what to do about the Successor axiom once it reaches 999....9999#318; Correcting Math
in [Logic]
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From: jmfbahciv on 25 Jan 2010 09:00 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. /BAH
From: Nam Nguyen on 25 Jan 2010 09:35 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.
From: Nam Nguyen on 25 Jan 2010 21:05 Jesse F. Hughes wrote: > Nam Nguyen <namducnguyen(a)shaw.ca> writes: > >> 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? > > 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" 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). > > 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").
From: Nam Nguyen on 25 Jan 2010 21:59 Jesse F. Hughes wrote: > Nam Nguyen <namducnguyen(a)shaw.ca> writes: > >> Jesse F. Hughes wrote: >>> Nam Nguyen <namducnguyen(a)shaw.ca> writes: >>> >>>> 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? >>> 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! >
From: Nam Nguyen on 25 Jan 2010 22:56
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. 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? 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. > 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. > > Proofs: > > Nam's definition was > > Finite(x) <-> ~(Ey( y <= x) /\ AyEz[(y <= x) -> (z < y)]) > > Since x <= x, this amounts to > > Finite(x) <-> ~AyEz[(y <= x) -> (z < y)] > <-> EyAz[(y <= x) & ~(z < y)] > <-> EyAz[(y <= x) & (y <= z)] > > If (Ey)(Az)(y <= z), then clearly (Ey)(Az)(y <= x & y <= z), so > > (Ey)(Az)(y <= z) -> (Ax)Finite(x). > > Conversely, suppose (Ax)Finite(x). Thus, > > (Ax)(Ey)(Az)[(y <= x) & (y <= z)]. > > From this, it follows that (Ey)(Az)(y <= z). So, we see > > (Ey)(Az)(y <= z) -> (Ax)Finite(x). > > The proof that > > ~(Ex)(Ay)( x <= y ) <-> (Az)Infinite(z) > > is similarly easy. > > Footnotes: > [1] Assuming that < is linear, for the claim > (Az)Finite(z) -> (Ex)(Ay)( x < y ). |