# Difference between revisions of "Talk:1208: Footnote Labyrinths"

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:There are differences in interpretation here. If we write "foo<sup>3<sup>6</sup></sup>", is it equal to "foo<sup>1<sup>1<sup>2</sup></sup></sup>" or "foo<sup>3<sup>1<sup>1<sup>2</sup></sup></sup></sup>"? I assumed the former and you assumed the latter. My reasoning is that footnotes modify their arguments and not themselves. [[User:Alpha|Alpha]] ([[User talk:Alpha|talk]]) 17:44, 6 May 2013 (UTC) | :There are differences in interpretation here. If we write "foo<sup>3<sup>6</sup></sup>", is it equal to "foo<sup>1<sup>1<sup>2</sup></sup></sup>" or "foo<sup>3<sup>1<sup>1<sup>2</sup></sup></sup></sup>"? I assumed the former and you assumed the latter. My reasoning is that footnotes modify their arguments and not themselves. [[User:Alpha|Alpha]] ([[User talk:Alpha|talk]]) 17:44, 6 May 2013 (UTC) | ||

− | Shouldn't 5 be true (because 6 is actually 1<sup>3</sup>; therefore 5 is true<sup>1<sup>3<sup>3</sup></sup></sup>; so the 2 is ignored regardless the truth of 3) and 3 is not true? Sebastian --[[Special:Contributions/178.26.118.249|178.26.118.249]] 18:35, 6 May 2013 (UTC) | + | Shouldn't 5 be true (because 6 is actually 1<sup>3</sup>; therefore 5 is true<sup>2<sup>1<sup>3<sup>3</sup></sup></sup></sup>; so the 2 is ignored regardless the truth of 3) and 3 is not true? Sebastian --[[Special:Contributions/178.26.118.249|178.26.118.249]] 18:35, 6 May 2013 (UTC) |

## Revision as of 18:36, 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 : a^{bc} = a^{(bc)} = a^{b^c} (confer the definition of a gogol = 10^100 = 10^{102}, and a gogolplex = 10^gogol = 10^{(10100)}, not 10^110. So since 1^2= 1, No^{12} really means No^{1}. 192.54.145.66 (talk) *(please sign your comments with ~~~~)*

- Yes, so "no
^{1}" means to ignore the "no" and the answer for the second explanation is "we found evidence for the data." By the way, it's spelled "googol." Alpha (talk) 17:51, 6 May 2013 (UTC)

- 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". -- Urah (talk) *(please sign your comments with ~~~~)*

- Yes, but at each stage you may "
*toggle between interpreting nested footnotes as footnotes on footnotes and interpreting them as exponents (minus one, modulo 6, plus 1).*" That is, a^{23}may*either*be read as "apply note 8 (=2mod6) to text*a*", or as "apply note 3 to text "2", then the result to text*a*". 192.54.145.66 (talk)*(please sign your comments with ~~~~)* - There are differences in interpretation here. If we write "foo
^{36}", is it equal to "foo^{112}" or "foo^{3112}"? I assumed the former and you assumed the latter. My reasoning is that footnotes modify their arguments and not themselves. Alpha (talk) 17:44, 6 May 2013 (UTC)

Shouldn't 5 be true (because 6 is actually 1^{3}; therefore 5 is true^{2133}; so the 2 is ignored regardless the truth of 3) and 3 is not true? Sebastian --178.26.118.249 18:35, 6 May 2013 (UTC)