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We can define a field F with the following properties:

  • Binary operations + (addition) and ⋅ (multiplication)
  • Commutativity
  • Associativity
  • Identities
  • Inverses
  • Distributivity

Now, the additive inverse existence condition can be equivalently replaced with the statement ∀a∈F 0a=0, because each one is a consequence of the other.

The proof (assuming 0=0a): $$0=0a=(1-1)a=a+(-1a)$$ Which means that for every a there exists an additive inverse, more precisely it is -1a.

But now take the set {0,1} with logical conjunction for ⋅; and inclusive disjunction for +. (so that 1+1=1)

  • 1 doesn't have an additive inverse, as 1+1=1+0=1≠0
  • concurrently, 0⋅0=0 and 0⋅1=0; so this indeed appears to be a field by the alternative condition.

Which is a contradiction.

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The error is in my proof that 0=0a →-1a=-a, because -1∉{0,1}.

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