4 ms·

That's either a nice troll or you don't know what you think you know!

by Almaviva 16y ago

cstross 16y ago
Possibly. What do you think I don't know?
- RiderOfGiraffes 16y ago
  Interesting question. The problem is that you've mentioned there being more than one transfinite number, when this question is actually about convergence. It appears that you've simply gone "Alleged paradox - infinity - must be ignorant about Cantor and the uncountables." In fact, Cantor and the uncountables have nothing to do with it. If you think otherwise then I'd be interested in seeing a more complete explanation of your comment.
  - cstross 16y ago
    Well, there's a transfinite implicit in the series of exponentiations -- we have aleph-0 exponentiations, don't we? And if we knock off that first one, we've got (aleph-0 minus 1) exponentiations. (NB: I don't understand your use of the term "convergence" -- must be either something I've forgotten in the third of a century since I studied maths, or something above the level I reached. (Sub A-level.) (See also: Dunning-Kruger effect.))
    - JoachimSchipper 16y ago
      Technically, f(x) converges to y as x -> infinity if for every epsilon there is an X such that |f(x) - y| < epsilon for all x > X ("the error becomes arbitrarily small"). For instance, 1/x converges to 0 as x -> infinity (pick X = 1/epsilon), but the sequence 0, 1, 0, 1, 0, ... does not converge to any y (pick epsilon < 1/2 to see this). In particular, the sequence r, r^r, r^r^r, ..., where r^2 = 2, does converge (to 2; the first terms are approximately 1.41, 1.63, 1.76, ...); the sequence 2, 2^2, 2^2^2, ..., however, does not (the first terms are 2, 4, 16, ... - and it only grows faster from there.)
    - gjm11 16y ago
      Convergence is what happens when you have a sequence of numbers, and they get closer and closer to some fixed number. Some sequences converge and some don't, and part of what's going on with this "paradox" is that one of the sequences it implicitly asks you to consider doesn't converge. Yes, there's a transfinite number implicit in the sequence of exponentiations, namely aleph-0. (It might be better to say omega-0 -- an infinite ordinal rather than an infinite cardinal -- but that's a technicality that doesn't really matter here.) But the problem with the "proof" doesn't have anything to do with the existence of infinite cardinals larger than aleph-0.
  - pbhjpbhj 16y ago
    It's 4am here and I need some sleep - choose 2 numbers from within the convergent range and do the same slight of hand with them. Clearly we can argue that x^^\inf = 4 has no convergence and so makes no sense but what if the equations are both in the converging region x € [e^−e, e^1/e] about [0.066, 1.445]: {1} x^^\inf = 2 we've shown in the OP that x^2 = 2 is a solution. Consider, {2} x^^\inf = e^(2/e) similarly to the OP we do {3} x^ (x^^\inf) = x^ (e^(2/e)) but this is just {4} x^^\inf = x^ (e^(2/e)) and substituting on the left from {2} {5} e^(2/e) = x^ (e^(2/e)) gives us (I think) {6} x = e^ (2e^ (-1-(2/e))) ; ~~1.42267 Hence we've shown that {7} 2 ~~ 1.423 Which isn't as neat as 2==4 but equally troublesome. There is no issue now with convergence is there (or there shouldn't be if I chose my numbers correctly)? --- Notations ^^ : ascii version of Knuths up-arrow notation \inf : first countable infinity, aleph-0 € : in the range, it's really a Euro symbol, but close enough ~~ : approximately equal to === Solution y = x^2 = 9 give solution (3,9) y = x^2 = 4 gives solution (2,4) 9=4? They're just different points on the same curve ... doh! We're not showing that 9=4 we're showing that (3,9) and (2,4) are point solutions of y = x^2. In the same way we have above 2 solution of y = x^^\inf.
- jules 16y ago
  That this doesn't have much at all to do with there being multiple transfinite numbers.
- india 16y ago
  No infinity other than Aleph-naught appears to be involved in this. Why do you think the existence of other aleph numbers matters here?

That's either a nice troll or you don't know what you think you know!

cstross 16y ago

Possibly. What do you think I don't know?

RiderOfGiraffes 16y ago
Interesting question. The problem is that you've mentioned there being more than one transfinite number, when this question is actually about convergence. It appears that you've simply gone "Alleged paradox - infinity - must be ignorant about Cantor and the uncountables." In fact, Cantor and the uncountables have nothing to do with it. If you think otherwise then I'd be interested in seeing a more complete explanation of your comment.
- cstross 16y ago
  Well, there's a transfinite implicit in the series of exponentiations -- we have aleph-0 exponentiations, don't we? And if we knock off that first one, we've got (aleph-0 minus 1) exponentiations. (NB: I don't understand your use of the term "convergence" -- must be either something I've forgotten in the third of a century since I studied maths, or something above the level I reached. (Sub A-level.) (See also: Dunning-Kruger effect.))
  - JoachimSchipper 16y ago
    Technically, f(x) converges to y as x -> infinity if for every epsilon there is an X such that |f(x) - y| < epsilon for all x > X ("the error becomes arbitrarily small"). For instance, 1/x converges to 0 as x -> infinity (pick X = 1/epsilon), but the sequence 0, 1, 0, 1, 0, ... does not converge to any y (pick epsilon < 1/2 to see this). In particular, the sequence r, r^r, r^r^r, ..., where r^2 = 2, does converge (to 2; the first terms are approximately 1.41, 1.63, 1.76, ...); the sequence 2, 2^2, 2^2^2, ..., however, does not (the first terms are 2, 4, 16, ... - and it only grows faster from there.)
  - gjm11 16y ago
    Convergence is what happens when you have a sequence of numbers, and they get closer and closer to some fixed number. Some sequences converge and some don't, and part of what's going on with this "paradox" is that one of the sequences it implicitly asks you to consider doesn't converge. Yes, there's a transfinite number implicit in the sequence of exponentiations, namely aleph-0. (It might be better to say omega-0 -- an infinite ordinal rather than an infinite cardinal -- but that's a technicality that doesn't really matter here.) But the problem with the "proof" doesn't have anything to do with the existence of infinite cardinals larger than aleph-0.
- pbhjpbhj 16y ago
  It's 4am here and I need some sleep - choose 2 numbers from within the convergent range and do the same slight of hand with them. Clearly we can argue that x^^\inf = 4 has no convergence and so makes no sense but what if the equations are both in the converging region x € [e^−e, e^1/e] about [0.066, 1.445]: {1} x^^\inf = 2 we've shown in the OP that x^2 = 2 is a solution. Consider, {2} x^^\inf = e^(2/e) similarly to the OP we do {3} x^ (x^^\inf) = x^ (e^(2/e)) but this is just {4} x^^\inf = x^ (e^(2/e)) and substituting on the left from {2} {5} e^(2/e) = x^ (e^(2/e)) gives us (I think) {6} x = e^ (2e^ (-1-(2/e))) ; ~~1.42267 Hence we've shown that {7} 2 ~~ 1.423 Which isn't as neat as 2==4 but equally troublesome. There is no issue now with convergence is there (or there shouldn't be if I chose my numbers correctly)? --- Notations ^^ : ascii version of Knuths up-arrow notation \inf : first countable infinity, aleph-0 € : in the range, it's really a Euro symbol, but close enough ~~ : approximately equal to === Solution y = x^2 = 9 give solution (3,9) y = x^2 = 4 gives solution (2,4) 9=4? They're just different points on the same curve ... doh! We're not showing that 9=4 we're showing that (3,9) and (2,4) are point solutions of y = x^2. In the same way we have above 2 solution of y = x^^\inf.
jules 16y ago
That this doesn't have much at all to do with there being multiple transfinite numbers.
india 16y ago
No infinity other than Aleph-naught appears to be involved in this. Why do you think the existence of other aleph numbers matters here?