The rational numbers
The equation 2 · x = 1 has no solution within the integers. (Why not?) One therefore
extends the number system ZZ one more time to allow for “fractions”. The resulting
number system is that of the rational numbers Q.
I This note shows how this is done.
Theorem 1
Consider the relation R on the set ZZ × (ZZ \ {0}) = {(p, q) | p, q ∈ ZZ and q 6= 0}
defined by (p1 , q1 )R(p2 , q2 ) if and only if p1 · q2 = p2 · q1 . Then R is an equivalence
relation.
Exercises
1. Prove Theorem 1.
We denote the set of equivalence classes of R by Q.
I The equivalence class [(p, q)] of
(p, q) will be denoted by p/q and called a rational number.
Be careful not to interpret the notation p/q as a division of integers until you have
completed the very last exercise of this note. At this point, the division of integers
does not exist. That is exactly why we introduce the rational numbers.
For example, 2/3 = [(2, 3)] = [(−10, −15)] = −10/ − 15.
2. Why do we not allow for rational numbers of the form p/0 ?
[Hint: examine your proof of Theorem 1 to answer this question. Show that the
relation R as defined above is not an equivalence relation on the set ZZ × ZZ.]
Definition
On Q
I we define
(1) [(p1 , q1 )] + [(p2 , q2 )] = [(p1 · q2 + p2 · q1 , q1 · q2 )]
(2) [(p1 , q1 )] · [(p2 , q2 )] = [(p1 · p2 , q1 · q2 )]
(
(3) [(p1 , q1 )] < [(p2 , q2 )] if and only if
(4) −[(p, q)] = [(−p, q)]
(5) [(p, q)]−1 = [(q, p)] if p 6= 0
p 1 · q2 < p 2 · q1
p 2 · q1 < p 1 · q2
in case
in case
0 < q 1 · q2
q1 · q2 < 0
Exercises
3. Show that addition, multiplication, order, negation, and taking reciprocals of
rational numbers is well-defined. That is, show that these “definitions” do not
depend on the particular choice of representatives.
As was the case for the integers, there is a list of basic properties of rational numbers
from which all other properties can be deduced. Here is that list:
[Keep in mind that every rational number a ∈ Q
I is to be represented by an equivalence
class a = p/q = [(p, q)] with p, q ∈ ZZ and q 6= 0.]
Exercises
4. Let a, b, c ∈ Q.
I Show that
(i) a + b = b + a
(ii) (a + b) + c = a + (b + c)
(iii) a · b = b · a
(iv) (a · b) · c = a · (b · c)
(v) (a + b) · c = (a · c) + (b · c)
5. a + (0/1) = a for all a ∈ Q.
I
6. a + (−a) = 0/1 for all a ∈ Q.
I
7. a · (1/1) = a for all a ∈ Q.
I
8. a · a−1 = 1/1 for all a ∈ Q
I with a 6= 0/1.
9. For all a, b ∈ Q,
I a < b if and only if 0/1 < b − a.
Here, we write b − a for b + (−a) as we did before.
10. For each a ∈ Q
I exactly one of the following is true: either 0/1 < a, or 0/1 < −a,
or 0/1 = a.
11. If a, b ∈ Q
I with 0/1 < a and 0/1 < b, then 0/1 < a + b and 0/1 < a · b.
If you tried to solve Exercises 11 through 18 on the integers without directly referring
to equivalence classes, then you already have a proof of most of the following exercises.
Exercises
12. a · (0/1) = 0/1 for all a ∈ Q.
I
13. −(−a) = a for all a ∈ Q.
I
14. (a−1 )−1 = a for all a ∈ Q
I with a 6= 0/1.
15. (−1/1) · a = −a for all a ∈ Q.
I
16. If a ∈ Q
I with a 6= 0/1, then 0/1 < a · a.
17. If a, b, c ∈ Q
I with a < b, then
(i) a + c < b + c
(ii) a · c < b · c, provided 0/1 < c
(iii) b · c < a · c, provided c < 0/1
18. ≤ is a linear order on Q
I (where a ≤ b stands for “a < b or a = b”).
19. If a, b, c ∈ Q
I with a + c = b + c, then a = b.
20. If a, b, c ∈ Q
I with a · c = b · c and c 6= 0/1, then a = b.
Just like the natural numbers IN can be regarded as a subset of the integers ZZ, the
integers ZZ can be regarded as a subset of the rational numbers Q,
I namely the set
{p/1 | p ∈ ZZ}. This is a consequence of the following theorem. We will therefore no
longer distinguish between the equivalence class p/1 = [(p, 1)] (which is a set of pairs
of integers) and the integer p.
Theorem 2
The function f : ZZ → Q
I given by f (p) = p/1 has the following properties:
(i) If p1 , p2 ∈ ZZ with p1 < p2 , then f (p1 ) < f (p2 ). In particular, f is injective.
(ii) f (p1 + p2 ) = f (p1 ) + f (p2 ) for all p1 , p2 ∈ Q.
I
(iii) f (p1 · p2 ) = f (p1 ) · f (p2 ) for all p1 , p2 ∈ Q.
I
In short, the function f is a bijection between the set ZZ of integers and the subset
{p/1 | p ∈ ZZ} of Q
I which preserves all relevant properties.
Exercises
21. Prove Theorem 2.
22. Writing a ÷ b for a · (b−1 ) when a, b ∈ Q
I and b 6= 0/1, show that we have
p/q = (p/1) ÷ (q/1) for all p, q ∈ ZZ with q 6= 0.