Introduction to Mathematical Analysis I - Second Edition

20 1.4 ORDERED FIELD AXIOMS Proof: (a) Suppose x + y = x + z . Adding − x (which exists by axiom (2d) ) to both sides, we have ( − x )+( x + y ) = ( − x )+( x + z ) . Then axiom (1a) gives [( − x )+ x ] + y = [( − x )+ x ] + z . Thus, again by axiom (2d) , 0 + y = 0 + z and, by axiom (1c) , y = z . (b) Since ( − x )+ x = 0, we have (by uniqueness in axiom (2d) ) − ( − x ) = x . The proofs of (c) and (d) are similar. (e) Using axiom (2e) we have 0 x = ( 0 + 0 ) x = 0 x + 0 x . Adding − ( 0 x ) to both sides (axiom (2d) ) and using axioms (1a) and (1c) , we get 0 = − ( 0 x )+ 0 x = − ( 0 x )+( 0 x + 0 x ) = ( − ( 0 x )+ 0 x )+ 0 x = 0 + 0 x = 0 x . That 0 x = x 0 follows from axiom (2b) . (f) Using axioms (2c) and (2e) we get x +( − 1 ) x = 1 x +( − 1 ) x = ( 1 +( − 1 )) x . From axiom (2d) we get 1 +( − 1 ) = 0 and from part (e) we get x +( − 1 ) x = 0 x = 0. From the uniqueness in axiom (2d) we get ( − 1 ) x = − x as desired. (g) Using axioms (2e) and (1c) we have xz + x ( − z ) = x ( z + ( − z )) = x 0 = 0. Thus, using axiom (2d) we get that x ( − z ) = − ( xz ) . The other equality follows similarly. (h) From x > 0, using axioms (3c) and (1c) we have x + ( − x ) > 0 + ( − x ) = − x . Thus, using axiom (2d) , we get 0 > − x . The other case follows in a similar way. (i) Since z < 0, by part (h) , − z > 0. Then, by axiom (3d) , x ( − z ) < y ( − z ) . Combining this with part (g) we get − xz < − yz . Adding xz + yz to both sides and using axioms (1a) , (3c) , (1b) , and (1c) we get xy = ( − xz + xz )+ xy = − xz +( xz + xy ) < − xy +( xz + xy ) = − xy +( xy + xz ) = ( − xy + xy )+ xz = xz . (j) Axiom (2c) gives that 1 6 = 0. Suppose, by way of contradiction, that 1 < 0. Then by part (i) , 1 · 1 > 0 · 1. Since 1 · 1 = 1, by axiom (2c) and 0 · 1 = 0 by part (e) , we get 1 > 0 which is a contradiction. It follows that 1 > 0. Note that we can assume that the set of all natural numbers is a subset of R (and of any ordered field, in fact) by identifying the 1 in N with the 1 in axiom (2c) above, the number 2 with 1 + 1, 3 with 1 + 1 + 1, etc. Furthermore, since 0 < 1 (from part (j) of the previous proposition), axiom (3c) gives, 1 < 2 < 3, etc (in particular all these numbers are distinct). In a similar way, can include Z and Q as subsets. We say that a real number x is irrational if x ∈ R \ Q , that is, if it is not rational. Definition 1.4.1 Given x ∈ R , define the absolute value of x by | x | = ( x , if x ≥ 0; − x , if x < 0 .