (1) yields a proof of (2) by modus tollens. According to you, however, if (2) is true (and I assume we all know that (2) was proved by Wiles), then (1) is meaningless. Yet, no one here balked at the claim that (1) could be used to prove (2) (once (3) was proved). No one here had any trouble understanding what (1) means. Everyone in the thread accepted this form of mathematical argument as beyond suspicion -- although the claim that Simon actually proved (1) is regarded as doubtful. So, you're just plain wrong. These statements that you call meaningless occur in ordinary mathematical reasoning all the time. Footnotes: [1] Message id <1917288606.455209.1269716329839.JavaMail.root(a)gallium.mathforum.org>, in the thread "Another Proof of Fermats Last Theorem". [2] Message id <50f09d88-a96b-464c-aec5-be000f0be40d(a)x23g2000prd.googlegroups.com>. [3] Message id <e6768d43-7706-41f4-bff8-8e666d693c3a(a)j21g2000yqh.googlegroups.com>. -- "There's lots of things in this old world to take a poor boy down. If you leave them be, you can save yourself some pain. You don't have to live in fear, but you best have some respect, For rattlesnakes, painted ladies and cocaine." -- Bob Childers
From: Transfer Principle on 30 Mar 2010 15:42 On Mar 26, 9:55 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Mar 26, 12:20 am, Transfer Principle <lwal...(a)lausd.net> wrote: > > proof of 0.999...=1 to which I linked explicitly lists the > > Replacement Schema as one of the axioms that is used > > in the proof. > Replacement isn't needed to prove .999... = 1. With David Ullrich's > help, I posted a rigorous proof (in Z-regularity) of .999... = 1 in a > thread for that purpose a few years ago. > That 1 is the multiplicative identity does not require replacement. As usual, I was unable to find Ullrich's proof via a Google search, but I did go back to the Metamath proof and find where the Replacement Schema was used. Let's begin by trying to proof that 1 is the multiplicative identity of N, in other words: AneN (n*1 = n) So we begin by noting that we are considering N to be the set of nonzero finite ordinals, omega\{0}. So we use the recursive definition of multiplication of ordinals to obtain: AneN (n*1 = n*0 + n) AneN (n*1 = 0 + n) But hold on a minute. Multiplication of ordinals is defined using transfinite recursion -- which requires the Replacement Schema! At this point, one might point out that we only need to define the operations on the _finite_ ordinals, and so we ought to use finite, rather than _trans_finite, recursion. But finite recursion -- though it apparently avoids Replacement Schema -- requires the Axiom of _Infinity_ in its proof. It may seem ironic that _finite_ recursion requires Infinity, yet _trans_finite recursion doesn't, but this is confimed at the following two links: http://us.metamath.org/mpegif/inf3.html http://us.metamath.org/mpegif/mmnotes2004.txt Thus, by using transfinite recursion rather than finite recursion, the Metamath proof of 1 as the multiplicative identity avoids the Axiom of Infinity. We know that mathematicians prefer parsimony with regards to the axioms used in a proof. So if T is a theory and phi is a axiom (that is undecidable in T), then given two proofs of a statement, one in T and the other in T+phi, the proof in T is usually preferable. If the proof in T is only slightly longer than the proof in T+phi, then the proof in T is still preferable. Only if the proof in T is significantly longer (say 10, or 100 times as long) than the proof in T+phi will mathematicians consider assuming phi to make the proof shorter. But what if there's a statement that's not provable in T, but we have a choice between two axioms (or schemata) to add to T in order to produce a proof? For "1 is the multiplicative identity" appears not to be provable in Z-Infinity (assuming it's consistent), but we can find a proof in either Z (using finite induction) or ZF-Infinity (using transfinite induction instead). Which proof is considered preferable? On one hand, the goal isn't to define N, but _R_. One can avoid Infinity when trying to prove "n*1 = n for all natural numbers," but good luck trying to prove "r*1 = r for all _real_ numbers" (or even anything else about _R_) in ZF-Infinity. Since we're going to use Infinity to define R anyway, there's no real point in avoiding the axiom when discussing the properties of N. Thus, it's better to give the proof avoiding Replacement Schema instead -- and this is exactly what MoeBlee and Ullrich did -- work in Z(-Regularity -- we already know how to avoid Regularity, so I won't keep writing "-Regularity" each time I refer to a theory). On the other hand, we already know that some so-called "cranks" and finitists don't work in ZF(C) -- and we already know that the majority of them find _Infinity_, not _Replacement_, objectionable. So in deference to the finitist, we ought to avoid Infinity in as many proofs as possible, even if it requires using other axioms and schemata such as Replacement. Thus, to such posters, the proof in ZF-Infinity would be preferable to the proof in Z. But of course, a(n ultra)finitist has no business even referring to 0.999... in the first place, unless to denote only _finitely_ many nines after the decimal point. A similar debate often occurs when discussing Q -- can a finitist or "crank" work with rational numbers? On one hand, rationals are often defined as equivalence classes of ordered pairs, so 1/2 would be defined as {(1,2), (2,4), (3,6), (4,8), ...} -- this is evidently an infinite set, so Infinity is necessary. But on the other hand, we can define rationals to be only a single ordered pair representing the lowest- terms fraction, so 1/2 would be defined as (1,2), which can be defined without the Axiom of Infinity. Metamath uses the former definition, for why should one have to keep dividing by the gcd of the numerator and denominator to get a fraction into lowest terms just to avoid Infinity, when the ultimate goal is to define not Q, but _R_, where we'd use Infinity anyway? But the finitist/"crank" trying to avoid Infinity at all costs will use the latter definition of rational number. Returning to N, since Metamath has already defined via transfinite recursion using Replacement, it might as well use it in its proof of "1 is the multiplicative identity" and all that. But MoeBlee and Ullrich were explicitly working in Z(-Regularity), and so they're going to use the construction of R that doesn't require transfinite induction or the Replacement Schema. I'm not sure whether MoeBlee and Ullrich wish to avoid Replacement in the same way that a finitist/"crank" wishes to avoid Infinity. It may be that MoeBlee and Ullrich are open to using Replacement when it is absolutely inevitable, but as long as they can avoid the schema -- and one can construct R without it -- they'll do so. (And in order to reduce my habit of lumping posters together, it could be that MoeBlee and Ullrich are divided on this issue -- perhaps one of them will use Replacement when it's necessary, and the other will avoid it like the plague, just as a finitist/"crank" avoids Infinity.) And therefore, even without finding MoeBlee's or Ullrich's proof, I can figure out what their proof is. We basically take the Metamath proof of "0.999...=1" and replace all instances of transfinite induction with finite induction. Then voila -- we'll have a proof of "0.999...=1" which doesn't require Replacement and thus works in Z(-Regularity). QED
From: Transfer Principle on 30 Mar 2010 15:57
On Mar 26, 10:04 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Mar 26, 12:42 am, Transfer Principle <lwal...(a)lausd.net> wrote: > > as powerful > This cries for a definition or explanation of what is meant by 'as > powerful'. In mathematical logic we have various notions such as > 'intepretability' and 'conservative extension'. But some of your > comments seem not to use 'as powerful' in such technical senses > (otherwise some of your questions in this regard would be non- > starters). Marshall Spight was the first to use the phrase in this thread. I only use the phrase in reponse to Spight's post. Here is a quote from the post from earlier in this thread, back on the 2nd of March, at 4:11PM Greenwich time: "One thing that the cranks and crankophiles never understand is that the systems they come up with add a lot of complexity while actually removing functionality or utility. To do so merely to avoid some counterintuitive but harmless property (such as vacuous truth, in Newberry's case) is a huge waste of time. Less powerful; more work to use: that's a crank theory for you." So Spight criticizes so-called "crank" theories as less powerful than some other theory -- presumably the standard theory (ZFC). My goal, therefore, is to find a theory that's _as_ powerful as ZFC, so that Spight would have less reason to criticize it. In order for me to reduce my habit of misinterpreting other posters, it would be far better for Spight himself to state what exactly he meant when he wrote "less powerful." That way, we can agree on what criteria a theory needs to satisfy in order for it to be "as powerful" as ZFC, without "removing functionality or utility." |