Timeline for Check the Regularity of the folowing Languages
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 31, 2016 at 16:18 | history | edited | J.-E. Pin | CC BY-SA 3.0 | deleted 203 characters in body |
| Oct 31, 2016 at 13:51 | comment | added | J.-E. Pin | Well, if $n = 0$, $(a^{pq})^n = a^0$. If $n = 1$, $(a^{pq})^n = a^{pq}$ and if $n \geqslant 2$, then $(a^{pq})^n = a^{pqn}$. Thus $L_2 = \{a^0\} \cup \{a^r \mid r \text{ is a product of at least 2 numbers } \geqslant 2\}$. Now, what is the definition of a prime number? | |
| Oct 31, 2016 at 13:38 | comment | added | Akhil Nadh PC | Sir, How did you find complement of $L_{2}$ ? I couldn't understand how did you get it as $P =\{ a^{p} | p$ $is$$ prime \} $ | |
| Oct 29, 2016 at 14:13 | history | edited | J.-E. Pin | CC BY-SA 3.0 | added 1 character in body |
| Oct 29, 2016 at 9:07 | history | answered | J.-E. Pin | CC BY-SA 3.0 |