Section 1.2 - Properties of the Real Numbers
1. The real number system, R, is a complete ordered field.
2. A field F is a nonempty set together with two operations, + and · (addition and multiplication), which satisfy the following:
A1 The operations + and · are binary operations, i.e. if a, b ∈ F , then a + b ∈ F and
a · b ∈ F.
A2 For a, b, c ∈ F , we have (a + b) + c = a + (b + c) and (a · b) · c = a · (b · c). (Associative)
A3 For a, b ∈ F , we have a + b = b + c and a · b = b · a. (Commutative)
A4 For a, b, c ∈ F , we have a · (b + c) = (a · b) + (a · c). (Distributive Property)
A5 There exist 0, 1 ∈ F such that 0 + a = a and 1 · a = a for all a ∈ F . (Identities)
A6 For all a ∈ F , there exists an element −a ∈ F for which a + (−a) = 0. (Additive
inverse)
A7 For each a ∈ F (a 6= 0), there exists an element a−1 for which a·a−1 = 1. (Multiplicative
inverse)
3. Theorem: Let F be a field. Then the additive and multiplicative identities are unique.
pf. Assume 0 6= 0̄ ⇒ 0 + 0̄ = 0 and 0̄ + 0 = 0̄. (Multiplicative identity is an exercise.)
4. Theorem: Let F be a field. Then the additive inverse and multiplicative inverse of an element
are unique.
pf. Let a ∈ F , a 6= 0. Suppose b and b̄ are multiplicative inverses of a. (Additive inverse is
an exercise.)
5. Theorem: Let F be a field. Then a · 0 = 0 for every a ∈ F .
pf. a · a = a · (a + 0) = a · a + a · 0. Add −(a · a) to both sides.
6. Theorem: Let F be a field and a, b ∈ F . Then
(a) a · (−b) = (−a) · b = −(a · b).
(b) −(−a) = a.
(c) (−a) · (−b) = a · b.
pf.
(a) We know 0 = a · 0 = a · b + a · (−b). Thus a · (−b) and −(a · b) are both additive inverses
of a · b. Works since additive inverse is unique. (Similar for other equality)
(b) −(−a) is additive inverse of (−a). Also a + (−a) = 0. (Additive inverse unique)
(c) 0 = (−a) · 0 = (−a) · (b + (−b)) = (−a) · b + (−a) · (−b). Thus (−a) · (−b is an additive
inverse of (−a) · b. So is a · b.
We will now write a · b = ab and a + (−b) = a − b.
7. Let F be a field. Then F is an ordered field if it satisfies the additional axiom:
A8 There is a nonempty subset P of F (called the positive subset) for which
i. If a, b ∈ P , then a + b ∈ P (additive closure)
ii. If a, b ∈ P , then ab ∈ P (multiplicative closure)
iii. For any a ∈ F , exactly one of the following holds: a ∈ P , −a ∈ P , a = 0 (law of
trichotomy).
For real number P is the set of positive numbers.
8. Let F be an ordered field; let P be the positive subset of F ; and let a, b ∈ F . We say a < b
(or b > a) if b − a ∈ P . We say a ≤ b if a < b or a = b.
9. This induces an ordering on the real numbers. If a, b ∈ F , then either b − a ∈ P (b > a),
−(b − a) = a − b ∈ P (b < a), or b − a = 0 (a = b) by the law of trichotomy.
10. Theorem: Let F be an ordered field. For a, b, c ∈ F the following hold:
(a) If a < b, then a + c < b + c.
(b) If a < b and b < c, then a < c. (< is transitive.)
(c) If a < b and c > 0, then ac < bc.
(d) If a < b and c < 0, then bc < ac.
(e) If a 6= 0, then a2 > 0.
pf. Uses Axiom A8.
(a) b − a ∈ P ⇒ b − a = b + c − a − c ∈ P ⇒ b + c > a + c.
(Note: We haven’t shown that −a − c = −(a + c); Exercise 2)
(b) (b − a) ∈ P , (c − b) ∈ P ⇒ (c − b) + (b − a) = c − a ∈ P ⇒ a < c.
(c) (b − a) ∈ P , c ∈ P ⇒ (bc − ac) ∈ P .
(d) Proof similar to (c).
(e) Either a > 0 ⇒ a ∈ P ⇒ a · a = a2 ∈ P or a < 0 ⇒ −a ∈ P ⇒ (−a) · (−a) = a2 ∈ P .
11. An interval of real numbers is a set A containing at least two numbers such that r, s ∈ A
with r < s and if t is a number such that r < t < s, then t ∈ A.
12. An interval is one of the following sets: R = (−∞, ∞), or
(i) (a, b) = {x| a < x < b}
(v) (−∞, a) = {x| x < a}
(ii) [a, b] = {x| a ≤ x ≤ b}
(vi) (−∞, a] = {x| x ≤ a}
(iii) [a, b) = {x| a ≤ x < b}
(vii) (a, ∞) = {x| x > a}
(iv) (a, b] = {x| a < x ≤ b}
(viii) [a, ∞) = {x| x ≥ a},
where a < b.
13. Define N, Z, Q, irrational numbers
14. We assume that xa+b = xa xb ; xa−b = xa /xb (a, b ∈ Z); and that laws of arithmetic are valid.
15. Theorem: Let x be a positive real number, and let n be a positive integer. Then there is a
unique positive number y such that y n = x.
√
16. Let x be a positive real number and n a positive integer. Then x1/n = n x is that unique
positive real number for which (x1/n )n = x. Thus, for p, q ∈ Z+ , xp/q = (x1/q )p .
17. Theorem: Let x be a positive real number, and let s1 and s2 be positive rational numbers
with s1 < s2 . Then
(a) xs1 < xs2 if x > 1.
(b) xs1 > xs2 if 0 < x < 1.
We can’t prove everything.
18. Theorem: Let x and y be positive real numbers with x < y, and let s be a positive rational
number. Then xs < y s .
Again, we can’t prove everything. √We can make similar conclusions for negative rational
numbers, but we must know what 2 3 means if we’re going to talk about f (x) = 2x .
19. Mathematical Induction (Induction Principle) Suppose that there is a statement P (n) for
which
(a) P (1) is true.
(b) For any positive integer k, if P (k) is true, then P (k + 1) is also true.
Then P (n) is true for every positive integer n.
20. Example: Show that for any positive integer n
n
X
j=1
j2 =
n(n + 1)(2n + 1)
.
6
21. Theorem: For positive integers m and j with j ≤ m
¶
¶ µ
µ
¶ µ
m
m
m+1
.
=
+
j
j
j−1
22. Theorem (Binomial Theorem): Let a and b be real numbers, and let m be positive integer.
Then
¶
m µ
X
m
m
(a + b) =
aj bm−j .
j
j=0
Don’t prove.
23. Theorem (Define absolute value): For a, b ∈ R the following hold:
(a) |a| ≥ 0 and |a| = 0 ⇔ a = 0.
(b) |a| = | − a|.
(c) −|a| ≤ a ≤ |a|.
(d) |ab| = |a| · |b|.
(e) 1/|b| = |1/b| if b 6= 0.
(f) |a/b| = |a|/|b| if b 6= 0.
(g) |a| < b ⇔ −b < a < b.
(h) |a + b| < |a| + |b| (triangle inequality).
(i) ||a| − |b|| ≤ |a − b|.
Prove part (h) using −|a| ≤ a ≤ |a| and −|b| ≤ a ≤ |b|, adding, and using part (g).
24. If X is a set and d : X × X → R is a function satisfying
(a) d(a, b) ≥ 0 and d(a, b) = 0 only if a = b,
(b) d(a, b) = d(b, a), and
(c) d(a, c) ≤ d(a, b) + d(b, c),
then d is called a metric on X.
Note that d : R × R → R defined by d(a, b) = |a − b| is a metric.
25. Example: Show that the function d is a metric where
d(x, y) =
|x − y|
.
1 + |x − y|