To actually construct the rationals
from the integers Z, let S = {(a, b): a, b ∈ Z and b ≠ 0}. Define an equivalence relation “~” on S by (a, b) ~ (c, d) iff a d = b c. Then define the set
of rational numbers to be the set of equivalence classes corresponding to ~. Let the equivalence class determined by the ordered pair (a, b) be denoted by [a/b]. Then [a/b] is what we usually think of as the fraction a/b. For a, b, c, d ∈ Z with b ≠ 0 and d ≠ 0, we define addition and multiplication in
by
We say that [a/b] is positive if a b ∈ N. Since a, b ∈ Z with b ≠ 0, this is equivalent to requiring a b > 0. The set of positive rationals is denoted by
+, and we define an order “<>
by
x <>∈
+.
(a) Verify that ~ is an equivalence relation on S
(b) Show that addition and multiplication are well-defined. That is, suppose
[a/b] = [ p/q] and [c/d ] = [r/s]. Show that
[(ad + bc)/bd ] = [( ps + qr)/qs] and [ac/bd] = [ pr/qs].
(c) For any b ∈ Z \ {0}, show that [0/b] = [0/1] and [b/b] = [1/1].
(c) For any b ∈ Z \ {0}, show that [0/b] = [0/1] and [b/b] = [1/1].
(d) For any a, b ∈
with b ≠ 0, show that [a/b] + [0/1] = [a/b] and [a/b]⋅[1/1] = [a/b]. Thus [0/1] corresponds to zero and [1/1] corresponds to 1.
(e) For any a, b ∈
with b ≠ 0, show that [a/b] + [(− a)/b] = [0/1] and [a/b]⋅[b/a] = [1/1].
(f) Verify that the set
with addition, multiplication, and order as given above satisfies the axioms of an ordered field.