Maybe “0.999… does not equal 1” implicitly defines (not redefines) what a limit is in the stupid numbers system. If so, can someone produce a theorem that shows that this is indeed the case and that this definition of limit applies throughout the stupid number system. If so, I think maybe we could show that the stupid number system corresponds to some system of the hyperreal numbers (there may be more than one version of the hyperreals depending on who you listen to). The hyperreal numbers are well developed and their properties are well known.
Will there be a single clear-cut definition for series convergence? The partial sums of 0.999… are all less than 1-ε.
It is still the case that lim[sub]n → ∞[/sub] 1 - 10[sup]-n[/sup] = 1 whether you use standard or non-standard analysis.
A non-standard definition of the limit of a sequence is that lim[sub]n → ∞[/sub]a[sub]n[/sub] = L if and only if a[sub]N[/sub] is infinitesimally close to L for any “infinite” (i.e., bigger than any standard) natural number N. (Here L and a[sub]n[/sub] are real numbers.) A sequence converges if and only if it is a Cauchy sequence, etc. Non-standard calculus is a different language, but it does not change the truth of any facts about the standard real numbers; that’s the whole point.
What is true, as you point out, is that the non-standard real number defined by (0.9, 0.99, 0.999, …) [this is just a number, not a sequence; see below] is strictly less than 1. How much less? By (0.1, 0.01, 0.001, …) = 10[sup]-ω[/sup], where ω = (1, 2, 3, …). This is a non-standard real number which is less than any positive standard real number. The use of sequences in this paragraph has to do with the construction of non-standard numbers as an ultrapower, and should not be confused with any other mention of sequences or series in this post!
Based on the OP, I’d say that we can define stupid numbers to be exactly equivalent to real numbers except that we define the “…” and " ̅ " symbols to have no general meaning and define the specific formations “0.999…” and "0.9 ̅ " to both be exactly equal to the integer 12,345,679, at which point we go home for the day. If somebody whines that undefining the … and ̅ symbols leaves a hole in the general system we can just define some other notation to express the idea and placate them that way.
Besides this making stupid numbers a demonstrably coherent and workable system, it also means that 0.999… * 8 = 98765432.
Given the origin of Axiom of Stupidty from the 0.999… does not equal 1 deniers, it should be clear that it is expressed in base 10. I am quite surprised that no one has brought up the horrors that arise if we try to express the Axiom of Stupidity in a base other than 10. For all we know so far, there is no such thing as using a base to express numbers in the stupid number system.
It’s probably a really bad (and stupid) idea to state an axiom using any base other than maybe binary. I know how to express 0.999… in non-decimal bases in the real numbers but I am having difficulty figuring out what it means in other bases in the stupid number system (if it means anything at all).
Obviously I meant to say “0.999… equals 1 deniers” in the previous post.
I’m no sure why we should prefer binary over other notations, especially since decimal points in binary make my head hurt, but I believe that 1 = .1111… is the comparable situation in binary. In non-stupid math anyway.
I present for your evaluation:
Ynnad’s Third Conjecture Concerning the Stupid Numbers
- Ynnad’s First Conjecture Concerning the Stupid Numbers (see previous post) is false, and
- Ynnad’s Second Conjecture Concerning the Stupid Numbers (see previous post) is false, and
- The set of the stupid numbers contains exactly three elements, the Additive Identity, hereinafter expressed as “0”, the Multiplicative Identity, hereinafter expressed as “1”, and 0.999… , i.e. S = {0, 0.999…, 1}.
- The binary operation of addition is only defined if at least one of the elements used in the operation is 0.
- The binary operation of multiplication is only defined if at least on of the elements used in the operation is 1.
Can anyone prove that this conjecture is true?
I’m genuinely not sure what you are asking about, but, yes, a field can have exactly three elements. I’ll leave working out the addition and multiplication tables as an exercise.
Hey, think about it… If it ain’t equal to unity, well, it ain’t unity.
Done. Them’s my thoughts.
Any integer has two representations in any fractional base representation. 13[sub]10[/sub] = 14 = 13.888888888… in base 9.
Irrational numbers, e.g. pi=3.14159… , have only a single representation in any base. Rational numbers have one or two representations depending on the base. 1/9 = 0.11111… in base 10 but has two representations in base 3: 0.01000… and 0.0022222…
Continued-fraction representation is a good representation when you want to represent reals as sequences of integers uniquely.
On-line you can read a letter from Cantor to Dedekind when the former was just discovering his cardinal numbers. He mentions the double representation as a minor stumbling-block in completing a key theorem about the cardinality of the continuum.