Timeline for A question about a theorem in ONAG by Conway
Current License: CC BY-SA 4.0
10 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Aug 16, 2022 at 18:45 | comment | added | Wouter Zandsteeg | I also came up with a way to avoid what I think is using the theorem itself, but it doesn't involve the latter four inequalities at all. For the first one I did: $x^R+y\leq x+z$, so none of $x^R+y\geq (x+z)^R$, and thus $x^R+y \ngeq x^R+z$, but from $y\geq z$ and induction follows $x^R+y \geq x^R+z$, so a contradiction. The third one is similar. The second one implies by $y^R\leq z$ using the first part of the proof, but $y\geq z$ implies $y^R\nleq z$, so a contradiction. The fourth one is similar. | |
Aug 16, 2022 at 18:45 | comment | added | Wouter Zandsteeg | I think I understand the structure of the proof, but to get from the first four inequalities to the latter four, I can't see an easy way without using the theorem itself. I do understand the rest though. | |
Aug 16, 2022 at 18:02 | comment | added | Apollo | Thanks. A typical argument by contradiction in (structural) induction goes: suppose that the proposition does not hold. Then take a least counter-example X. In our case, this means all elements of X do satisfy the proposition. Then prove that this implies a contradiction, hence the proposition holds for X, and as X was a least counterexample, the proposition must hold generally. Here, he is proving $x+y\ngeq x+z$ implies $y\ngeq z$ (the converse direction) by contradiction. | |
Aug 16, 2022 at 17:20 | history | edited | LSpice | CC BY-SA 4.0 |
Display equations
|
Aug 16, 2022 at 16:25 | comment | added | Wouter Zandsteeg | I edited the question to make it more readable and clear what I don't understand | |
Aug 16, 2022 at 16:25 | history | edited | Wouter Zandsteeg | CC BY-SA 4.0 |
added 551 characters in body
|
Aug 16, 2022 at 15:47 | comment | added | Apollo | As far as the question goes, I think he's just doing a typical inductive argument by contradiction - choose a least putative counter-example (so that all subgames satisfy the induction hypothesis) and show it leads to a contradiction. | |
Aug 16, 2022 at 15:43 | comment | added | Apollo | It's much better if you actually type out the formulae into the question, rather than providing an ephemeral link to an image. | |
S Aug 16, 2022 at 15:25 | review | First questions | |||
Aug 16, 2022 at 15:38 | |||||
S Aug 16, 2022 at 15:25 | history | asked | Wouter Zandsteeg | CC BY-SA 4.0 |