POST ANSWERS IN WHISPERS PLEASE. You can whisper a message by entering my username in "Whisper your comment to" above the post field.
(1) See my signature. The goal is to tell me ALL the errors. Look closely! Also, paradoxical answers are not accepted. (2) A: There is a list of 2015 statements: 1: Exactly 1 statement on this list is true. 2: Exactly 2 statements on this list are true. etc. 2015: Exactly 2015 statements on this list are true.
B: What happens if you change "Exactly" to "At most"? C: What if you change "Exactly" to "At least"? D,E,F: What if you change "True" to "False" in A, B, and C?
NOTE: I don't know the answer to problem 2, as I heard them as 2000 and the answer may be different for odd vs. even. (For example, if we have statements 0..n-1 where statement i says "Statement (i+1) mod n is false", it's valid for n even but a paradox if n is odd.)----------------- 3.14159265358979323846264338327950288... Nothing Here
Dude, if you know the answers, whisper them. There's no need to play tricks on ME.----------------- 3.14159265358979323846264338327950288... Nothing Here
I dont understand the second one, but as for the sig thing, meh. its easy. (nice try mathdude, i see what you did there with that one error :P)----------------- puzzled
On the second one, the first statement is asserting that only one of the statements is true. The 42nd statement says there are exactly 42 true statements.
Then, I discuss 5 variations where you change "Exactly" to "At most", then "At least", and if "true" is replaced with "false".----------------- 3.14159265358979323846264338327950288... Nothing Here
For each of these, consider a generalized list of n statements, then state the n = 2015 case.
A. No two statements can be simultaneously true. Therefore at most one statement is true. So, either no statements are true or only the first is. n = 2015: Only the first statement is true, or all of them are false.
B. The truth of the kth statement implies the n-k following. For k < n-k+1 this is a contradiction, so for k ≥ (n+1)/2 the kth statements are true and the rest false. - For n = 2m+1 odd this suggests the first m are false and the remaining m+1 are true. - For n = 2m even this has the first m are false with the last m true. But the mth statement is false, so there is a contradiction. n = 2015: Statements 1-1007 are false and 1008-2015 are true.
C. Trivially all statements can be false. The truth of the kth statement implies the k-1 preceding. Hence, each statement implies itself. Therefore for any k between 0 and n inclusive, the first k statements are true and the rest are false. n = 2015: There are 2016 solutions as described above.
D. No two statements can be simultaneously true. Then at most one is true; if none are true the last statement is a contradiction, while if one is true it must be the second-last. n = 2015: Only statement 2014 is true.
E. The last statement is true. This implies the previous statement, and so on. Therefore all statements are true. n = 2015: All statements are true.
F. The truth of the kth statement implies the k-1 preceding. For k > n-k this is a contradiction, so for k ≤ n/2 the kth statements are true and the rest false. - For n = 2m+1 odd this suggests the first m are true and the remaining m+1 are false. But the m+1th statement is false, so there is a contradiction. - For n = 2m even this suggests that the first m are true with the last m false. n = 2015: There is a contradiction (no solution).
Do tell me if I've made any logical leaps.----------------- Hopefully PA is inconsistent.