what to do about the Successor axiom once it reaches 999....9999#318; Correcting Math
in [Logic]
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From: Nam Nguyen on 25 Jan 2010 23:14 Jesse F. Hughes wrote: > Nam Nguyen <namducnguyen(a)shaw.ca> writes: > >> Jesse F. Hughes wrote: >>> Nam Nguyen <namducnguyen(a)shaw.ca> writes: >>> >>>> Jesse F. Hughes wrote: >>>>> Nam Nguyen <namducnguyen(a)shaw.ca> writes: >>>>> >>>>>> You've challenged few of us to come up with the definition of "finite- >>>>>> number" or else you'd go on with your ignorant babbling. So here they are, >>>>>> the definition of properties Finite(x) and Infinite(x): >>>>>> >>>>>> P(x) <-> Ey[y <= x) >>>>>> (*)P(x) <-> P(x) /\ AyEz[(y <= x) -> (z < y)] >>>>>> Finite(x) <-> ~(*)P(x) >>>>>> Infinite(x) <-> ~Finite(x) >>>>>> >>>>>> Can you quit babbling now? >>>>> I'm afraid I don't quite get it. >>>> I've never excluded the possibility that it's not what I really meant >>>> to come up with. I was just demonstrating to AP that there's always >>>> _technical_ definition that would not require _physics limitation_ >>>> such as 10^500: after all mathematics is abstract. >>>> >>>> Now let's re-examine what I had and what I intended. What I intended: >>>> >>>> Infinite(x) <-> "x is greater than infinitely many numbers" >>>> Finite(x) <-> "x is not infinite". >>>> >>>> Assuming we're talking about formal system in L(T) = L(<,...) where "..." >>>> means optional and where '<' has the same semantics like that in L(PA). >>>> >>>> >> P(x) <-> Ey[y <= x) >>>> >>>> means "x is greater than or equal to one number" >>> Which is trivially true for every x. >>>> and >>>> >>>> >> (*)P(x) <-> P(x) /\ AyEz[(y <= x) -> (z < y)] >>>> >>>> means, if my interpretation of the expression correct, "x is greater >>>> than infinitely many numbers" which is really Infinite(x). Finite(x) >>>> then is just ~Infinite(x). >>> That's not what it means at all. >>> >>> AyEz((y <= x) -> (z < y)) >>> >>> is false for *every* ordinal! It's false in any context in which < >>> has a least element. >>> >>> Thus, this clause doesn't do what you want it to do. >>> >>>>> A number is infinite iff (*)P(x) holds, right? >>>>> >>>>> That is, iff Ey[y <= x] & AyEz[(y <= x) -> (z < y)]. >>>>> >>>>> Now, I assume that we know that (Ax)( 0 <= x ), right? >>>>> >>>>> As well, (Az)~(z < 0), right? >>>>> >>>>> Thus, for all x, ~Ez(( 0 <= x ) -> (z < 0)) and hence >>>>> >>>>> ~AyEz(( y <= x ) -> ( z < y )). >>>>> >>>>> Hence, for all x, ~Infinite(x) and thus Finite(x) <-> P(x). (Note >>>>> that P(x) is trivially true, as well, since x <= x.) >>>>> >>>>> Am I missing something here? You seem to have defined Finite(x) as >>>>> the always true predicate. >>>> As I've just alluded, the definitions would work even in a language >>>> *more general* than L(PA). So yes, if you consider a formal system T >>>> of which the naturals is (supposed to be) a model then it's true >>>> all natural numbers are finite in this definition. (Remember in AP's >>>> dubious definition, 10^500 + 1 wouldn't be finite). But my definition >>>> would work for real numbers (perhaps with very little or no "tweaking"). >>> You've missed my point. Finite(x) is true for every x in an ordered >>> set with least element -- including those x's which are above an >>> infinite number of guys. >> No I think you missed mine. I'm not trying to assert anything about >> an truths or provability at all. Those would require what formal >> systems you've chosen to discuss. >> >> All what I'm doing here is _pure definition_ based on the "semantic" >> we'd typically think '<' has. That's all. If you want to discuss about >> anything being true or provable, using my (or for that matter using AP's) >> definitions you'd being talking _beyond_ just definitions! > > Your pure definition does not do anything useful. It does not express > what you wanted it to express. > > Using your pure definition in the context of, say, ordinals, we can > easily show > > Every x is finite -- in particular, w, w*w, 2^w and so on are all > finite. > > In the context of, say, integers, rationals, reals, and so on, > > Every x is infinite. In particular (integer, rational, real) 0, 1/2 > and pi are infinite. (Note: natural number 0 is finite, as are all > ordinals, according to your definition.) > > Your definition does not work, Nam. It does not yield a single useful > distinction. In particular: > > (Ex)(Ay)( x <= y ) <-> (Az)Finite(z) [1] > > In other words, if < has a bottom element, then *every* element of the > order is "finite". Moreover, if < has no bottom, then *every* element > of the order is "infinite". Thus, as consequence, > > (Ex)Finite(x) <-> (Ax)Finite(x). > > All of this is provable assuming only that < is a linear order. Your > "pure definition based on the 'semantic' [sic]" of < is just > poppycock. Sorry. Sorry Jesse. You seemed to underestimate the power of mathematical encoding here.
From: Nam Nguyen on 25 Jan 2010 23:46 Jesse F. Hughes wrote: > Nam Nguyen <namducnguyen(a)shaw.ca> writes: > >> Jesse F. Hughes wrote: >>> Your pure definition does not do anything useful. It does not express >>> what you wanted it to express. >>> >>> Using your pure definition in the context of, say, ordinals, we can >>> easily show >>> >>> Every x is finite -- in particular, w, w*w, 2^w and so on are all >>> finite. >>> >>> In the context of, say, integers, rationals, reals, and so on, >>> >>> Every x is infinite. In particular (integer, rational, real) 0, 1/2 >>> and pi are infinite. (Note: natural number 0 is finite, as are all >>> ordinals, according to your definition.) >>> >>> Your definition does not work, Nam. It does not yield a single useful >>> distinction. >> You got to be precise Jesse: "does not work" in _all_ contexts? >> Could you find a natural numbers being "infinite" using my >> definition? Don't you remember in AP's definitions, 10^500 + 1 is >> an infinite number? > > So, you're really trying to pretend that your nonsensical definition > was nonsensical on purpose, intended only to work in settings in which > > (Ax)Finite(x) > > is true. Is this your story? It's actually your story of attacking me Jesse. Everything is in the ng and _the fact_ is I was responding to AP challenging me to come up with a better than physics-10^500 definition of "finite number" for the naturals. > >> For the nth time, I just wanted to point out why his definition doesn't >> make mathematical sense in the natural number, which is (iirc) the context >> of what his challenging me et al is. >> >> Do I want to spend time to come up with a more perfect definitions of >> "finite-number"? No. That's not my intention, in responding to AP. > > I don't get it. Did you screw up on accident or on purpose? I already told you the context I had responded to AP before you even uttered a word here. I also told you I'm not interested in spending more time to come up with better definition. (Note I never claimed my is the best or perfect). Why you got to have a b.s. fight here is really beyond me, Jesse. > Of > course, I have my own theories, but I'm sure your explanation is > mighty enlightening. You forgot something Jesse: there's such a thing as semantic transformation, *translation between expressions of different languages*. The question you should have investigated is my definitions is _semantically sound_. But I guess you were so occupied with truth/provability or obsessed with the desire to attack people that you don't see it. > > You *really* have trouble simply saying, "You're right, this > definition didn't work as I thought it did. My mistake." Honestly, > it was a little mistake, after all, but your response has turned this > trivial observation that your definition don't work into a needlessly > long discussion. You're just b.s. Jesse and in a sense not much different from AP's attacking. First you said: >>> Your definition does not work, Nam. when I asked for precise clarification: >> You got to be precise Jesse: "does not work" in _all_ contexts? >> Could you find a natural numbers being "infinite" using my >> definition? Don't you remember in AP's definitions, 10^500 + 1 is >> an infinite number? You didn't answer even one question (and these are simple technical question!). And now you just have all-hell-break-loose personal attacks like: > You *really* have trouble simply saying, "You're right, this > definition didn't work as I thought it did. My mistake." (What mistake anyway when you refused my simple questions). It's a shame that your respond here is at crackpot level. >
From: jmfbahciv on 26 Jan 2010 09:12 Nam Nguyen wrote: > Marshall wrote: >> On Jan 25, 6:35 am, Nam Nguyen <namducngu...(a)shaw.ca> wrote: >>> jmfbahciv wrote: >>>> Nam Nguyen wrote: >>>>> Archimedes Plutonium wrote: >>>>>> David R Tribble wrote: >>>>>>> Archimedes Plutonium wrote: >>>>>>>> The Peano Axioms are flawed and inconsistent because >>>>>>>> they require a Successor Axiom which builds this set >>>>>>>> {0, 1, 2, 3, 4, 5, 6, . . . . , 9999....9999} >>>>>>> Unfortunately, you have never demonstrated how this can >>>>>>> be so. Specifically, you've never explained what happens >>>>>>> in the ". . . ." ellipses following the '6'. >>>>>> Your juvenile mind has accepted 1 + 1 + 1+ . . . + 1 diverging to >>>>>> infinity >>>>>> so that means it is not a finite-number, yet simultaneously >>>>>> your juvenile mind thinks {0, 1, 2, 3, . . . . } are all finite >>>>>> numbers. >>>>>> How come you hold such simultaneous contradictory beliefs? >>>>>>> You have also never given a good explanation of what the >>>>>>> "..." in the last number is supposed to mean. Is it 500 >>>>>>> '9' digits in a row? Or 10^500 digits? Or an unending sequence >>>>>>> of digits? Or what? >>>>>> You have never given any definition of finite-number versus infinite >>>>>> number? You said you would, but apparently you fail at this also. >>>>> You've challenged few of us to come up with the definition of "finite- >>>>> number" or else you'd go on with your ignorant babbling. So here they >>>>> are, >>>>> the definition of properties Finite(x) and Infinite(x): >>>>> P(x) <-> Ey[y <= x) >>>>> (*)P(x) <-> P(x) /\ AyEz[(y <= x) -> (z < y)] >>>>> Finite(x) <-> ~(*)P(x) >>>>> Infinite(x) <-> ~Finite(x) >>>>> Can you quit babbling now? >>>> <snort> After pigs and dinosaurs fly. I enjoyed the useful >>>> tidbits. Thanks. >>> Apparently you didn't know birds are dinosaurs. What a shame. >> >> I think you missed the fact that he was honestly thanking you. >> The "snort" was for the idea that AP would ever stop babbling, >> which you have to admit is at least modestly far-fetched. > > I do sincerely apologize to jmfbahciv and all for my shortcoming here. That's OK. A misreading of a short snort is not a shortcoming, especially if you need to translate. :-) > Thanks, Marshall. yep. Thanks, Marshall. /BAH
From: Nam Nguyen on 27 Jan 2010 00:46 Jesse F. Hughes wrote: > Nam Nguyen <namducnguyen(a)shaw.ca> writes: > >> Jesse F. Hughes wrote: >>> Nam Nguyen <namducnguyen(a)shaw.ca> writes: >>> >>>> Jesse F. Hughes wrote: >>>>> Nam Nguyen <namducnguyen(a)shaw.ca> writes: >>>>> >>>>>> Jesse F. Hughes wrote: >>>>>>> Nam Nguyen <namducnguyen(a)shaw.ca> writes: >>>>>>> >>>>>>>> You've challenged few of us to come up with the definition of "finite- >>>>>>>> number" or else you'd go on with your ignorant babbling. So here they are, >>>>>>>> the definition of properties Finite(x) and Infinite(x): >>>>>>>> >>>>>>>> P(x) <-> Ey[y <= x) >>>>>>>> (*)P(x) <-> P(x) /\ AyEz[(y <= x) -> (z < y)] >>>>>>>> Finite(x) <-> ~(*)P(x) >>>>>>>> Infinite(x) <-> ~Finite(x) >>>>>>>> >>>>>>>> Can you quit babbling now? >>>>>>> I'm afraid I don't quite get it. >>>>>> I've never excluded the possibility that it's not what I really meant >>>>>> to come up with. I was just demonstrating to AP that there's always >>>>>> _technical_ definition that would not require _physics limitation_ >>>>>> such as 10^500: after all mathematics is abstract. >>>>>> >>>>>> Now let's re-examine what I had and what I intended. What I intended: >>>>>> >>>>>> Infinite(x) <-> "x is greater than infinitely many numbers" >>>>>> Finite(x) <-> "x is not infinite". >>>>>> >>>>>> Assuming we're talking about formal system in L(T) = L(<,...) where "..." >>>>>> means optional and where '<' has the same semantics like that in L(PA). >>>>>> >>>>>> >> P(x) <-> Ey[y <= x) >>>>>> >>>>>> means "x is greater than or equal to one number" >>>>> Which is trivially true for every x. >>>>>> and >>>>>> >>>>>> >> (*)P(x) <-> P(x) /\ AyEz[(y <= x) -> (z < y)] >>>>>> >>>>>> means, if my interpretation of the expression correct, "x is greater >>>>>> than infinitely many numbers" which is really Infinite(x). Finite(x) >>>>>> then is just ~Infinite(x). >>>>> That's not what it means at all. >>>>> >>>>> AyEz((y <= x) -> (z < y)) >>>>> >>>>> is false for *every* ordinal! It's false in any context in which < >>>>> has a least element. >>>>> >>>>> Thus, this clause doesn't do what you want it to do. >>>>> >>>>>>> A number is infinite iff (*)P(x) holds, right? >>>>>>> >>>>>>> That is, iff Ey[y <= x] & AyEz[(y <= x) -> (z < y)]. >>>>>>> >>>>>>> Now, I assume that we know that (Ax)( 0 <= x ), right? >>>>>>> >>>>>>> As well, (Az)~(z < 0), right? >>>>>>> >>>>>>> Thus, for all x, ~Ez(( 0 <= x ) -> (z < 0)) and hence >>>>>>> >>>>>>> ~AyEz(( y <= x ) -> ( z < y )). >>>>>>> >>>>>>> Hence, for all x, ~Infinite(x) and thus Finite(x) <-> P(x). (Note >>>>>>> that P(x) is trivially true, as well, since x <= x.) >>>>>>> >>>>>>> Am I missing something here? You seem to have defined Finite(x) as >>>>>>> the always true predicate. >>>>>> As I've just alluded, the definitions would work even in a language >>>>>> *more general* than L(PA). So yes, if you consider a formal system T >>>>>> of which the naturals is (supposed to be) a model then it's true >>>>>> all natural numbers are finite in this definition. (Remember in AP's >>>>>> dubious definition, 10^500 + 1 wouldn't be finite). But my definition >>>>>> would work for real numbers (perhaps with very little or no "tweaking"). >>>>> You've missed my point. Finite(x) is true for every x in an ordered >>>>> set with least element -- including those x's which are above an >>>>> infinite number of guys. >>>> No I think you missed mine. I'm not trying to assert anything about >>>> an truths or provability at all. Those would require what formal >>>> systems you've chosen to discuss. >>>> >>>> All what I'm doing here is _pure definition_ based on the "semantic" >>>> we'd typically think '<' has. That's all. If you want to discuss about >>>> anything being true or provable, using my (or for that matter using AP's) >>>> definitions you'd being talking _beyond_ just definitions! >>> Your pure definition does not do anything useful. It does not express >>> what you wanted it to express. >>> >>> Using your pure definition in the context of, say, ordinals, we can >>> easily show >>> >>> Every x is finite -- in particular, w, w*w, 2^w and so on are all >>> finite. >>> >>> In the context of, say, integers, rationals, reals, and so on, >>> >>> Every x is infinite. In particular (integer, rational, real) 0, 1/2 >>> and pi are infinite. (Note: natural number 0 is finite, as are all >>> ordinals, according to your definition.) >>> >>> Your definition does not work, Nam. It does not yield a single useful >>> distinction. In particular: >>> >>> (Ex)(Ay)( x <= y ) <-> (Az)Finite(z) [1] >>> >>> In other words, if < has a bottom element, then *every* element of the >>> order is "finite". Moreover, if < has no bottom, then *every* element >>> of the order is "infinite". Thus, as consequence, >>> >>> (Ex)Finite(x) <-> (Ax)Finite(x). >>> >>> All of this is provable assuming only that < is a linear order. Your >>> "pure definition based on the 'semantic' [sic]" of < is just >>> poppycock. Sorry. >> Sorry Jesse. You seemed to underestimate the power of mathematical >> encoding here. > > > Oh, I see. You have a very clever mathematical encoding in which < > does *not* stand for the usual less-than relationship, but instead > stands for some other relation which magically makes your definitions > do what they're supposed to do. > > I think I underestimated your capacity for spurious self-defense. I had this definition: >>>>>>>> (*)P(x) <-> P(x) /\ AyEz[(y <= x) -> (z < y)] Clearly we can't talk about the truth or validity of (*)P(x) in, say, T = {(*)P(x) /\ ~(*)P(x)}. Also clearly is the _fact_ that (*)P(x) has a (typical) semantic/meaning. Could you tell me and others *just* the typical meaning of this formula (i.e. without mentioning anything about truth), and why '<' "does *not* stand for the usual less-than relationship"?
From: Nam Nguyen on 27 Jan 2010 01:30
Aatu Koskensilta wrote: > Nam Nguyen <namducnguyen(a)shaw.ca> writes: > >> I had this definition: >> >>>>>>>>>> (*)P(x) <-> P(x) /\ AyEz[(y <= x) -> (z < y)] >> Clearly we can't talk about the truth or validity of (*)P(x) in, say, >> T = {(*)P(x) /\ ~(*)P(x)}. Also clearly is the _fact_ that (*)P(x) has >> a (typical) semantic/meaning. Could you tell me and others *just* the >> typical meaning of this formula (i.e. without mentioning anything >> about truth), and why '<' "does *not* stand for the usual less-than >> relationship"? > > (*) does not appear to be an explicit definition at all. Somehow I've just had an odd feeling my use of of the typical notation '(*)' might have clouded the definition (but I'm not so sure). Let me clarify the definition: P(x) <-> Ey[y <= x] Infinite(x) <-> P(x) /\ AyEz[(y <= x) -> (z < y)] How would it be the case this definition "does not appear to be an explicit definition at all"? > If we take > quantifiers to range over naturals there's a very simple definition of > finiteness: > > x = x That's right I did think about this, in English: "All naturals are finite". But let's not forget the context of defeating AP's crazy claim that his definition (being a finite natural is being <= 10^500) is the best. Would this be better than his? In a sense it's properly not: at least he had an explanation that physically there seems to be some limit. If we force him to accept "All naturals are finite", we'd have no way to explain him why, based on the definition's own meriy, and he'd still go on rambling nonsense. Imho, > > If, on the other hand, we wish to say something more informative, and > seek a definition of finiteness that works in a more general setting, > your definitions (those that I could make anything of, in any case) are > inadequate, as Jesse notes. For the record, I've never disputed that. I actually have no illusion that it's difficult (if not impossible) to define "finite numbers" for *all kinds* of numbers one could come up. But my intention was only to show him that there's *at least one* way to define finite numbers without depending on a physics-limitation. Math after all should be abstract and not concrete. Naturally. > > As a closing plea, could you two perhaps trim the by now quite excessive > quotes when replying? Sure. |