5 ms·

Same result. Public share https://claude.ai/share/19fd1a34-d63b-4a16-8d83-60d5b79e7747 https://claude.ai/share/19fd1a34-d63b-4a16-8d83-60d5b79e7747 It did the

Same result. Public share https://claude.ai/share/19fd1a34-d63b-4a16-8d83-60d5b79e7747 https://claude.ai/share/19fd1a34-d63b-4a16-8d83-60d5b79e7747 It did the multiple verification sequence before expanding to internet search where it found this thread.

minimaxir 3mo ago
> So the conjecture that survived Keller, Abhyankar, Moh's degree-100 verification, and five-plus published wrong proofs appears to have died via tweet during the World Cup final.
- rvz 3mo ago
  ..and (you guessed it) before GTA 6.
- FacelessJim 3mo ago
  I lol’d. I didn’t know Fable was this sassy
  - eru 3mo ago
    Oh, Fable can be surprisingly witty and sassy, when you get them in the mood.
moffkalast 3mo ago
I like how first it's amazed and doesn't believe it, accepts it, then realizes you not only stole it from twitter but that it's its own proof lmao.
kzrdude 3mo ago
Which exact model does claude use here?
- drcongo 3mo ago
  > the "fable" in that tweet is Claude Fable, i.e., this model Looks like this was also Fable.
nxtfari 3mo ago
Everyone using Claude Fable to verify this proof is so funny. If you read the definition of the Jacobian Conjecture and (I am not exaggerating this) have passed a college Calc 3 class, you can just verify the proof yourself in 30 seconds. The problem was very hard to solve but the counterexample is very easy to verify! --- edit, adding an explanation: To summarize it, the conjecture says if you have any multi-variable polynomial function that maps an input to an output in the same dimensional space (take for example: F = (x+2, y+2), which maps 2D space into another 2D space), AND that function has a constant-valued non-zero Jacobian determinant, THEN the conjecture is that the polynomial has an inverse, meaning basically you can find a polynomial that turns the output space back into the input space. Fable provided the example polynomial (which was very hard to do) and the coordinates which if you plug into it, results in two points being mapped to the same output point. This means that the polynomial can't be inverted, because if you have that output point, how do you know which input point it came from? You can just plug in the two coordinates it gave into the equation and verify that you get the same output point from both. That's the contradiction of the conjecture and it takes 30 seconds. --- Something something outsourcing of thinking something.
- mswphd 3mo ago
  verification also requires computing the jacobian. an undergraduate-level exercise, but harder than verifying p(pt1) == p(pt2).
- mw888 3mo ago
  Something something pedantry.
cwillu 3mo ago
“I'll admit the sequence on my end was genuinely disorienting: I verified the determinant three separate ways looking for the error, verified the evaluations twice, ran out of subtleties to check, and only then searched and discovered that the reason it holds up is that it's apparently my own homework — the "fable" in that tweet is Claude Fable, i.e., this model, working with Alpöge.”
jacv 3mo ago
[flagged]

minimaxir 3mo ago

> So the conjecture that survived Keller, Abhyankar, Moh's degree-100 verification, and five-plus published wrong proofs appears to have died via tweet during the World Cup final.

rvz 3mo ago
..and (you guessed it) before GTA 6.
FacelessJim 3mo ago
I lol’d. I didn’t know Fable was this sassy
- eru 3mo ago
  Oh, Fable can be surprisingly witty and sassy, when you get them in the mood.

moffkalast 3mo ago

I like how first it's amazed and doesn't believe it, accepts it, then realizes you not only stole it from twitter but that it's its own proof lmao.

kzrdude 3mo ago

Which exact model does claude use here?

drcongo 3mo ago
> the "fable" in that tweet is Claude Fable, i.e., this model Looks like this was also Fable.

nxtfari 3mo ago

Everyone using Claude Fable to verify this proof is so funny. If you read the definition of the Jacobian Conjecture and (I am not exaggerating this) have passed a college Calc 3 class, you can just verify the proof yourself in 30 seconds. The problem was very hard to solve but the counterexample is very easy to verify! --- edit, adding an explanation: To summarize it, the conjecture says if you have any multi-variable polynomial function that maps an input to an output in the same dimensional space (take for example: F = (x+2, y+2), which maps 2D space into another 2D space), AND that function has a constant-valued non-zero Jacobian determinant, THEN the conjecture is that the polynomial has an inverse, meaning basically you can find a polynomial that turns the output space back into the input space. Fable provided the example polynomial (which was very hard to do) and the coordinates which if you plug into it, results in two points being mapped to the same output point. This means that the polynomial can't be inverted, because if you have that output point, how do you know which input point it came from? You can just plug in the two coordinates it gave into the equation and verify that you get the same output point from both. That's the contradiction of the conjecture and it takes 30 seconds. --- Something something outsourcing of thinking something.

mswphd 3mo ago
verification also requires computing the jacobian. an undergraduate-level exercise, but harder than verifying p(pt1) == p(pt2).
mw888 3mo ago
Something something pedantry.

cwillu 3mo ago

“I'll admit the sequence on my end was genuinely disorienting: I verified the determinant three separate ways looking for the error, verified the evaluations twice, ran out of subtleties to check, and only then searched and discovered that the reason it holds up is that it's apparently my own homework — the "fable" in that tweet is Claude Fable, i.e., this model, working with Alpöge.”

jacv 3mo ago

[flagged]