The answer is not one. The true correct answer that is in my paper that just went through peer review and accepted as correct is the hyperreal number 'p-theta' that I invented that is the absolute identity.
Let me explain. 1 is the multiplicative identity, where 1k = k
However, let c denote p-theta. c is the ultimate identity so not only would ck = k but also k + c = 0.
This has a lot of implications when you think about polynomial and non polynomial time when you c time to your algorithm. As it is the absolute identity, if you add c time to your algorithms time, the algorithms time would not changed.
>>7793253 You require the Axiom of choice just to define them, which is another way of saying that you never explicitly give the sequence for anything (except a few token examples) but instead just say "if it doesn't exist then my system doesn't make sense, so I'll just wish it into existence!".
The rationals are a subset of the algebraic numbers which are a subset of the definable numbers, which are a subset of the reals. The definable numbers are countably infinite. Every transcendental number you have ever seen is a definable number. Try giving me a sequence to any undefinable number. I claim that at best you will only ever be able to construct sequences for definable numbers and thus will never actually work with the uncountable reals..
>>7793618 You have to disprove my claim that "you cannot write an undefinable each real using this method and this it doesn't work for all reals". In other words do it for at least one undefinable real.
>90% of /sci/'s users can't join any kind of math conversation that goes beyond high school algebra >However, they are all really eager to pretend to be smart so they sit down, all day, waiting for a post to be about really simple maths that even they can understand and then they can post about their opinion all day long >Proofs that 0.999...= 1 are literally high school algebra >So that big majority all gets pulled in, ready to begin their cries for attention >Huge fight about who deserves to be right. Again, just to get attention. Probably to brag to their retard friends about how they have SERIOUS mathematical debates online, like th enlightened fedoras they are. >Thread dies >Everyone sits down once again waiting for another low level math thread to start
I bet no one even has an opinion. They just want the /sci/ experience without putting in the effort of learning analysis or abstract algebra.
>>7795452 It's not divisive, it's le epic trole. The entire point is that arguments both for and against are based on completely different definitions of .999..., which is exploitable for lulz. If we require it to be real the nested sphere theorem gives equality If we define it as the hyperreal 1-h (h is the hyperreal element) then clearly equality doesn't hold.
It's exactly like "is 1 prime?" Depends. Are we defining "prime" as usual to specifically exclude 1 so we can have the fundamental theorem of arithmetic and the results built on it?
>>7793253 Choose a programming language and index all Turing machines (or executable program) by [math]i[/math]. Let [math]h(i,s)[/math] be [math]0[/math] or [math]1[/math], depending on whether the machine with index [math]i[/math] halts before [math]s[/math] steps. For any given pair of numbers, you can indeed compute [math]h(i,s)[/math] by just running the program and wait [math]s[/math] time steps.
Define a sequence of rationals by having the n'th number given by the following sum (where you run through i and run it to step [math] s=n-i [/math])
But computing limit n to infinity (and thus s to infinity for all i) requires knowledge whether Turing machines ever halt, which we know to be impossible since the 30's. Thus this real number (insert Wildberger smirk here) is not computable.
And most real numbers are worse, really, because this number is at least definable. Most reals aren't even that.
>>7795601 >The entire point is that arguments both for and against are based on completely different definitions of .999..., which is exploitable for lulz. There is no sane definition of 0.999... besides that of the real number equal to 1.
>If we define it as the hyperreal 1-h But that makes no sense. The complex numbers, the reals, or some subset of those is the default assumption in any sort of mathematical question. It's like going into a discussion where people are arguing about basic arithmetic and claiming 1+1=0 because "well, technically, in one particular nonstandard number system, this is valid". In the real numbers that everyone uses, 0.999...=1.
>It's exactly like "is 1 prime?" Depends. Are we defining "prime" as usual to specifically exclude 1 so we can have the fundamental theorem of arithmetic and the results built on it? Exactly. The only sensible way to answer is to assume people are using standard definitions for all the terms involved, and respond "no, 1 isn't prime". Otherwise you become the guy asking "well how are we defining '1'? What about '=' and '+'?"
>>7795676 Without AC you can only give constructions for definable reals.
>>7795450 I don't claim they exist. I only claim that the definable reals exist, they are countably infinite, and they are a subset of the reals (they satisfy the properties of the reals).
A claim that the reals are uncountably infinite implies that there exist undefinable reals.
A definable real it's defined as follows. Choose a finite alphabet (eg, English+Math, LaTeX, or even the union of all human alphabets plus any symbols you can come up with in your lifetime). Then we say that a word or sentence in the alphabet is a string of symbols (note that we don't distinguish between words and sentences because our alphabet may include a space). The set of all finite words over our alphabet is countable. >Simply write all the words of length 1 followed by all the words of length 2 followed by all the words of length 3 and so on..
As a corollary we have that the sets of all theorems, definitions, proofs, and so on are countable.
Furthermore, recall that in formal logic we do not allow infinite sentences. Hence if we regard an axiomatic system over a logic as a formal language over an alphabet then the set of all sentences in the axiomatic system is countable. This means that with regards to any mathematical foundation built on logic (e.g. ZFC set theory, where Dedekind cuts and Cauchy sequences live) we have only a countable number of possible sentences. Therefore the number of definable real numbers in such a system must be countable.
>>7796933 It's true that there are only countably many definable reals... [math]from \ the \ perspective \ of \ one \ outside \ the \ given \ model \ of \ set \ theory [/math].
But truth is undefinable, and this is why in reality there need not even be a set of definable reals, and even if there were there certainly there need not be a bijection between the set of definable reals and the integers (even though such a bijection certainly exists outside the model). This is how, even if all reals are definable, the reals remain uncountable.
So your claim >that the reals are uncountably infinite implies that there exist undefinable reals need not be true.
Furthermore, it isn't true. There exists a model in which all reals are definable (necessarily from outside the model), but in which, of course, the reals are uncountable: choose your favorite transitive model and consider [math]L_\alpha[/math] where [math]\alpha[/math] is the least ordinal such that [math]L_\alpha \models ZF[/math] (and therefore, as Gödel showed, also [math]ZFC[/math] ). One can show that all ordinals in this model are definable (even though, of course, from inside the model there are class-many ordinals), and thus all reals are definable.
>>7796933 (>>7798245) And the reals are, of course, uncountable by for example the Cantor diagonalization argument.
But the point is that even though the reals are uncountable, it is possible for all reals to be definable. This is not a contradiction because the bijection exists outside the model, as by Tarski's Theorem on the Undefinability of Truth truth is.. undefinable within the model.
Furthermore, since you take issue with AC, note that the model of set theory [math]L_\alpha[/math] I provided is such that all reals are definable, yet AC still holds.
Moreover, if you are a constructivist (i.e. you believe in the axiom of constructibility), you must also believe in AC, as AC is a consequence of [math]V = L [/math].
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