I don’t really follow …999999999 + 1 = …000000000 . To me the carry must be dealt with and I would say that …99999 + 1 results in 1000...000 which is greatly larger than …0000.
I see surreals everywhere but engaging with this does tickle a confusino I had with them. In a sense ω is like the unity of transfinite numbers but it is surrouded by ω+2,ω+1, ω−1 and ω−2 if you count up to ω from 0 why you don’t stop at the previous −1 ones? 9 has a number at digit 1, 90 has the 9 in the digit 2. With …9999 all the finite number indexed digits are full of 9. When you add 1 to the thing and addition starts to carry it can’t land on any finite index, yet it must go somewhere. So in the spirit of transfinite induction, the carry lands in the digit ω with 1 leaving all the finite indexes 0. This would seem to be independent of the base used.
I am bit unsure on reading whether uncountable integers are supposed to start with periodic integers or only with non-periodic integers. Rational numbers have the same cardinality as integers so being able to express fractions doesn’t get the party started yet.
What Viliam is looking at here doesn’t (I think) have much to do with surreal numbers.
I don’t understand the question “why you don’t stop at the previous −1 ones?”. I think there may be a false assumption built into it. You don’t “count up to ω from 0″, in the surreal numbers. You do do that, in some sense, in the ordinals, but there there isn’t a ω−1.
Perhaps there might be a variant of p-adic numbers where you allow a digit for each ordinal position rather than each non-negative integer position. But that seems like it might be problematic because there are too many ordinals; for instance, in ZF set theory with the usual conventions there are no functions from the ordinals to {0,1,...,p-1} to be these numbers. (You could consider partial functions, with the convention that anything not specified thereby is zero. But then e.g. you no longer have a representation for −1.)
There are only countably many p-adic numbers with periodic digit sequence. There are uncountably many once you fill in the gaps and allow non-periodic sequences. (Just like with ordinary decimals.)
It might be a little informal in my head but I liken that you get the ordinary finite integers from a successor function and the finite integers get their birthdays by finite induction by being constructed from the previous birthday. So each of that steps seem like “+1”. Then when you do the first transfinite induction it feels like “+1″ “real hard”. And when you have calculations like ω∗1=ω that can seem like it correspond to the operation of “+1, omega times”
As gjm said, this article assumes that there are only digits at integer-numbered positions, and whatever gets pushed to infinity position is effectively lost (because there is no such position). Having digits at surreal-integer-numbered positions seems also interesting, but it’s not what I was trying to do.
I am bit unsure on reading whether uncountable integers are supposed to start with periodic integers or only with non-periodic integers.
Only with non-periodic, because… what you said. Otherwise, there is a finite periodic part (i.e. the sequence that gets repeated is finite), optionally followed by a finite part at the end (and optionally a finite decimal part), that is only countably many options.
I get that …9999 + 1 = …000 goes to an interesting direction and …9999 + 1 = 100...000 goes to an interesting direction. And we can assume/axiom into those directions. But it can also be taken as a claim. One could explore what “2+2=5” implies but another natural reaction is also to be disinterested because one asssume things in opposition to that. So is there multiple ways to make formal or concretise the informal intuitions and does that say something interesting how addition works?
I don’t really follow …999999999 + 1 = …000000000 . To me the carry must be dealt with and I would say that …99999 + 1 results in 1000...000 which is greatly larger than …0000.
I see surreals everywhere but engaging with this does tickle a confusino I had with them. In a sense ω is like the unity of transfinite numbers but it is surrouded by ω+2,ω+1, ω−1 and ω−2 if you count up to ω from 0 why you don’t stop at the previous −1 ones? 9 has a number at digit 1, 90 has the 9 in the digit 2. With …9999 all the finite number indexed digits are full of 9. When you add 1 to the thing and addition starts to carry it can’t land on any finite index, yet it must go somewhere. So in the spirit of transfinite induction, the carry lands in the digit ω with 1 leaving all the finite indexes 0. This would seem to be independent of the base used.
I am bit unsure on reading whether uncountable integers are supposed to start with periodic integers or only with non-periodic integers. Rational numbers have the same cardinality as integers so being able to express fractions doesn’t get the party started yet.
What Viliam is looking at here doesn’t (I think) have much to do with surreal numbers.
I don’t understand the question “why you don’t stop at the previous −1 ones?”. I think there may be a false assumption built into it. You don’t “count up to ω from 0″, in the surreal numbers. You do do that, in some sense, in the ordinals, but there there isn’t a ω−1.
Perhaps there might be a variant of p-adic numbers where you allow a digit for each ordinal position rather than each non-negative integer position. But that seems like it might be problematic because there are too many ordinals; for instance, in ZF set theory with the usual conventions there are no functions from the ordinals to {0,1,...,p-1} to be these numbers. (You could consider partial functions, with the convention that anything not specified thereby is zero. But then e.g. you no longer have a representation for −1.)
There are only countably many p-adic numbers with periodic digit sequence. There are uncountably many once you fill in the gaps and allow non-periodic sequences. (Just like with ordinary decimals.)
ω has birthday ω and ω−1 has a birthday of ω+1.
It might be a little informal in my head but I liken that you get the ordinary finite integers from a successor function and the finite integers get their birthdays by finite induction by being constructed from the previous birthday. So each of that steps seem like “+1”. Then when you do the first transfinite induction it feels like “+1″ “real hard”. And when you have calculations like ω∗1=ω that can seem like it correspond to the operation of “+1, omega times”
As gjm said, this article assumes that there are only digits at integer-numbered positions, and whatever gets pushed to infinity position is effectively lost (because there is no such position). Having digits at surreal-integer-numbered positions seems also interesting, but it’s not what I was trying to do.
Only with non-periodic, because… what you said. Otherwise, there is a finite periodic part (i.e. the sequence that gets repeated is finite), optionally followed by a finite part at the end (and optionally a finite decimal part), that is only countably many options.
I get that …9999 + 1 = …000 goes to an interesting direction and …9999 + 1 = 100...000 goes to an interesting direction. And we can assume/axiom into those directions. But it can also be taken as a claim. One could explore what “2+2=5” implies but another natural reaction is also to be disinterested because one asssume things in opposition to that. So is there multiple ways to make formal or concretise the informal intuitions and does that say something interesting how addition works?
Some thought experiments have later some surprising use e.g. in physics. Other thought experiments don’t. It is probably difficult to predict.