Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with surreal-numbers
Search options answers only
not deleted
user 18060
For questions about the surreal numbers, which are a real-closed ordered proper-class-sized field that contains both the real numbers and the ordinal numbers. Thus they contain both infinite numbers (including the ordinals, but also infinite numbers like ω-1 and sqrt(ω)) and infinitesimal numbers (like 1/ω). They can also be identified with a subclass of two-player partisan games.
2
votes
What surreal numbers are representable by Red-Blue Hackenbush games?
If you only allow connected graphs, and in particular do not allow ordinal numbers, the answer is no, unless you assign surreal numbers to Hackenbush games by a means other than induction.
First it's …
- 148k
20
votes
Accepted
Are Conway's omnific integers the Grothendieck group of the ordinals under commutative addit...
There is an obvious extension of Cantor normal form to the Grothendieck group of the ordinals. Then the standard argument that $\sqrt{x}$ does not lie in the ring $\mathbb Z[x]$ applies to $\sqrt{\ome …
- 148k