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 x ≠ 0 and xy = xz, then y = z. (iv) If x ̸ = 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 and z < 0, then xz > 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) From x > 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. Adding xz+ 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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