ad = cb (1) cf = ed (2) adf = cbf (3) cf b = edb (4)


 Jonah Morton
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1 10 Most proofs re left s reding exercises. Definition Z = Z {0}. Definition Let be the binry reltion defined on Z Z by, b c, d iff d = cb. Theorem is n equivlence reltion on Z Z. Proof. We just check tht is trnsitive. So suppose tht, b c, d nd c, d e, f. Then d = cb (1) cf = ed (2) Multiplying (1) by f nd (2) by b, we obtin df = cbf (3) cf b = edb (4) Hence df = edb. Since d 0, the Cncelltion Lw implies tht f = eb. Hence, b e, f. Definition The set Q of rtionl numbers is defined by ie. Q is the set of equivlence clsses. Q = Z Z / Nottion For ech, b Z Z, the corresponding equivlence clss is denoted by [, b ]. Next we wnt to define n ddition oppertion on Q. [Note tht b + c d d + cb =. bd This suggests we mke the following definition.] Definition We define the binry opertion + Q on Q by [, b ] + Q [ c, d ] = [ d + cd, bd ]. Remrk Since b 0 nd d 0 we hve tht bd 0 nd so d + cb, bd Z Z. 2006/10/25 1
2 Lemm Q is welldefined. Theorem For ll q, r, s Q, we hve tht q + Q r = r + Q q q + Q (r + Q s) = (q + Q r) + Q s. Definition 10.9 (Identity element for + Q ). 0 Q = [ 0, 1 ]. Theorem () For ll q Q, q + Q 0 Q = q. (b) For ny q Q, there exists unique r Q such tht q + Q r = 0 Q. Proof. () Let q = [, b ]. Then q + Q 0 Q = [, b ] + Q [ 0, 1 ] = [ b, b 1 ] = [, b ] = q. To show tht there exists t lest one such element, consider r = [, b ]. Then q + Q r = [, b ] + Q [, b ] = [ b + ( )b, b 2 ] = [ 0, b 2 ] Since 0 1 = 0 b 2, we hve 0, b 2 = 0, 1. Hence q + Q r = [ 0, b 2 ] = [ 0, 1 ] = 0 Q As before, simple lgebr shows tht there exists t most one such element. Definition For ny q Q, q is the unique element of Q such tht q + Q ( q) = 0 Q. Definition We define the binry opertion Q on Q by q Q r = q + Q ( r). 2006/10/25 2
3 Next we wnt to define multipliction opertion on Q. [Note tht b c d = c bd. This suggests we mke the following definition.] Definition We define the binry opertion Q on Q by [, b ] Q [ c, d ] = [ c, bd ]. Remrk Since b 0 nd d 0m wee hve tht bd 0 nd so c, bd Z Z. Lemm Z is welldefined. Theorem For ll q, r, s Q, we hve tht q Q r = r Q q (q Q r) Q s = q Q (r Q s) q Q (r + Q s) = (q Q r) + Q (q Q s) Definition (Identity element for Q). 1 Q = [ 1, 1 ]. Theorem () For ll q Q, q Q 1 Q = q. (b) For every 0 Q q Q, there exists unique r Q such tht q Q r = 1 Q. Proof. () Let q = [, b ]. Then q Q 1 Q = [, b ] Q [ 1, 1 ] = [ 1, b 1 ] = [, b ] = 1 Q (b) Suppose tht q = [, b ] [ 0, 1 ]. Then 0 nd so b, Z Z. Let r = [ b, ]. Then q Q r = [, b ] Q [ b, ] = [ b, b ] = [ 1, 1 ] = 1 Q. 2006/10/25 3
4 To see tht there is t most one such r Q, suppose tht lso tht q Q r = 1 Q. Then r = r Q 1 Q = r Q (q Q r ) = (r Q q) Q r = (q Q r) Q r = 1 Q Q r = r. Definition For ny 0 Q q Q q 1 = 1 Q. q Q, q 1 is the unique element of Q such tht Finlly we wnt to define n order reltion on Q. [Note tht if b, d > 0, then b < c d iff d < cb. Note tht [, b ] = [, b ], so ech q Q cn be represented s [, b ], where b > 0. This suggests tht we mke the following definition.] Definition Suppose tht r, s Q nd tht r = [, b ] nd s = [ c, d ], where b, d > 0. Then r Q s iff d < cb. Lemm < Q is welldefined. Theorem < Q is liner order on Q. Definition If q Q, then q is positive iff 0 Q < Q q. q is negtive iff q < Q 0 Q. Definition If q Q, then the bsolute vlue of q is q = q if q is negtive = q otherwise. Remrk Clerly Z is not literlly subset of Q. However, Q does contin n isomorphis copy of Z. Definition Let E : Z Q be the function defined by E() = [, 1 ]. 2006/10/25 4
5 Theorem E is n injection of Z into Q which stisfies the following conditions for ll, b Z E( + b) = E() + Q E(b) E(b) = E() Q E(b) E(0) = 0 Q nd E(1) = 1 Q. < b iff E() < Q E(b). Nottion From now on, we write +,, <, 0, 1 insted of + Q, Q, < Q, 0 Q, 1 Q. 2006/10/25 5
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