Re: infinity
- From: stephen@xxxxxxxxxx
- Date: Fri, 30 Sep 2005 18:31:22 +0000 (UTC)
Tony Orlow <aeo6@xxxxxxxxxxx> wrote:
> stephen@xxxxxxxxxx said:
>> Tony Orlow <aeo6@xxxxxxxxxxx> wrote:
>> > Actually, I believe I said that the definition of an infinite set as one with a
>> > surjection into a proper subset is a generally good definition. If, however,
>> > you are using natural numbers and restrict them to finite values, you limit the
>> > infinity of the set.
>>
>> I never mentioned surjections or proper subsets. I was talking
>> only about last elements. You agreed that that was a good
>> definition for 'when finiteness is thrown into the mix',
>> which still makes no sense. Even if you were talking
>> about the surjection definition of infinite, your statement
>> becomes:
>> 'the definition of an infinte set as one with a surjection
>> into a proper subset is a generally good definition except
>> when you throw the fact that some sets do no have a surjection
>> with a proper subset into the mix'
>>
> Oh, don't be dense. Didn't I already explain this? Read carefully. The set of
> naturals as defined is infinite, but requires infinite values. When you
> "prove" they are all finite, you are affecting the definition of the set. All
> sets are defined by their elements. So, if you restrict the set to only finite
> values and exclude infinite values, you have decreased the size of the set for
> sure, compared to the set including infinite values. In fact, when you restrict
> your elements to only finite values, you reduce the size of the set to a finite
> number of elements. That is what i mean by "throwing finiteness into the mix".
All finite naturals are finite, by definition. Do you really think
the set of finite naturals contains elements that are not finite
naturals?
We all agree that there is no last finite natural (well, you
sometimes agree to that), so by the simple naive concept
of "infinite = unending", the finite naturals are infinite.
By the bijection definition of infinite, the finite naturals
are infinite. You claim this is a good definition, except
when it is not, which is really not much of an explanation.
<snip>
>> >>
>> >> No, you have a line with some arbitrary finite length. Everybody
>> >> else agrees that the length of the line is not finite.
>> >> You are the only one who says its finite, according to your
>> >> definition of finite. Claiming that this is a contradiction
>> >> is nonsense.
>> > The value range is finite.
>>
>> Only according to you your definitions. Your definitions
>> are the source of all your contradictions.
> Not internal contradictions, but contradictions with standard theory, which
> contradicts enough other things that I really can't worry too much about that.
There is nothing in standard theory that says the range of
the finite naturals is finite. Stop claiming that this
notion of yours has any support in standard mathematics.
>>
>> > The number of elements per unit of value range is
>> > finite. If you divide the first by the second, you get a finite number of
>> > elements. In my system, which is better than your crappy old one, the concept
>> > that you can divide a finite by a finite and get an infinite is nonsense.
>>
>> As it is in standard mathematics.
> Except that that is what you are doing when you have a sequence of elements
> where no element is infinitely before or after any other, and claim there are
> an infinite number of them. you are taking a finite range, dividing it by a
> finite unit, and declaring an infinite result.
Only according to your definitions of range. The contradictions
you keep harping on about are in your definitions. According
to the standard definitions, you cannot divide a finite quantity
by a non-zero finite quantity and get an infinite result. Stop
claiming otherwise.
>>
>> >>
>> >> >> I suppose this is referring to your nonsense about S^L.
>> >> > So, you think N=S^L is nonsense? That has nothing to do with quantitative
>> >> > systems like the naturals so, no, that is not what I am talking about. But, you
>> >> > really do not understand the use of that formula? Ugh!
>> >>
>> >> I really do not understand how you use it.
>>
>> > Then you are prematurely calling it nonsense. Do you consider everything you
>> > don't understand to be nonsense?
>>
>> > Let's give anexample. A long while back, in another thread and space, I was
>> > asked about the size of the set of all decimal naturals only containing 1's and
>> > 0's. Let us define N as the number of naturals and say there are the same
>> > number of naturals when expressed in decimal. There are then log10(N) digits
>> > required to have N unique decimal strings, since with x digits and 10 symbols,
>> > we have 10^x strings, and 10^log10(N) is N. That is the first application of
>> > N=S^L. Now, we consider that we have log10(N) digits, but only 2 symbols, 0 and
>> > 1, so S=2. So, the number of strings we have in the set is 2^log10(N). That is
>> > the second application of N=S^L. Does that help any?
>>
>> No, because you are making up a whole bunch of mathematics here.
> Yes, I am. Is that really a problem for you? Do you hate new math?
>> log10(N) is not well defined for infinite numbers, and N must
>> be an infinite number if it is the number of naturals.
> N is more like a variable which can take on infinite values. Infinities are
> relative. log10(N) is a formula on that variable. In order to produce any N
> number of strings, we need strings of log10(N) length, since our S is 10. This
> holds for all N finite and infinite.
You have not defined what sort of values N can have. It
does not matter if it is a variable or not. What matters
are the properties of the actual elements that N can
have a value of. Just saying log10(N) is meaningless
until you tell me something about what N actually is.
>> You
>> are taking some simple little fact that everyone agrees with,
>> and then drawing all sorts of unfounded conclusions by
>> applying standard definitions to your non-standard definitions.
>> It does not make sense.
> Ah, but it does. Constant relationships such as this can be taken to infinity,
> essentially the same as inductive proof. There is no reason why, as n goes to
> infinity, the number of strings of length n should obey any other formula than
> that which applies to well-known finite string lengths. Is there?
Yes, there are lots of reasons. Here is one. Every
natural number can be represented in base 2, base 3, base 10,
or whatever base you want. The number of strings of length
n or less in base 2 is
sum 2^k 0<k<n
The number of strings of length n or less in base 3 is
sum 3^k 0<k<n
For any given finite n,
sum 2^k 0<k<n < sum 3^k 0<k<n
This formula applies to all finite lengths n. However
the number of natural numbers is independent of the base,
and your insistence that finite and infinite must behave
the same leads to the conclusion that there are more
base 3 representations than base 2 representations.
>>
>> <snip>
>>
>> >> >> >>
>> >> >> >> Anyway, as you have been repeatedly told, the 'range' of the
>> >> >> >> finite naturals is infinite according to the standard definitions.
>> >> >> > The diameter, you mean.
>> >> >>
>> >> >> No, I meant the 'range'. I put it in quotes because it
>> >> >> is not a technical definition, but it is essentially what
>> >> >> you have been talking about.
>> >> > The range does not have a "standard definition", and is not infinite. You are
>> >> > talking about standardly defined "diameter", as if a set has a center anyway.
>> >> > What a dumb term! ;)
>> >>
>> >> More word games from Tony. What does it matter whether
>> >> it is called 'range' or 'diameter'. Graphs also have
>> >> diameters, but they do not have centers. I know what
>> >> the concept that you are trying to discuss is, and
>> >> noone but you thinks that it is finite for the natural numbers.
>> > Yes, I know, and that is problem.
>>
>> Why is that a problem? Nobody else thinks the range of the
>> finite naturals is finite, so nobody else thinks that according
>> to the standard definitions a finite number divided by a finite
>> number is infinite. Where is the problem?
> The problem is that the reange IS finite.
Only according to you.
> You agree that there is no infinite
> difference between any pair of finite naturals, and that there is 1 unit of
> difference between any two neighboring ones. Since no infinite differnce
> exists, the greatest possible difference cannot be infinite, and is therefore
> finite.
No, the greatest possible differences does not exist, and is therefore
neither finite nor infinite. It does not exist. It does not
have any properties.
> Within that finite value range between any two elements, one can only
> fit a finite number of unit differences. There is no sequence within the set
> containing an infinite number of elements, including the entire set.
This is only according to your contradictory definitions.
Nobody else thinks the range of the finite naturals is finite.
Claiming that there is a contradition in the standard definitions
because they contradict your ideas is nonsense.
>>
>> Alternating series can converge to all sorts of values. This is
>> a standard result of series.
> They can converge to finite values only if the alternating terms have a limit
> of zero at n=oo. That is a standard rule.
> http://mathworld.wolfram.com/DivergenceTests.html
Yes. But they can converge to all sorts of finite values.
http://www.maths.abdn.ac.uk/~igc/tch/ma2001/notes/node51.html
Note the last sentence. Of course you will claim it is
nonsense, but it is part of standard mathematics.
<snip>
>> > yes, so, why should you object to alternative definitions for "infinite"?
>>
>> I would object to alternative definitions for "infinite set"
>> because that phrase is already in use. Insisting that
>> in mathematics "infinite set" should mean something other
>> that what it already means is just confusing and contrary.
>> And as this whole discussion has been about whether or
>> not the set of finite numbers is infinite or not, you
>> can't claim that you were discussing some other context
>> for the word "infinite".
> True. I think the current definition of "infinite set" stinks. It's over the
> hill. Washed up. Caput!
Well you have not made a very good case.
<snip>
>> >>
>> >> You have proven that for finite S, and finite L, S^L is
>> >> finite. However, what you have been claiming is that
>> >> the sum of S^L for all finite L is finite.
>> > The only way that could NOT be the case is if there are an infinite number of
>> > finite naturals. You know my position on that.
>>
>> Yes, but you keep claiming that there is something wrong with
>> the standard definitions. One of your first arguments that
>> there must only be a finite number of finite numbers used
>> that S^L idea.
> Yes, in terms of the digital representations of the naturals. The standard
> definition strays too far from the original concept of set size as "number of
> elements" and ends up contradicting it.
What original concept of "set size as number of elements?"
Where is this original concept written down? Presumably
it predates Cantor, because you reject his definition
of "set size as number of elements". To Whose definition
are you referring?
>>
>> Besides, claiming that sum S^L for finite L is a finite
>> number leads to immediate contradictions for any
>> normal definition of 'finite number'. Of course you
>> have your 'tenuously existing unspecific' numbers to
>> help you avoid such problems, but only you seem to
>> know how to use them.
> LOL!! My tenuously existing finite number is your aleph_0, which you barely
> know what to do with. My definition of finite immediately contradicts yours,
> because you have defined it in a way that contradicts quantitative notions of
> infinity by concentrating solely on iterative notions.
No it is not. aleph_0 is not a natural number. You have
been repeatedly told this. You keep claiming that your
tenuously existing finite number is a natural number.
Therefore whatever it is, it cannot be aleph_0.
Of course you think the set of finite naturals
contains things that are not finite naturals, so you
probably will not understand the reason that your
tenuously existing finite number cannot be aleph_0.
>>
>> >> You
>> >> have never proven the latter, and all your "proofs"
>> >> just assume there is a finite number of finite numbers.
>> > No, that is also proven in the form that no finite natural has an infinite
>> > number of predecessors in the order of the set, and that no infinite number of
>> > finite intervals can fit in any finite interval as defined by the value range
>> > of the finite naturals as the largest possible difference within the set. It's
>> > not assumed, but derived. You just don't like what it's derived from.
>>
>> The number of predecessors is irrelevant to the standard definition
>> of infinite. All the finite numbers cannot fit into any finite
>> interval according to the standard definitions. Again you
>> are claiming that the standard definitions are wrong because
>> they disagree with your private definitions.
> Because they violate almost every intuitive notion one generally would have
> about infinite sets and sets in general.
I think the most intuitive notion about infinite sets is
that they keep going on forever and ever. The standard
definitions are quite consistent with that intuitive notion.
You on the other hand refuse to talk about a set until
someone has told you where it ends, which is completely
contraty to the intuitive notion of infinite sets never
ending.
>>
>> >> Nowhere in standard mathematics does anybody assume
>> >> there is only a finite number of finite numbers.
>> >> That is not part of any well established system.
>> >> It is only part of your nonsense.
>> > It is part of my system, which works, better than the standard one.
>>
>> Works? How does it work? Where does it work?
> See above, below, and to either side.
Nope. I do not see any evidence of your "theory" "working".
<snip>
>>
>> I did not claim 'superiority'. I claim it was more established.
> So what?
You claimed that set theory contradicted well established
areas of mathematics. That's what.
<snip>
>> >>
>> >> Any proof of this assertion?
>> > Proof of a prevalence of infirmation-oriented people opposing the standard
>> > theory? No, just seems to be my experience. Do a survey if you like. How many
>> > anti-Cantorians are involved in information-related fields?
>>
>> None that I know of. The only anti-Cantorians I know of are
>> people like you who post rants on Usenet. Some of them
>> make a bit more sense than you do. The ones who make
>> the most sense reject avoid infinities as much as possible.
>> I do not agree with their position, but I can see the
>> motivation for it.
> Yes, I am not a typical objector. Most say there is only one infinity, or no
> infinity, or only potential infinity. Most of the motivation comes from the
> bizarre conclusions of set theory which people naturally reject, plus the fact
> that infinity is considered unfathomable, and people like their mysteries. I on
> the other hand, rather like solving mysteries, and never anticipate that I will
> run out of them, so don't worry about that.
There is no mystery to mathematical infinity. It is quite
well defined.
> I just went out for a cigarette, and a thought occurred to me about one of the
> very basic ways set theory fails to work with other math. A line segment is a
> set of points. Line segments have a size, which is their length, the distance
> from one endpoint to the other, in finite units. One line segment can be twice
> as long as another, and is generally considered to have twice the size. Yet,
> when set theory deals with this set of points we call a line segment, it
> declares the sizes of those two sets to be the same. Why does set theory not
> agree on this most basic level with basic definitions in geometry? Why does it
> not even try?
Who said that the length of a line is related to the number
of points in the line? Is that an axiom of yours?
I have a dvd and a cd here. They are both the same size.
The dvd can store 4.7 gigabytes, and the cd can only 700
megabytes. So the dvd is nearly 7 times as big as the
cd. But how can this be if they are the same size?
<snip>
>> >>
>> >> There is no end to an infinite process. You are confusing
>> >> potential and actual infinities.
>> > I am talkign about infinite values generated in the set.
>>
>> 'generated in the set'? What does that mean? Sets are not
>> usually 'generated'. Time is not a consideration. The
>> set of naturals is assumed to exist, in its entirety.
>> It is created in one step by an axiom.
> A recursive deinition which, like inductive proof, has an implicit infinite
> number of iterations. You know just what I mean. If there are an infinite
> number of elements, they are the result of an infinite number of iterations
> each of which includes an increment.
That is not how sets are defined. You may think that is
how other people think about it, but they do not.
>>
>> >>
>> >> > It
>> >> > disagrees with N=S^L, which implies that an infinite set of strings on a finite
>> >> > alphabet must contain strings of infinite length.
>> >>
>> >> No, it does not imply that. There are an infinite number of finite
>> >> values. That is what the standard definitions say. Therefore,
>> >> according to the standard definitions,
>> >> sum S^L for all finite L
>> >> is an infinite sum of finite values, which clearly is not finite.
>> >> I know you think there are only a finite number of finite values,
>> >> but that says nothing about whether there are inconsistencies
>> >> in the standard definitions.
>> > But, according to that theorem, there are a finite number of finite strings. My
>> > systems disagrees with yours, but is internally consistent, and externally
>> > consistent as well, which is just as important, ultimately.
>>
>> No, according to that theorem the number of finite strings is
>> larger than any finite number.
> It is larger than any GIVEN finite number, but then, for any given finite
> number, there is another finite number larger, which is not infinite. So that
> doesn't make it infinite.
>> The sum S^L for all finite L
>> is not a finite number according to the standard definitions.
> That is only because you assume an infinite number of finite natural lengths.
No, I prove there are an infinite number of finite natural lengths.
Let F be the set of finite natural lengths. F is infinite
if there exists a bijection from F to a proper subset of F.
Consider the function f(x)=2x. This maps F to f(F).
For every finite natural x, 2x is a finite natural.
Therefore f(F) is a subset of F. There is no finite
natural x such that 2x=1, so 1 is not a member of f(F)
and f(F) is a proper subset of F. If x<>y, then 2x<>2y,
so f is a bijection from F to f(F). Therefore, F
is infinite.
>> The sum diverges. There is no finite number equal to the
>> sum, and there is no finite number larger than the sum.
> Your insistence doesn't make you any less wrong.
So once again Tony is claiming that there exists sum finite
number M such that
M = sum S^L for all finite L.
Well M is finite, so
M = S^M + sum S^L for all finite L<>M.
M > S^M
So according to Tony there is some finite number M,
that is greater than S^M for any S>=1.
>> >> It does not disagree with that at all. It really does not
>> >> address that question however, until you define multiplicative
>> >> inverses in terms of sets.
>> > Then waht does it have to say?
>>
>> First define multiplicative inverses for sets. What
>> is the multiplicative inverse of {1, 3, 9 }? Off the
>> top of my head I do not of any set operation that
>> remotely resembles an inverse operation, but perhaps
>> someone somewhere has thought of one.
> So, then, nothing?
It it does not say anything, it does not contradict it.
>>
>> Of course 1/0=oo is just an axiom in the extended reals.
>> So we can just make up an equivalent axiom once
>> we make up a definition for 'division'.
> Consider it already done.
Then what is the problem?
<snip>
>> >> >> Whoever said it meant that? What does 'so large that all
>> >> >> finites are relatively nothing' even mean? Can you express
>> >> >> that in first order logic?
>> >> > No, I can say that for finite x, lim(n->oo: x/n)=0 and lim(n->0: x/n)=oo.
>> >>
>> >> Yes. You can say that. But of course, if you look at the
>> >> definitions of lim(n->oo : x/n) there is not actually
>> >> any mention of oo at all.
>> > Except where it says "oo".
>>
>> That is not in the definition. The definition does not
>> include oo at all.
> No, just what you're trying to define.
So according to you, all definitions refer to themselves?
You are just being stupid at this point. The definition
does not mention oo.
>>
>> >>
>> >> > I can say 1/0=oo/1.
>> >>
>> >> You can state that as an axiom. I believe that is
>> >> how it is handled in the extended reals.
>> > What a coincidence! Another area of math my system agrees with. Whaddya know?
>>
>> You can't always be wrong Tony.
> I can't afford to be wrong around here very often.
You must be in serious debt then by this point, given
how often you are wrong around here.
>> >> But they are relatively nothing, and you were claiming
>> >> that finite sets where not 'relatively nothing' compared
>> >> to infinite sets.
>> > No, they ARE relatively nothing, but absolutely something. Adding a finite
>> > number to an infinite doesn't make the infinite look any bogger on its own
>> > scale, but it is bigger nonetheless.
>>
>> So you claim that 1/oo equals 0, because 1 is "relatively nothing"
>> compared to infinity, but because it is not "absolutely nothing"
>> 1+1/oo > 1, because adding a "relative nothing" must make it
>> bigger nonetheless, so it follows that 0>0.
> It follows that the unit infinitesimal is infintiesmally larger than absolute
> zero. 0.000...001>0.000...000
>>
>> I know you will start talking about infinitesimals,
>> but you are the one claiming that 1/oo = 0.
> That is using 0 as the unit infinitesimal. I don't seem to have a good symbol
> for it. I guess I should make one up. Anyway, I am sure you don't disagree that
> dividing a unit segment into an infinite number of equal pieces yield pieces
> with no length?
Of course I do not disagree with that. You however are the one claiming
that the equal pieces really do have a length. What are those
infinitesmals measuring afterall if there is not length?
<snip>
>> >> Consider f(x)=x+1. This is a function from N, to f(N),
>> >> also called the image of N under f.
>> >>
>> >> We all agree that if x is a finite natural, then x+1 is
>> >> a finite natural. This means that f(N) is a subset of N.
>> >> Every element in the image is a finite natural.
>> >>
>> >> We all agree that there does not exist an x in the
>> >> natural numbers such that x+1=0. Therefore f(N)
>> >> does not contain 0, and f(N) is a proper subset of N.
>> > Sure, you add 1 to everything, eliminating 0 as an element, so you appear to
>> > have a proper subset which is the same size. Is it really a proper subset? It
>> > seems so, only because of the lack of a last element.
>>
>> It is clearly a proper subset. The only way it could
>> not be a proper subset if there exists a finite x
>> such that x+1 was not a finite natural number.
>> We all agree that for each finite natural x, x+1
>> is a finite natural. That is part of the definition.
>> An immediate conclusion of this is that yes, there
>> is not last element.
>>
>> > If you had any concept of
>> > the last element, then you would see that, whatever it is, it is now 1 greater.
>> > Perhaps you should define that the largest finite is the predecessor of 0. That
>> > would kind of make sense.
>>
>> What is wrong with not having a last element? Given that
>> 'infinite' and 'endless' are synonyms in certain contexts,
>> that fact that an infinite set does not have a last element
>> seems to make a whole lot of sense to me.
> And yet, with that approach, a proper subset is the same "measure" as the
> superset, which violates the notion that taking away from something makes it
> less.
You are the one who insists that finite things should be
relatively nothing compared to infinite things.
> We need s system that is more consistent with the finite case and with
> general understanding. From your perspective, adding 1 to every element
> produces exactly the same set as removing the bottom element. One the one hand
> we would have the same size set in any finite case, and in the second we would
> have a smaller set in any finite case. So, your set is both smaller AND the
> same size? I'm sorry, it doesn;t work for me.
Yes, we know it does not work for you Tony, but that
is not anyone's problem other than your own.
<snip>
>> >>
>> >> Is flat space true or not? I really do not know what you
>> >> mean by 'true'. I once directed you towards the axioms
>> >> of set theory to ask you which ones were 'wrong', but
>> >> you apparently never found the time to do so.
>> > No, I started to, but life is busy. I'll dig that up at some point.
>>
>> > I didn't call flat space "true" I said that Euclid's fifth is true of flat
>> > space. It's not true in curved space. Euclid, of course, was studying flat
>> > space. No one considers Euclid's set of axioms to be universally true any more,
>> > but only pertaining to a certain type of situation.
>>
>> Which can be said of any axiom. Axioms are true in the situations
>> in which they are used, by assumption. 'Flat space' is defined as a
>> situation where Euclid's fifth is true. It is a tautology.
>> Correspondence with the physical world is not part of the picture.
> Apparently not for set theorists. In real science, theories that fail to
> predict reality are trashed.
Whoever said mathematics was a real science? What parts
of reality does your theory make predictions about? Give
me an example of a real physical prediction that requires
your infinite finite naturals.
<snip>
>>
>> The first time you introduced your S^L argument was as one
>> of three "proofs" that the set of finite naturals must
>> be finite. It is all in google. For example,
>> see
>> http://groups.google.com/group/sci.math/msg/f625bc7afe8ec1bf?dmode=source&hl=en
>>
>> This is not your first mention of it, but you are clearly
>> proving that the set of finite naturals is finite
>> by proving that the set of finite strings is finite.
> Yes, through a direct correspondence between the two.
Yes, but your proof that there were only a finite number
of finite strings uses the "fact" that there are only
a finite number of finite numbers. You cannot prove
the that there are a finite number of finite numbers
by proving that if there are a finite number of finite numbers,
then there are only a finite number of finite strings.
That is what you were repeatedly trying to do.
<snip>
>> >> In the spirit of Jesse Hughes:
>> >> --
>> >> Then you have an "unbounded" set. But this term, "unbounded",
>> >> is simply part of the Cantorian mind-bend, as I have come to realize.
>> >> This distinction between "unbounded" and "infinite", when both mean
>> >> "without end or bound", is purely artificial, and only serves to
>> >> cover up the inconsistencies in the theory as a whole. -Tony Orlow
>>
>> > Huh!!! I remember saying that. It sounds so very backwards now. Of course, I
>> > have been using those two words and distinguishing them lately, haven't I? And
>> > yet, you have been objecting to my distinction, saying the one implies the
>> > other. And yet, you don't really think they are the same, do you? What
>> > distinction did I think you were making between them back then? Times do
>> > change. I am learning a lot! Okay, I know you don't believe that.... :D
>>
>> I think you have to be careful when using 'unbounded' and 'infinite'.
>> The word 'unbounded' does not make sense when applied to
>> an individual. Saying that 7 is unbounded is nonsense.
>> Saying that x is unbounded, where x is a variable, means
>> that x can take on any value, no matter how large.
>> When talking about sets you have to make it clear
>> whether 'unbounded' refers to the set, or the elements
>> in the set. I believe in normal mathematical usage,
>> 'unbounded' or 'bounded' never is used to refer to a set,
>> but to the elements in the set. So if someone says
>> a set is 'infinite but bounded', they mean the
>> set is infinite, but all the elements are smaller
>> than some finite bound.
>>
>> You have been saying a set is 'finite but unbounded',
>> which is a contradiction according to standard
>> definitions.
> Well, it sounds like standard definitions just won't get me the results I
> desire, will they? Maybe I should use a different approach......
You are welcome to use a different approach. However,
do not use standard terminology if you mean something
non-standard by it, and do you claim that the
standard approach is contradictory if it disagrees
with your approach. Noone is going to object
if you use your own clearly defined terminology and stop making
false claims about standard mathematics.
Stephen
.
- References:
- Re: infinity
- From: aeo6
- Re: infinity
- From: Tony Orlow
- Re: infinity
- Prev by Date: Re: x^3-x=6*y^3
- Next by Date: Re: infinity
- Previous by thread: Re: infinity
- Next by thread: Re: infinity
- Index(es):
Relevant Pages
|
Loading
