Download 7. Rationals

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VAN WYK’S MATH 315
7. The Rationals (10/19/10)
In Section 5.2, we built the integers as equivalence classes of an equivalence relation on
ordered pairs of natural numbers; the equivalence relation identified ordered pairs with a
common difference between their first and second components. In this section, we build the
rationals as equivalence classes of an equivalence relations on ordered pairs of integers; the
equivalence relation we will use identifies ordered pairs with a common quotient of their first
and second components.
Since we have already built the integers, we no longer need to consider individual integers
as subsets of N × N. We have all the usual properties of Z (except for order) at our disposal.
Definition 7.1. Let (a, b), (c, d) ∈ Z × (Z − {0}). Define (a, b) ∼ (c, d) provided ad = bc.
Notice the similarity between this definition and Definition 5.1 on N × N; both are of the
form (a, b) ∼ (c, d) provided a ∗ d = b ∗ c for a binary operation ∗.
Theorem 7.1. The relation ∼ in Definition 7.1 is an equivalence relation on Z × (Z − {0}).
For example, (5, 3) ∼ (10, 6) and (2, 5) ∼ (6, 15). (You should be thinking ab when you see
(a, b); notice the second component is never 0.) In the first example, the quotient between
each first component and second component is 53 , while in the second example, the quotient
between each first component and second component is 25 . This is how we construct the
rationals.
Definition 7.2. The set of equivalence classes in Definition 7.1 is the set of rationals,
denoted Q.
Consider the equivalence class
[(5, 3)] = {. . . , (−15, −9), (−10, −6), (−5, −3), (5, 3), (10, 6), (15, 9) . . . }.
We identify this with the rational number 53 . Similarly, we identify the equivalence class
[(2, 5)] with the rational number 52 . We will write 0 for [(0, 1)] and 1 for [(1, 1)]; notice 1 &= 0.
See Problem 2.
As with the integers, we want to define the usual operations of addition and multiplication
on the rationals, but since we will be trying to multiply entire equivalence classes using
representative elements, we must show these operations are well-defined before we can do
anything else.
THE REAL NUMBER SYSTEM
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How should we try to add? Suppose we have the following identifications of rationals:
x ↔ (a, b) and y ↔ (c, d). Then when we compute x + y, we get
a c
+
b d
ad + bc
=
bd
! (ad + bc, bd).
(a, b) + (c, d) !
How should we try to multiply? Using the same identifications as above, when we compute
x · y, we get
a c
·
b d
ac
=
bd
! (ac, bd).
(a, b) · (c, d) !
We must show both of these binary operations are independent of choices of representatives
from the equivalence classes.
Theorem 7.2. Assume (a1 , b1 ) ∼ (a2 , b2 ) and (c1 , d1 ) ∼ (c2 , d2 ), where each ordered pair lies
in Z × (Z − {0}). Then
(1) (a1 d1 + b1 c1 , b1 d1 ) ∼ (a2 d2 + b2 c2 , b2 d2 ), and
(2) (a1 c1 , b1 d1 ) ∼ (a2 c2 , b2 d2 ).
Proof. We’ll prove Theorem 7.2 (2) and leave part 1 as an exercise.
We know that b1 d1 &= 0, b2 d2 &= 0, a1 b2 = b1 a2 , and c1 d2 = d1 c2 . So
a1 c1 b2 d2 = (a1 b2 )(c1 d2 ) = (b1 a2 )(d1 c2 ) = b1 d1 a2 c2 , which is what we need.
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So, we can now define addition of rationals as [(a, b)] + [(c, d)] = [(ad + bc, bd)] and
multiplication of rationals as [(a, b)] · [(c, d)] = [(ac, bd)]. We will use these definitions to
prove the familiar properties of addition and multiplication hold for Q. And what are these
familiar properties? They are contained in Definition 6.1, the definition of a ring.
Theorem 7.3. The set of rationals, Q, under the operations of addition and multiplication
above, form a commutative ring with unity. The element 0 serves as the additive identity,
and the element 1 serves as the multiplicative identity.
Proof. Let q, r, s ∈ Q, where (a, b) ∈ q, (c, d) ∈ r, and (e, f ) ∈ s.
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VAN WYK’S MATH 315
(1) + is associative.
(q + r) + s ! ((a, b) + (c, d)) + (e, f )
= (ad + bc, bd) + (e, f )
= ((ad + bc)f + bde, bdf )
= (adf + bcf + bde, bdf )
= (adf + b(cf + de), bdf )
= (a, b) + (cf + de, df )
= (a, b) + ((c, d) + (e, f ))
! q + (r + s).
(2) + has an identity element. Show 0 = [(0, 1)] serves as the additive identity.
(3) Every element has an inverse under +. Show −q = [(−a, b)] serves as the
additive inverse of q.
(4) + is commutative.
q + r ! (a, b) + (c, d)
= (ad + bc, bd)
= (cb + da, db)
= (c, d) + (a, b)
! r + q.
(5) · is associative. Exercise.
(6) The left and right distributive laws hold. We will show the left distributive law
holds; once (7) is proven, the right distributive law will follow.
q(r + s) ! (a, b)((c, d) + (e, f ))
= (a, b)(cf + de, df )
= (acf + ade, bdf )
∼ (b(acf + ade), b(bdf ))
= (acbf + bdae, bdbf )
= (ac, bd) + (ae, bf )
= (a, b)(c, d) + (a, b)(e, f )
! qr + qs.
Since (a, b)((c, d) + (e, f )) ∼ (a, b)(c, d) + (a, b)(e, f ), it follows that q(r + s) = qr + qs.
Notice it is not necessary to have (a, b)((c, d) + (e, f )) = (a, b)(c, d) + (a, b)(e, f ).
THE REAL NUMBER SYSTEM
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(7) · is commutative. Exercise.
(8) · has an identity. Show 1 = [(1, 1)] serves as the multiplicative identity.
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The usual notation for rationals, ab for [(a, b)], gives the usual formulas for addition and
multiplication of rationals:
a c
ac
• · =
b d
bd
ad + bc
a c
• + =
b d
bd
The integers under the usual operations of addition and multiplication, and the rationals
under the same operations, both form a commutative ring with 1. But there is a property
that the rationals possess that the integers do not: solutions to equations of the form ax = 1
for every element a. The integers only have such a solution for a = ±1, while the rationals
have a solution for all nonzero a:
Theorem 7.4. Every nonzero rational number has a multiplicative inverse in Q.
Proof. Let q ∈ Q and let (a, b) ∈ q. Since q is a nonzero rational, a is a nonzero integer (see
Problem 2). So (b, a) ∈ Z × (Z − {0}). Since
(a, b)(b, a) = (ab, ba)
∼ (1, 1)
! 1,
[(b, a)] is the multiplicative inverse of [(a, b)].
In terms of the usual notation,
b
a
"
is the multiplicative inverse of ab ; the former exists exactly
when a is nonzero.
Commutative rings with this property have a special name.
Definition 7.3. A field is a commutative ring with 1 in which every nonzero element has
a multiplicative inverse.
Recall an integral domain is a commutative ring with 1 that has no zero divisors. An
obvious question to ask is: what is the relationship between integral domains and fields?
Theorem 7.5. Every field is an integral domain.
Proof. Since both fields and integral domains are commutative rings with 1, we only need to
show that the property of every nonzero element having a multiplicative inverse guarantees
there are no zero divisors.
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VAN WYK’S MATH 315
Let F be a field (under the operations + and ·), let a, b ∈ F , and assume ab = 0 where
a, b &= 0. Since a &= 0, a has a multiplicative inverse, a−1 (this is standard notation for
multiplicative inverses in an algebra class). So a−1 (ab) = a−1 · 0, so b = 0. This contradicts
the assumption that b &= 0. Thus F has no zero divisors, and is an integral domain.
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At this point, you should be curious about the converse of Theorem 7.5. See Problem 9.
Finally, Problem 8 in Section 5.2 identifies each natural number n with the integer
[(n + 1, 1)]. In a similar way, the set of integers can be identified with a subset of Q; see
Problem 10. In particular, the additive inverse 0 of Z is identified with the additive inverse
0 of Q, and 1 in the former is identified with 1 in the latter.
Problems. 1. Prove Theorem 7.1.
2. Prove (a, b) ∈ 0 iff a = 0, (a, b) ∈ 1 iff a = b, and 1 &= 0.
3. Prove Theorem 7.2 (1).
4. Prove Theorem 7.3 (2).
5. Prove Theorem 7.3 (3).
6. Prove Theorem 7.3 (5).
7. Prove Theorem 7.3 (7).
8. Prove Theorem 7.3 (8).
9. Show the converse of Theorem 7.5 is false by considering Z.
10. Define ψ : Z → Q by ψ(a) = [(a, 1)] (i.e., a )→ a1 ). Prove ψ is an injection. (Note: It is
also true that for all a, b ∈ Z, ψ(a+b) = ψ(a)+ψ(b) and ψ(ab) = ψ(a)ψ(b), two properties
that might look familiar to some of you.)