Math 316, Intro to Analysis the Order of the Real Numbers. the Field

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Math 316, Intro to Analysis The order of the real numbers. The field axioms are not enough to completely describe R. An algebraist will know that Q or Z/2 are also fields. More properties are needed to completely describe R. 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 a ∈ 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 is nonempty by assumption, we let p ∈ P . In case (1) 1 = 0. Then using The axiom of the multiplicative identity and from the previous lecture, p = 1 · p = = . This contradicts the trichotomy law because

In case (2) −1 ∈ P . Then using a result about ﬁelds ( from

last time) −p = · p. Since −1 ∈ P we see by closure under multiplication that

. This contradicts 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? You might have been expecting something involving the symbol “<.” 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

1 2

Let’s begin by proving some properties which inequality should have. Indeed, these properties are enough to completely understand <. (Challenge problem.) 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 6= 0 then a2 := a · a 0. (The notation a2 := a · a is telling you that a2 is deﬁned to be a · a.) We/You will prove (1), (2), and (3). The Proofs of (4) and (5) are left for you to work out.

Proof. We’ll start with the proof of the ﬁrst claim: for all a, b, c ∈ F if a < b then a + c b + c.

Consider any a, b, c ∈ F . Assume that . According to Deﬁnition 3, this means that .

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 . 3

Thus we see that and by Deﬁnition 3, we conclude that .

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. n Definition 5. Let F be a field. For any x ∈ F and any natural number n ∈ N we define x recursively by the rule x1 = x and xn = x · xn−1 if n > 1. (This is a more rigorous way of n times saying that xn = x · ... · x. Recurive definitions will be used throughout this class.) 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 xn (2) Let x ∈ F and n be natural number. If 0 < x < 1 then 1 xn (3) Let x = 1 and n be a natural number. Then 1 xn In order to prove this we will need to recall the proof technique called mathematical induction. (Whenever you see a recursive definition, induction might be worth considering.)

Proof technique. Let P (n) be a statement which makes sense for every natural number n ∈ N. (For example P (n) might be “3n − 1 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. Notice that this proof technique only works if you want to prove something for every n ∈ N. 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 x1 1. 4

n n+1 Inductive step Consider any n ∈ N. Assume x 1. We must show that x 1.

n Thus, by the principle of mathematics induction, If x = 1 then x 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 x1 1.

n n+1 Inductive step Consider any n ∈ N. Assume x 1. We must show that x 1.

n Thus, by the principle of mathematics induction, If x > 1 then x 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 < x1 < 1.

n Inductive step Consider any n ∈ N. Assume that 0 < x < 1. We must show that 0 < xn+1 < 1.

Thus, by the principle of mathematics induction, If 0 < x < 1 then 0 < xn < 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. 5

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 6= 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 . By the trichotomy law, there are three cases: (0) a = 0 or (1) a > 0 or (2) a < 0. In each case we must conclude that .

In case (0), a = 0, so that −|a| = , a = , and |a| = . Thus, the

inequality we need to check is ≤ ≤ , which is a truism.

In case (1) a > 0 so that |a| = , −|a| = , and the claimed inequality amounts to the inequalities and . One of these (the 1st / 2nd) is a truism since a = a. To see the other inequality, notice that since a > 0, −a 0. By transitivity, 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 1st / 2nd) is a truism since −a = −a. To see the other inequality, notice that since a ≤ 0, −a 0. By the transitivity law, then a − a, completing the proof in case (2). 6

Proof of claim 6. Let F be an ordered ﬁeld and consider any a, b ∈ F . We must prove that |a| ≤ b if and only if −b ≤ a ≤ b. (⇒) Assume that |a| ≤ b. We must show that −b ≤ a and a ≤ b. Since |a| ≥ 0 and b ≥ |a| transitivity implies b . 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. To see the second, notice that as b ≥ 0, −b 0. By assumption in this case 0 a. Transitivity now implies that , 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 inequalities. On the other hand, since b 0 a, we can

use transitivity to conclude that 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

st nd which is precisely the (1 or 2 )which one? of the assumed inequalities. In the second case |a| = and the inequality to be proven becomes .

Multiplying by −1 we see that this is equivalent to which is precisely the (1st or 2nd) of the assumed inequalities. Thus, in either case we conclude that |a| ≤ b. This completes the proof. Proof of the triangle inequality. Let F be an ordered ﬁeld and a, b ∈ F . We wish to show that |a + b| ≤ |a| + |b|. We start by using claim (3) to see that −|a| a |a| and −|b| b |b|. Adding these inequalities together and using the additivity property of Theorem 4 we see that . Factoring a negative sign out of the leftmost expression gives us . By Claim (6) this is equivalent to , which is exactly what we set out to prove.