Difference between revisions of "Talk:1208: Footnote Labyrinths"
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Explaination is wrong : a<sup>b<sup>c</sup></sup> = a<sup>(b<sup>c</sup>)</sup> = a<sup>b^c</sup> (confer the definition of a gogol = 10^100 = 10<sup>10<sup>2</sup></sup>, and a gogolplex = 10^gogol = 10<sup>(10<sup>100</sup>)</sup>, not 10^110. So since 1^2= 1, No<sup>1<sup>2</sup></sup> really means No<sup>1</sup>. | Explaination is wrong : a<sup>b<sup>c</sup></sup> = a<sup>(b<sup>c</sup>)</sup> = a<sup>b^c</sup> (confer the definition of a gogol = 10^100 = 10<sup>10<sup>2</sup></sup>, and a gogolplex = 10^gogol = 10<sup>(10<sup>100</sup>)</sup>, not 10^110. So since 1^2= 1, No<sup>1<sup>2</sup></sup> really means No<sup>1</sup>. | ||
+ | |||
+ | == Question, alternative explination == | ||
+ | |||
+ | I wasn't really satisfied with the whole discarding of the infinite loop, so I worked through the problem seperately using the nested footnotes. Then, when we hit the infinite loop I split between the two possible answers (either the infinite loop ends on true or false). As I read it, they both get the same answer: | ||
+ | |||
+ | no (3) | ||
+ | no (not true (5)) | ||
+ | no (not true (true (2 < 6 < 3)) | ||
+ | no (not true (true (2 < 6 < (not true)))) | ||
+ | no (not true (true (2 < (actually 1 < 2 < 2 (not true 3 < 2))))) | ||
+ | no (not true (true (2 < (actually 1 < 2 < 2 (not true (5))))) | ||
+ | |||
+ | Split! | ||
+ | If 6 is false (infinite loop possibility) | ||
+ | no (3 < 5 < 2) | ||
+ | no (not true (7)) - meaningless, so discard | ||
+ | no (not true) | ||
+ | |||
+ | If 6 is true (infinite loop possibility) | ||
+ | no (3 < 5 < 1 < 2 < 2) | ||
+ | no (3 < 5 < 1 < 4) | ||
+ | no (3 < 5 < 1) | ||
+ | no (3) | ||
+ | no (not true) | ||
+ | |||
+ | Both lead to the answer "... experiments to observe this and we found evidence for it in our data". |
Revision as of 15:03, 6 May 2013
Way to nerd-snipe me, Randall. Alpha (talk) 04:52, 6 May 2013 (UTC)
In the nested-footnotes interpretation, 5 has to be ignored: The 6 must be true, and the 6 says that it’s “actually a 1”, but with footnote 2+2 which says “ibid.” and thus equals footnote 3, which is true. So 6 really does mean actually a 1, which leaves 5 to be ignored. --77.186.8.191 10:47, 6 May 2013 (UTC)
The footnote for 6 is actually 1 to the 2 to the 2 Schmammel (talk) 12:36, 6 May 2013 (UTC)
Explaination is wrong : abc = a(bc) = ab^c (confer the definition of a gogol = 10^100 = 10102, and a gogolplex = 10^gogol = 10(10100), not 10^110. So since 1^2= 1, No12 really means No1.
Question, alternative explination
I wasn't really satisfied with the whole discarding of the infinite loop, so I worked through the problem seperately using the nested footnotes. Then, when we hit the infinite loop I split between the two possible answers (either the infinite loop ends on true or false). As I read it, they both get the same answer:
no (3) no (not true (5)) no (not true (true (2 < 6 < 3)) no (not true (true (2 < 6 < (not true)))) no (not true (true (2 < (actually 1 < 2 < 2 (not true 3 < 2))))) no (not true (true (2 < (actually 1 < 2 < 2 (not true (5)))))
Split! If 6 is false (infinite loop possibility) no (3 < 5 < 2) no (not true (7)) - meaningless, so discard no (not true)
If 6 is true (infinite loop possibility) no (3 < 5 < 1 < 2 < 2) no (3 < 5 < 1 < 4) no (3 < 5 < 1) no (3) no (not true)
Both lead to the answer "... experiments to observe this and we found evidence for it in our data".