4 ms·
Maths Where Fermat's Last Theorem Doesn't Hold
Are there maths that violate Fermat’s Last Theorem? I ask because of the following.
Let A,B,C, and K be elements of the natural numbers.
1. ∃A,B,C,K(K≤2∧A^K+B^K=C^K )
2. ∃A,B,C,K(K≤2∨A^K+B^K=C^K )
3. ∃A,B,C,K(K>2→A^K+B^K=C^K )
Conclusion: ∃A,B,C,K(K>2→A^K+B^K=C^K )
Premise 1 is true because at K=1 and K=2 there does exist A,B, and C such that A^K+B^K=C^K . Premise 2 is true because of Premise 1. Premise 3 is true since it is logically equivalent to Premise 2. Therefore the conclusion must be true. But this seems contrary to Fermat’s Last Theorem though. So it is either the case that this argument is invalid, one of the premises is wrong, or there exists maths where Fermat’s Last Theorem doesn’t hold.
- josephcsible 3y ago"∃A,B,C,K(K>2→A^K+B^K=C^K )" is true but is not contrary to Fermat's Last Theorem. The K that must exist will just never be greater than 2.
- al2o3cr 3y agoShort short version: if the left-hand side of → is false, then the result of → is ALWAYS TRUE. This is tricky enough compared to normal intuition about "implication" that there's a name for it: https://en.wikipedia.org/wiki/Paradoxes_of_material_implication https://en.wikipedia.org/wiki/Paradoxes_of_material_implicat...
- josephcsible 3y agoOP already knows this, though, since that's how you get from 2 to 3!