Maths Where Fermat's Last Theorem Doesn't Hold

3 points by Kabbalist ↗ HN
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.

3 comments

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"∃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.