20 1.4 Ordered Field Axioms Proposition 1.4.1 For x,y,z ∈R, the following hold: (i) If x+y =x+z, then y =z. (ii) −(−x) =x. (iii) If x̸ =0 and xy =xz, then y =z. (iv) If x̸ =0, then 1/(1/x) =x. (v) 0x =0=x0. (vi) −x = (−1)x. (vii) x(−z) = (−x)z =−(xz). (viii) If x >0, then−x <0; if x <0, then−x >0. (ix) If x <y andz <0, thenxz >yz. (x) 0<1. Proof: (i) Suppose x+y =x+z. Adding −x (which exists by axiom (A4)) to both sides, we have (−x)+(x+y) = (−x)+(x+z). Then axiom (A1) gives [(−x)+x] +y = [(−x)+x] +z. Thus, again by axiom (A4), 0+y =0+z and, by axiom (A3), y =z. (ii) Since (−x)+x =0, we have (by uniqueness in axiom (A4)) −(−x) =x. The proofs of (iii) and (iv) are similar. (v) Using axiom (D1) we have 0x= (0+0)x=0x+0x. Adding−(0x) to both sides (axiom (A4)) and using axioms (A1) and (A3), we get 0=−(0x)+0x =−(0x)+(0x+0x) = (−(0x)+0x)+0x =0+0x =0x. That 0x =x0 follows from axiom (M2). (vi) Using axioms (M3) and (D1) we get x+ (−1)x =1x+ (−1)x = (1+ (−1))x. From axiom (A4) we get 1+(−1) =0 and from part (v) we get x+(−1)x =0x =0. From the uniqueness in axiom (A4) we get (−1)x =−x as desired. (vii) Using axioms (D1) and (A3) and part (v) we have xz+x(−z) =x(z+(−z)) =x0=0. Thus, using axiom (A4) we get that x(−z) =−(xz). The other equality follows similarly. (viii) Fromx >0, using axioms (O3) and (A3) we have x+(−x) >0+(−x) =−x. Thus, using axiom (A4), we get 0>−x. The other case follows in a similar way. (ix) Since z <0, by part (viii), −z >0. Then, by axiom (O4), x(−z) <y(−z). Combining this with part (vii) we get −xz <−yz. Addingxz+yz to both sides and using axioms (A1), (O3), (A2), and (A3) we get yz = (−xz+xz)+yz =−xz+(xz+yz) <−yz+(xz+yz) =−yz+(yz+xz) = (−yz+yz)+xz =xz.
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