Timeline for A question about primitive recursive functions
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 20, 2013 at 15:30 | answer | added | Paul Taylor | timeline score: 3 | |
| Apr 18, 2013 at 14:10 | vote | accept | Sencodian | ||
| Apr 18, 2013 at 14:09 | vote | accept | Sencodian | ||
| Apr 18, 2013 at 14:10 | |||||
| Apr 18, 2013 at 14:09 | vote | accept | Sencodian | ||
| Apr 18, 2013 at 14:09 | |||||
| Apr 18, 2013 at 14:03 | comment | added | Joel David Hamkins | DK, you mean to say that they are not a group. Frank, the inverse of Ackermann is primitive recursive, but this is not a bijection. But you can fix it up via the even/odd trick as in my argument and also as in DK's link (and those arguments are fundamentally similar). | |
| Apr 18, 2013 at 13:39 | comment | added | Denis | Exercise 5.6 in this book claims that bijective primitive functions are a group, i.e. such a function $f$ exists: books.google.co.il/… | |
| Apr 18, 2013 at 13:37 | answer | added | Joel David Hamkins | timeline score: 10 | |
| Apr 18, 2013 at 13:15 | history | asked | Sencodian | CC BY-SA 3.0 |