Math 361 Construction of Q

10 Construction of Q Most proofs are left as reading exercises.

Denition 10.1. Z0 = Zr 0 . { } Denition 10.2. Let be the binary relation dened on Z Z0 by a, b c, d i ad = cb. h i h i Theorem 10.3. is an equivalence relation on Z Z0. Proof. We just check that is transitive. So suppose that a, b c, d and c, d e, f . Then h i h i h i h i ad = cb (1) cf = ed (2)

Multiplying (1) by f and (2) by b, we obtain

adf = cbf (3) cfb = edb (4)

Hence adf = edb. Since d = 0, the Cancellation Law implies that af = eb. Hence a, b e, f . 6 h i h i Denition 10.4. The set Q of rational numbers is dened by

Q = Z Z0/ ie. Q is the set of –equivalence classes.

Notation For each a, b Z Z0, the corresponding equivalence class is denoted by [ a, b ]. h i ∈ h i Next we want to dene an addition opperation on Q. [Note that a c ad + cb + = . b d bd This suggests we make the following denition.]

Denition 10.5. We dene the binary operation +Q on Q by [ a, b ] + [ c, d ] = [ ad + cd, bd ]. h i Q h i h i Remark 10.6. Since b = 0 and d = 0 we have that bd = 0 and so ad + cb, bd Z Z0. 6 6 6 h i ∈

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Lemma 10.7. +Q is well-dened. Theorem 10.8. For all q, r, s Q, we have that ∈

q +Q r = r +Q q

q +Q (r +Q s) = (q +Q r) +Q s. Denition 10.9 (Identity element for + ). 0 = [ 0, 1 ]. Q Q h i Theorem 10.10.

(a) For all q Q, q + 0 = q. ∈ Q Q (b) For any q Q, there exists a unique r Q such that q + r = 0 . ∈ ∈ Q Q Proof. (a) Let q = [ a, b ]. Then h i q + 0 = [ a, b ] + [ 0, 1 ] Q Q h i Q h i = [ a 1 + 0 b, b 1 ] h i = [ a, b ] h i = q.

To show that there exists at least one such element, consider r = [ a, b ]. Then h i q + r = [ a, b ] + [ a, b ] Q h i Q h i = [ ab + ( a)b, b2 ] h i = [ 0, b2 ] h i Since 0 1 = 0 b2, we have 0, b2 = 0, 1 . Hence h i h i q + r = [ 0, b2 ] Q h i = [ 0, 1 ] h i = 0Q As before, simple algebra shows that there exists at most one such element.

Denition 10.11. For any q Q, q is the unique element of Q such that ∈ q + ( q) = 0 . Q Q Denition 10.12. We dene the binary operation on Q by Q q r = q + ( r). Q Q

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Next we want to dene a multiplication operation on Q. [Note that a c ac = . b d bd This suggests we make the following denition.]

Denition 10.13. We dene the binary operation on Q by Q [ a, b ] [ c, d ] = [ ac, bd ]. h i Q h i h i Remark 10.14. Since b = 0 and d = 0m wee have that bd = 0 and so ac, bd Z Z0. 6 6 6 h i ∈ Lemma 10.15. is well-dened. Z Theorem 10.16. For all q, r, s Q, we have that ∈ q r = r q Q Q (q r) s = q (r s) Q Q Q Q q (r + s) = (q r) + (q s) Q Q Q Q Q

Denition 10.17 (Identity element for ). Q 1 = [ 1, 1 ]. Q h i Theorem 10.18.

(a) For all q Q, q 1 = q. ∈ Q Q (b) For every 0 = q Q, there exists a unique r Q such that q r = 1 . Q 6 ∈ ∈ Q Q Proof. (a) Let q = [ a, b ]. Then h i q 1 = [ a, b ] [ 1, 1 ] Q Q h i Q h i = [ a 1, b 1 ] h i = [ a, b ] h i = 1Q

(b) Suppose that q = [ a, b ] = [ 0, 1 ]. Then a = 0 and so b, a Z Z0. Let r = [ b, a ]. Then h i 6 h i 6 h i ∈ h i

q r = [ a, b ] [ b, a ] Q h i Q h i = [ ab, ba ] h i = [ 1, 1 ] h i = 1Q.

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To see that there is at most one such r Q, suppose that also that q r0 = 1 . Then ∈ Q Q r = r 1 Q Q = r (q r0) Q Q = (r q) r0 Q Q = (q r) r0 Q Q = 1 r0 Q Q = r0.

1 Denition 10.19. For any 0Q = q Q, q is the unique element of Q such that 1 6 ∈ q q = 1 . Q Q Finally we want to dene an order relation on Q. [Note that if b, d > 0, then a c < i ad < cb. b d Note that [ a, b ] = [ a, b ], so each q Q can be represented as [ a, b ], where b > 0. This suggestsh thati whe make ithe following∈denition.] h i

Denition 10.20. Suppose that r, s Q and that r = [ a, b ] and s = [ c, d ], where b, d > 0. Then ∈ h i h i r s i ad < cb. Q

Lemma 10.21.

Theorem 10.22.

Remark 10.25. Clearly Z is not literally a subset of Q. However, Q does contain an “isomorphis copy” of Z. Denition 10.26. Let E : Z Q be the function dened by → E(a) = [ a, 1 ]. h i

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Theorem 10.27. E is an injection of Z into Q which satises the following conditions for all a, b Z ∈ E(a + b) = E(a) + E(b) Q E(ab) = E(a) E(b) Q E(0) = 0 and E(1) = 1 . Q Q a < b i E(a) < E(b). Q

Notation From now on, we write +, , <, 0, 1 instead of + , , < , 0 , 1 . Q Q Q Q Q

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