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Fix 2012 A4 #255

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Fix 2012 A4 #255

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ocfnash
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@ocfnash ocfnash commented Jan 10, 2025

This was disprovable because of the n > 2 condition in the goal.
For example if we take:

  • $q = 101$
  • $r = 1$
  • $A = [0, 100]$
  • $B = [1, 2]$

Then $S = \{1, 2\}$ and there is no three-or-more term arithmetic progression whose intersection with $A$ is $S$.

This was disprovable because of the `n > 2` condition in the goal.
For example if we take:
 * $q = 101$
 * $r = 1$
 * $A = [0, 100]$
 * $B = [1, 2]$

Then $S = \{1, 2\}$ and there is no three-or-more term arithmetic
progression whose intersection with $A$ is $S$.
@ocfnash ocfnash force-pushed the ocfnash/fix_2012_a4 branch from bff7ce3 to c6be8d3 Compare January 10, 2025 16:44
(hT : T = {x : ℤ | ∃ b m : ℤ, (b : ℝ) ∈ Set.Icc (B 0) (B 1) ∧ x = b + m * q})
(hS : S = {a : ℤ | (a : ℝ) ∈ Set.Icc (A 0) (A 1) ∧ r * a ∈ T})
(prod_lt : (A 1 - A 0) * (B 1 - B 0) < q) :
IsFiniteAP {x | x ∈ S ∧ (x : ℝ) ∈ Set.Icc (A 0) (A 1)} :=
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This final goal line states that the A \cap S is an arithmetic progression, but the problem statement says S is A \cap X where X is a set of points of an (finite) AP. The latter implies the former but my feeling is that the goal as stated does not reflect the informal statement, what do you think?

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