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   (mod 3).
   Simple properties follow from the properties of a field!
  Theorem 4
  If x ∈ F where F is a field, 0x = 0.
                                                             1
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
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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.
                                                              √
                                                              3
Remark 13. In Assignment 2 Exercise 7, you will show that       2 ∈ R.
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