Timeline for An easy proof that $S(n)$ does not embed into $A(n+1)$?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Aug 9, 2022 at 15:27 | comment | added | LSpice | I think that this answers "does $S_n$ embed in $A_{n + 1}$ for all $n \ge 2$?" (in the negative), but it fails to prove "$S_n$ does not embed in $A_{n + 1}$ for all $n \ge 2$", of which the usual parenthesisation is "($S_n$ does not embed in $A_{n + 1}$) for all $n \ge 2$". | |
| Aug 9, 2022 at 13:48 | comment | added | Alex M. | Not quite: the problem could have easily been restated for $n \ge 3$. The point here is not to solve a problem, but to understand the deep reason for which a mathematical fact is true, which your solution does not do. In other words, your solution chooses to go for the "low-hanging fruit". | |
| Aug 9, 2022 at 13:37 | comment | added | Dario | Sure, in Rotman's exercise the question was posed for $n\geq 2$, so the counterexample with $n=2$ was sufficient. | |
| Aug 9, 2022 at 12:20 | review | Late answers | |||
| Aug 9, 2022 at 12:41 | |||||
| Aug 9, 2022 at 12:15 | comment | added | Alex M. | The problem is that you only attempt to prove that $S_2$ cannot be embedded in $A_3$, but maybe the embedding is true for $n \ge 3$. | |
| S Aug 9, 2022 at 12:02 | review | First answers | |||
| Aug 9, 2022 at 12:15 | |||||
| S Aug 9, 2022 at 12:02 | history | answered | Dario | CC BY-SA 4.0 |