User talk:Archgimp

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9999999... question
The bar notation isn't usually used except after the decmial point. However, if by $$\overline{9}$$ you mean an "infinite string of nines," then it could be defined to mean

$$9\sum_{i=0}^\infty 10^i = \lim_{n \to \infty} 9\sum_{i=0}^n 10^i$$

Clearly this limit does not exist; that is to say, the series is divergent, because the terms (9,99,999,9999,99999,...) become arbitrarily large. One could say that the limit of the sequence was infinity, in a sense, but this is not the standard language used in mathematics. Does that answer your question? -- SCZenz 22:47, 17 November 2006 (UTC)


 * I don't know what the phrase "the integers are divergent" means. It is true that sequences in the integers either eventually repeat the same number, and converge to that number, or do not converge at all.  For example:


 * $$a_n = (-1)^n$$ alternates between 1 and -1 and so does not converge
 * $$a_n = 10^n$$ if $$n < 1000$$, 3 if $$n \ge 1000$$ converges to 3
 * $$a_n = n$$ increases without bound, and so does not converge
 * It may be helpful for you to read some of the articles on sequence and the limit of a sequence.
 * Regarding your function, it could also be written $$f(n) = 10^{\log_{10}n - \left \lfloor \log_{10}n \right \rfloor}$$. It certainly does return a value in the interval [1,10), but it's not a bijection.  I'm not sure offhand if it's onto [1,10), but it certainly isn't one-to-one; f(100) = f(1000) = 1, for example.
 * Hopefully this helps. Let me know what still needs answering. -- SCZenz 21:27, 18 November 2006 (UTC)

Two quick comments

 * 1) All numbers represented in base n, for any n, contain the symbol "10" as their first two-digit symbol because it is defined as the symbol for 1*n+0. It doesn't seem too mysterious to me.
 * 2) You may be barking up the wrong tree in your efforts, if I understand what you're doing. You seem to be interested in proving that you can create a bijection between the integers and a subset of the rationals, and then arguing from there that the cardinality of the sets is different.  This second step won't work; for example, I can create a bijection between the natural numbers {1,2,3,...} and the subset of the natural numbers {3,4,5,...}, f(n) = n+2, but that doesn't mean that the two sets have different cardinality. -- SCZenz 19:34, 19 November 2006 (UTC)

Bijections and cardinality
I'm still a little confused about what you're trying to do, but perhaps this helps... Two sets have equal cardinality if there exists a bijection between them. However, here are two things you should note: Does that help? -- SCZenz 08:44, 20 November 2006 (UTC)
 * 1) It is false that if there exists a bijection between a set A and a non-trivial subset of set B, then A and B have different cardinality. (For example, there exists a bijection between the natural numbers and the positive rationals, but the two still have the same cardinality because there exists a different bijection between N and Q.  On the other hand, obviously sometimes the sets will have different cardinality after all&mdash;there's a bijection between N and Q but not between N and R&mdash;but to prove that you;'d have to show that no bijection exists, not that a particular bijection only goes to a subset.)
 * 2) The real numbers do not have the same cardinality as the natural numbers and rational numbers.  You cannot even create a bijection between the natural numbers and a finite interval [a,b] in R.