Lecture 4: The Characterization of the Real Numbers

18.100A: Complete Lecture Notes Lecture 4: The Characterization of the Real Numbers Question 1. Last time we stated that Q was an example of a field, but what is a field? Definition 2 (Field) A set F is a field if it has two operations: addition (+) and multiplication (·) with the following properties. A1) If x, y ∈ F then x + y ∈ F . A2) (Commutativity) ∀x, y ∈ F , x + y = y + x. A3) (Associativity) ∀x, y, z ∈ F , (x + y) + z = x + (y + z). A4) ∃ an element 0 ∈ F such that 0 + x = x = x + 0. A5) ∀x ∈ F , ∃y ∈ F such that x + y = 0. We denote y = −x. M1) If x, y ∈ F , then x · y ∈ F . M2) (Commutativity) ∀x, y ∈ F , x · y = y · x. M3) (Associativity) ∀x, y, z ∈ F , (x · y) · z = x · (y · z). M4) ∃1 ∈ F such that 1 · x = x = x · 1 for all x ∈ F . M5) ∀x ∈ F \ {0}, ∃x−1 such that x · x−1 = 1. D) (Distributativity) ∀x, y, z ∈ F , (x + y)z = xz + yz. These may seem like trivial properties, but consider the following non-example: Z. Z fails M5)– multiplicative inverses do not exist in the integers. Example 3 Here are two examples of fields: 1. Z2 = {0, 1} where 1 + 1 = 0. 2. Z3 = {0, 1, 2} with c := a + b (mod 3). In other words, 2+1=3=0 (mod 3) and 2 · 2 = 4 = 3 + 1 = 1 Simple properties follow from the properties of a field! Theorem 4 If x ∈ F where F is a field, 0x = 0. 1 (mod 3). Proof : If x ∈ F , then 0 = 0 · x − 0 · x = (0 + 0) · x − 0 · x = 0 · x + 0 · x − 0 · x = 0 · x. Definition 5 (Ordered field) A field F is an ordered field if F is also an ordered set with ordering < and i) ∀x, y, z ∈ F , x < y =⇒ x + z < y + z. ii) If x > 0 and y > 0 then xy > 0. If x > 0 we say x is positive, and if x ≥ 0 we say x is non-negative. Example 6 Q is an ordered field. A non-example would be Z2 . For instance, consider 0 < 1. If 0 < 1, then 1 + 0 < 1 + 1 = 0 =⇒ 1 < 0 which is a contradiction. If 1 < 0, then 0 = 1 + 1 < 0 + 1 =⇒ 0 < 1 which is a contradiction. Hence, Z2 is not an ordered field. Using the definition of an ordered field, one can easily prove all of the usual facts about inequalities. Theorem 7 If x > 0, then −x < 0 (and vice versa). Proof : If x ∈ F and x > 0, then by i), −x + x > −x =⇒ 0 > −x. One can see Proposition 1.1.8 [L] for a list of other simple inequality facts. Theorem 8 Let x, y ∈ F where F is an ordered field. If x > 0 and y < 0 or x < 0 and y > 0, then xy < 0. Proof : Suppose x > 0 and y < 0. Then, x > 0 and −y > 0. Hence, −xy = x(−y) > 0. Thus, xy < 0. If x < 0 and y > 0, then −x > 0 and y > 0 =⇒ −xy = (−x)y > 0 =⇒ xy < 0. Question 9. Is there a greatest lower bound property? For an ordered field F , if F has the least upper bound property then F has a greatest lower bound property. Theorem 10 Let F be an ordered field with the least upper bound property. If A ⊂ F is nonempty and bounded below, then inf A exists in F . Proof : Suppose A ⊂ F is nonempty and bounded below, i.e. ∃a ∈ F such that ∀x ∈ A, a ≤ x. Let B = {−x | x ∈ A}. Then, ∀x ∈ A, −x ≤ −a =⇒ −a is an upper bound for B. Since F has the least upper bound 2 property, ∃c ∈ F such that c = sup B. Then, ∀x ∈ A, −x ≤ c =⇒ ∀x ∈ A, −c ≤ x. Hence, −c is a lower bound for A. We have also shown that if a is a lower bound for A, then −a is an upper bound for B. Therefore, c ≤ −a since c = sup B =⇒ a ≤ −c. Hence, −c is the greatest lower bound for A. Real Numbers Theorem 11 There exists a "unique" ordered field, labeled R, such that Q ⊂ R and R has the least upper bound property. One can construct R using Dedekind cuts or as equivalence classes of Cauchy sequences. (We will define Cauchy sequences later in the course.) Theorem 12 ∃!r ∈ R such that r > 0 and r2 = 2. In other words, √ √ 2 ∈ R but 2 ∈ / Q. Proof : Let Ẽ = {x ∈ R | x > 0 and x2 < 2}. Then, since Ẽ is bounded above (by 2 for instance), we have that r := sup Ẽ exists in R. Then, one can show that r > 0 and r2 = 2. This is left as an exercise. We now prove uniqueness. Suppose that there is a r̃ > 0 with r̃2 = 2. Then, since (r + r̃) > 0, 0 = r2 − r̃2 = (r + r̃)(r − r̃) =⇒ r − r̃ = 0 =⇒ r = r̃. Remark 13. In Assignment 2 Exercise 7, you will show that 3 √ 3 2 ∈ R.