proof... A field consists of a set F, distinguished elements 0 ∈ F, 1 ∈ F, and functions + : F ×F → F , and · : F × F → F . We will write a + b for +(a, b) and ab for ·(a, b). These data are subject to the axioms:
i) ∀a,b,c∈F, (a+b)+c=a+(b+c) ii) ∀a∈F, 0+a=a
iii) ∀a∈F, ∃a ̃∈F, a ̃+a=0 iv) ∀a,b∈F, a+b=b+a
v) ∀a,b,c ∈ F, (ab)c = a(bc) vi) ∀a∈F, 1a=a
vii) ∀a∈F, ∼(a=0)⇒∃a′ ∈F,a′a=1 viii) ∀a, b ∈ F, ab = ba
ix) ∀a,b,c∈F, (a+b)c=ac+bc

Question

proof... A field consists of a set F, distinguished elements 0 ∈ F, 1 ∈ F, and functions + : F ×F → F , and · : F × F → F . We will write a + b for +(a, b) and ab for ·(a, b). These data are subject to the axioms:
i) ∀a,b,c∈F, (a+b)+c=a+(b+c) ii) ∀a∈F, 0+a=a
iii) ∀a∈F, ∃a ̃∈F, a ̃+a=0 iv) ∀a,b∈F, a+b=b+a
v) ∀a,b,c ∈ F, (ab)c = a(bc) vi) ∀a∈F, 1a=a
vii) ∀a∈F, ∼(a=0)⇒∃a′ ∈F,a′a=1 viii) ∀a, b ∈ F, ab = ba
ix) ∀a,b,c∈F, (a+b)c=ac+bc

anonymous · Answer

Consider the set F = {0, 1} with the following addition and multiplication tables:

anonymous · Answer

|dw:1351460297489:dw|