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Fields Let (F; +; ) be a …eld. For the sake of brevity, we will omit reference to the use of commutative axioms. The dot ‘’in multiplication is sometimes omitted. x2 is de…ned to be x x: Proposition 1.16(a). For any x 2 F we have 0 x = 0: (And hence 0 = 0 x:) Proof. By (D) and (A4), we have 0 x + 0 x = (0 + 0) x = 0 x: Since 0 x + 0 x = 0 + 0 x; Proposition 1.14(a) [i.e., x + y = x + z ) y = z] implies 0 x = 0. The number 1 is de…ned to be the additive inverse of 1. 2 Lemma. ( 1) = ( 1) ( 1) = 1: Proof. By de…nition, 0 = 1 + 1: By Proposition 1.16(a), Axioms (A5), (D), and (M4), we have 0 = 0 ( 1) = ( 1 + 1) ( 1) = ( 1) ( 1) + 1 ( 1) = ( 1) ( 1) + ( 1) : Now, since 1 + ( 1) = 0 = ( 1) ( 1) + ( 1) ; we conclude from Proposition 1.14(a) that 1 = ( 1) ( 1). Lemma. For any x 2 F we have x = ( 1) x: Proof. We have 0 = 0 x = ( 1 + 1) x = ( 1) x + 1 x = ( 1) x + x: Since x + x = 0 = ( 1) x + x; we conclude from Proposition 1.14(a) that x = ( 1) x. Lemma.1 For any x; y 2 F we have ( x) ( y) = xy: Proof. We compute 2 ( x) ( y) = ( 1) x ( 1) y = ( 1) xy = xy 2 since ( 1) = 1: The real …eld Theorem 1.20(a). If x; y 2 R and x > 0, then there is a positive integer n such that nx > y: Proof. See p. 9 of the book. Proposition. For any real number x there exists a greatest integer n0 with n0 x. 1I used this statement without proof in class. 1 Z. Proof. De…ne the set S = fn 2 Z : n xg, which we consider as a subset of Claim. S is nonempty and bounded from above. Proof of claim. (1) By Theorem 1.20(a), since 1 > 0 and x 2 R, there exists a positive integer n such that n = n 1 > x. That is, n < x; which says that n 2 S. So S is nonempty. (2). By Theorem 1.20(a) again, since 1 > 0 and x 2 R, there exists a positive integer m such that m = m 1 > x. Then, for each p 2 S we have p x < m: Thus m is an upper bound for S: This completes the proof of the claim. Now recall from Math 109 the following fact which is equivalent to the Principle of Mathematical Induction. Least Natural Number Principle. Each set of positive integers contains a least element. From this it can easily be proved that: ‘Greatest Negative Integer Principle.’ Each set of negative integers contains a greatest element. We can extend this to: Statement. Each set of integers which is bounded above contains a greatest element. The statement now implies that the set S has a greatest element, which we call n0 : By the de…nition of S, this means that n0 is the greatest integer with n0 x. Constructing the reals and Dedekind cuts Given r 2 Q, de…ne R = fp 2 Q : p < rg = ( 1; r). We say that a subset S of Q is a Dedekind cut if (1) If x 2 S and y 2 Q satisfy y < x, then y 2 S. (2) If x 2 S, then there exists z 2 S with z > x. Exercise: Determine where the hidden universal quanti…es are in the above conditions. Clearly if r 2 Q, then ( 1; r) is a Dedekind p cut. Here the Dedekind cut corresponding to 2. De…ne S Q the union of all nonpositive rationals and the positive rationals r with r2 < 2. What we showed in class on Monday Jan 5 is that S is a Dedekind cut. We also show that there is no rational number r0 with r02 = 2. Now, for each set of the form ( 1; r), where r 2 Q, the number r is the least upper bound for ( 1; r) : Lemma. S does not have a least upper bound in Q. Hence, S cannot be of the form ( 1; r), where r 2 Q. Proof. Suppose s 2 Q is an upper bound for S. Then s 2 = S since S has no greatest element. This implies that s2 > 2. But then we can …nd q 2 Q with q < s and q 2 > 2 (see p. 2 of the book). Thus s cannot be the least upper bound. 2