20 1.4 Ordered Field Axioms Proposition 1.4.1 For x,y,z ∈R, the following hold: (i) If x+y =x+z, theny =z. (ii) −(−x)=x. (iii) If x̸ =0 and xy =xz, theny =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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