Math 316, Intro to Analysis The order of the real numbers. The field axioms are not enough to give R, as an extra credit problem will show. Definition 1. An ordered field F is a field together with a nonempty subset P F (called the set of positive elements) such that (1) P is closed under addition: For all a, b P (2) P is closed under multiplication: For all a, b P (3) The trichotomy law: For each x F exactly one of the following holds: (a) (b) (c) Proposition 2 (1 is positive.). Let F be an ordered field and 1 be the multiplicative identity, then 1 P Proof. Suppose for the sake of contradiction that 1 / P. According to the trichotomy law then either (1) 1 = 0 or (2) 1 P. Since P, we let p P. In case (1) 1 = 0. Then using and from the previous lecture, p = 1 p = =, contradicting the trichotomy law because. In case (2) 1 P. Then using from Friday, p = p, Since 1 P we see by closure under multiplication that contradicting the trichotomy law because. The algebraic and order properties of R can be summarized as: Axiom. R is an ordered field. This doesn t look like an ordering does it? involving the symbol <. You might have been expecting something Definition 3. Let F be an ordered field and P be its set of positive elements. If a, b F then we say that a < b if, a > b if a b if, and a b if 1
2 Let s begin by proving some properties which inequality should have. Theorem 4 (Properties of <). Let F be an ordered field and a, b, c F. Then (1) (Additivity) If a < b then a + c b + c. (2) (Transitivity) If a < b and b < c then a c. (3) (Multiplicativity) If a < b and 0 < c then a c b c. (4) (Multiplicativity II) If a < b and c < 0 then a c b c. (5) If a 0 then a 2 := a a 0. 1 We (you) will prove (1), (2), and (3) Proof. We ll start with the proof of the first claim: That for all a, b, c F if a < b then a + c b + c. Consider any a, b, c F. Assume that that.. According to Definition 3, this means Thus, and by Definition 3,. This completes the proof of the first result. Next we prove the second claim. Suppose that a, b, c F, and. According to Definition 3, this means that and. Thus we see that and by Definition 3, we conclude that. 1 The notation a 2 := a a is telling you that a 2 is defined to be a a.
3 Now we will prove the third claim. Consider any a, b, c F. Assume and. According to Definition 3, this means that and. Thus we see that and by Definition 3, we conclude that. Our next goal is a discussion of the action of raising real numbers (or any element of an ordered field) to a natural number. Definition 5. Let F be a field. For any x F and any natural number n N we define x n recursively by x 1 = x and x n = x x n 1 if n > 1. 2 How should order behave under exponentiation? Theorem 6. Let F be an ordered field (1) Let x F and n be a natural number. If x > 1 then 1 x n (2) Let x F and n be natural number. If 0 < x < 1 then 1 x n (3) Let x = 1 and n be a natural number. Then 1 x n In order to prove this we will need to recall the proof technique called mathematical induction. 3 Proof technique. Let P (n) be a statement which makes sense for every natural number n N. (For example P (n) might be n is even. ) If you can prove that (Base Case) P (1) is true and (Inductive step) For all n N P (n) implies P (n + 1) then you have proven that P (n) is true for all n N. This proof technique cannot be used to prove something for all n R. Let s prove claims (3) and (1) of Theorem 6: Proof of claim 3. Let F be a field and x = 1. Base Case We must show that x 1 1. 2 This is a more rigorous way of saying that x n = x n times... x 3 Whenever you see a recursive definition, induction might be worth considering.
4 Inductive step Consider any n N. Assume that x n 1. We must show that x n+1 1. Thus, by the principle of mathematics induction, If x = 1 then x n 1 for all n N. That was the easiest claim, though. Let s see how well the proof generalizes to the first claim Proof of claim 1. Let F be an ordered field and x > 1. Base Case We must show that x 1 1. Inductive step Consider any n N. Assume that x n 1. We must show that x n+1 1. Thus, by the principle of mathematics induction, If x > 1 then x n 1 for all n N. If there is demand then we will go through the proof of claim 2. Either way, here is an outline: Proof of claim 2. Let F be an ordered field and 0 < x < 1. Base Case We must show that 0 < x 1 < 1. Inductive step Consider any n N. Assume that 0 < x n < 1. We must show that 0 < x n+1 < 1.
Thus, by the principle of mathematics induction, If 0 < x < 1 then 0 < x n < 1 for all n N. If we have time, then at this point I will encourage people to begin on today s homework. Next time we will pick up on the absolute value. Absolute value. Continuing from last time, let F be an ordered field. The following function should be familiar. Definition 7 (Absolute Value). The absolute value function on the ordered field F, is defined by if x 0 x = if x < 0 Let s begin by proving some basic (and hopefully familiar) properties of the absolute value. Theorem 8. Let F be an ordered field. let a, b F. Then (1) a 0 (2) a a (3) a a a (4) ab a b (5) If b 0 (b 1 ) ( b ) 1 (6) a b if and only if b a b. (7) The Triangle inequality. a + b a + b (8) a b a b The Triangle inequality is by far the most important (and hardest to prove) of these. We ll talk though the proof of (3) and (6) because we ll need them in the proof of the triangle inequality (7). Proof of claim 3. Let F be an ordered field. Let a F. There are three cases: (0) a = 0 or (1) a > 0 or (2) a < 0. In case (0), a = 0, so that a = and a =. Thus, the inequality we need to check is, which is a truism. 5 In case (1) a > 0 so that a =, and the claimed inequality amounts to the inequalities and. One of these (the 1 st / 2 nd ) is a truism since a = a. To see the other inequality, notice that since a > 0, a 0, by the trichotomy law. By the transitivity law, then a a, completing the proof in case (1). In case (2) a < 0 so that a =, and the claimed inequality amounts to the inequalities and. One of these (the 1 st / 2 nd ) is a
6 truism since a = a. To see the other inequality, notice that since a 0, a the transitivity law, then a a, completing the proof in case (1). 0. By Proof of claim 6. Let F be an ordered field and consider any a, b F. ( ) Assume that a b. We must show that b < a and a < b. Since a 0 and b a we have that b 0. Again we will make a case-wise attack. Either (1) a 0 or (2) a < 0. In case (1), a = so that the assumption a b becomes, which proves one of the needed inequalities. Since we have assumed a 0, a 0. Since b 0, it follows that b 0 a and by transitivity completing the proof in this case. In case (2), a = so that the assumption a b becomes. Multiplying both sides by 1 and using Theorem 4, claim (4) we see. This proves one of the desired inequlities. Since b 0, it follows that b 0 a and by transitivity completing the proof in this case. Thus, if a b then b a b. This proves one of the implications. ( ) Suppose that b a and a b. We must show that a < b. Again we make a case-by-case proof. Either (1) a 0 or (2) a < 0. In the first case a = and the inequality to be proven becomes which is precisely the (1 st or 2 nd ) of the assumed inequalities. In the second case a = and the inequality to be proven becomes which is precisely the (1 st or 2 nd ) of the assumed inequalities. Proof of the triangle inequality. Let F be an ordered field and a, b F. We wish to show that a + b a + b. According to claim (6) this is equivalent to proving that or by factoring out the negative sign that Now using claim (3) a a a and b b b. Adding these inequalities together and using the additivity property of Theorem 4 we see that, which we have already seen to be equivalent to the claimed inequality.