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Definition A.F.1(K,+,*,0,1) is a field if and only if the following conditions hold. A.F.A1 (K,+,0) is a group. A.F.A2 /\(x,y:-K) x+y = y+x A.F.A3 /\(x,y:-K) x*y:-K A.F.A4 (K\{0},*,1) is a group A.F.A5 /\(x,y:-K) x*y = y*x A.F.A6 /\(x,y,z):-K z*(x+y) = z*x+z*yTheorem A.F.2If (K,+,*,0,1) is a field and x:-K then x*0 = 0.ProofBy A.F.A6 x*0 + x*0 = x*(0+0). By A.F.A2 0+0 = 0. Thus x*0 + x*0 = x*0. Now by A.F.A2 and Theorem A.G.2 x*0 = 0.Definition A.F.3Let (K,+,*,0,1) be a field and x,y:-K. We define that y = -x if and only if y+x = 0.Remark A.F.4Definition A.F.3 is correct because of Theorem A.G.7.Theorem A.F.5If (K,+,*,0,1) is a field and x:-K then (-1)*x = -x.ProofFollow the calculations below. (-1)*x+x = x*(-1)+x*1 = x*(-1+1) = x*0 = 0. Thus by Definition A.F.3 (-1)*x = -x.Definition A.F.6Let (K,+,*,0,1) be a field and x,y:-K\{0}. We define that y = x^(-1) if and only if y*x = 1.Remark A.F.7Definition A.F.6 is correct because of Theorem A.G.7.Theorem A.F.8If (K,+,*,0,1) is a field, x,y:-K and x*y = 0 then x = 0 or y = 0.ProofAssume to the contrary that x != 0 and y != 0. By A.F.A4 there exist x^(-1) and y^(-1). By Theorem A.F.2 ( (y^(-1))*(x^(-1)) )*x*y = 0. By A.F.A4 ( (y^(-1))*(x^(-1)) )*x*y = (y^(-1))*y = 1. Thus 1 = 0. But by A.F.A4 1:-K\{0}. Thus 1 != 0. Contradiction. We have proved that x = 0 or y = 0.