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Math 309: Axioms and Definitions
We will assume the existence of the following sets of numbers. Notation of the form a ∈ S
means “a is a member of the set S.”
N = {1, 2, 3, ...} is the set of natural numbers
Z = {..., −2, −1, 0, 1, 2, ...} is the set of integers
na
o
Q=
| a, b ∈ Z and b 6= 0 is the set of rational numbers
b
R is the set of real numbers
The following axioms hold for the set of real numbers.
(1) Closure: a + b and ab are real numbers for all a, b ∈ R
(2) Commutative laws: a + b = b + a and ab = ba for all a, b ∈ R
(3) Associative laws: a + (b + c) = (a + b) + c and a(bc) = (ab)c for all a, b, c ∈ R
(4) Distributive law : a(b + c) = ab + ac for all a, b, c ∈ R
(5) Additive identity: a + 0 = a for all a ∈ R
(6) Additive inverse: for all a ∈ R there exists b ∈ R such that a + b = 0
(7) Multiplicative identity: a · 1 = a for all a ∈ R
(8) Multiplicative inverse: for all a ∈ R with a 6= 0 there exists c ∈ R such that ac = 1
(9) Multiplication by zero: a · 0 = 0 for all a ∈ R
(10) Zero product: for all a, b ∈ R, if ab = 0 then a = 0 or b = 0
(11) Cancellation of addition: for all a, b, c ∈ R, if a + b = a + c then b = c
(12) Cancellation of multiplication: for all a, b, c ∈ R with a 6= 0, if ab = ac, then b = c
(13) Trichotomy law : for all a ∈ R, exactly one of the following statements is true:
(i) a < 0
(ii) a = 0
(iii) a > 0
(14) Properties of inequality: for all a, b, c ∈ R,
(i) If a < b, then a + c < b + c
(ii) If a < b and c > 0, then ac < bc
(iii) If a < b and c < 0, then ac > bc
(iv) If a < b and b < c, then a < c
Note that each of the axioms on the previous page also holds if R is replaced by Q, and
all but (8) hold if R is replaced by Z. To distinguish R from Q, an additional axiom called
the Completeness Axiom is necessary. This is a topic for Math 312; we will not need it for
our work.
Definition 1. An integer n is even if and only if n = 2k for some integer k.
Definition 2. An integer n is odd if and only if n = 2k + 1 for some integer k.
Definition 3. An irrational number is a real number that is not rational. That is, a is
irrational if and only if a ∈ R and a 6∈ Q.