"prove that there are no non negative numbers less than 3"
Claude puts out this elaborate BS - whats funny is the pretension.
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This is trivial to prove by contradiction:
Assume there exists a non-negative number n < 3. Then either:
n = 0. But 0 is not less than 3, so this is a contradiction. 0 < n < 3. But the only numbers between 0 and 3 are 1 and 2, and neither of those are less than 3, so this is also a contradiction. Therefore, our assumption that such an n exists must be false. By contradiction, there are no non-negative numbers less than 3.
In formal logical terms:
Assume ∃n ∈ R, n ≥ 0 and n < 3 (There exists some non-negative number n less than 3) Then either: a) n = 0. But 0 ≥ 3 is false, so this is a contradiction. b) 0 < n < 3. The only numbers in (0,3) are 1 and 2, neither of which is less than 3, so this is also a contradiction. Therefore the assumption in (1) must be false. By contradiction, the statement ∀n ∈ R, n ≥ 0 → n ≥ 3 (For all non-negative numbers n, n is greater than or equal to 3) must be true. So the proof by contradiction shows that there are no non-negative numbers less than 3.
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boosters can hype all you want, but the damn thing cant compute. and please dont say - "but most humans will fail at this too"
meanwhile...: claude-coders working furiously to add this test case - lol.