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@DomonkosV DomonkosV commented Nov 19, 2025

fix #366

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Thanks for the contribution! Looks good to me after these comments have been addressed :)


@[category research open, AMS 11]
theorem erdos_125 :
({ x : ℕ | (digits 3 x) ⊆ {0, 1} } + { x : ℕ | (digits 4 x) ⊆ {0, 1} }).HasPosDensity ↔
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Suggested change
({ x : ℕ | (digits 3 x) ⊆ {0, 1} } + { x : ℕ | (digits 4 x) ⊆ {0, 1} }).HasPosDensity ↔
{ x : ℕ | (digits 3 x).toFinset ⊆ {0, 1} } + { x : ℕ | (digits 4 x).toFinset ⊆ {0, 1} }.HasPosDensity ↔

Apparently here {0, 1} is treated as a list, because List has an instance of the Insert class, which is used to implement the notation { } for writing e.g. {a, b}. So the statement currently says that x has first digit 0 and second digit 1 (i.e. x = 3 or x = 4). I've suggested a fix.

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The round brackets are mandatory for .HasPosDensity to take the full set.

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Ah right, I forgot you were doing addition inside the brackets, my bad!

Comment on lines 38 to 39
({ x : ℕ | (digits 3 x).toFinset ⊆ {0, 1} } + { x : ℕ | (digits 4 x).toFinset ⊆ {0, 1} }
).HasPosDensity ↔ answer(sorry) := by
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Suggested change
({ x : ℕ | (digits 3 x).toFinset ⊆ {0, 1} } + { x : ℕ | (digits 4 x).toFinset ⊆ {0, 1} }
).HasPosDensity ↔ answer(sorry) := by
theorem erdos_125 :
({ x : ℕ | (digits 3 x).toFinset ⊆ {0, 1} } +
{ x : ℕ | (digits 4 x).toFinset ⊆ {0, 1} }).HasPosDensity ↔ answer(sorry) := by

Final formatting suggestion :)

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Approved once this final formatting comment has been addressed :) Thanks a lot!

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Erdős Problem 125

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