The Field Q of Rational Numbers


 Ambrose Holmes
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1 Chapter 3 The Field Q of Ratioal Numbers I this chapter we are goig to costruct the ratioal umber from the itegers. Historically, the positive ratioal umbers came first: the Babyloias, Egyptias ad Grees ew how to wor with fractios, but egative umbers were itroduced by the Hidus hudreds of years later. It is possible to reflect this i the buildup of the ratioals from the atural umbers by first costructig the positive ratioal umbers from the aturals, ad the itroducig egatives (Ladau proceeds lie this i his Foudatios of Aalysis). While beig closer to history, this has the disadvatage of gettig a rig structure oly at the ed. 3. The Ratioal Numbers Let Z deote the rig of itegers ad cosider the set V {(r, s) : r, s Z, s 0} of pairs of itegers. Let us defie a equivalece relatio o V by puttig (r, s) (t, u) ru st. It is easily see that this is a equivalece relatio, ad we ow let [r, s] {(x, y) V : (x, y) (r, s)} deote the equivalece class of (r, s). Such a equivalece class [r, s] is called a ratioal umber, ad we ofte write r s istead of [r, s]. We deote by Q the set of all equivalece classes [r, s] with (r, s) V. We start studyig Q by realizig Z as a subset of Q via the map ι : Z Q defied by ι(r) [r, ]. The ι is ijective; i fact, assume that x, y Z are 3
2 such that ι(x) ι(y). The [x, ] [y, ], i.e., (x, ) (y, ), ad by defiitio of equivalece i V this meas x y, hece x y. We wat to have r s + t u ru+st su, so we are led to defie [r, s] [t, u] [ru + st, su] (3.) for r, s, t, u Z with s, u > 0. This is well defied ad agrees with additio o Z uder the idetificatio ι: i fact, ι(x) ι(y) [x, ] [y, ] [x + y, ] [x + y, ] ι(x + y). Thus it does ot matter whether we add i Z ad the idetify the result with a ratioal umber, or first view the itegers as elemets of Q ad add there. Next we defie multiplicatio of fractios by [r, s] [t, u] [rt, su]. (3.) This is motivated by r s t u rt su. Agai, multiplicatio is well defied ad agrees with multiplicatio o the subset Z Q: we have ι(x) ι(y) ι(xy) because ι(x) ι(y) [x, ] [y, ] by defiitio of ι [xy, ] by defiitio (3.) ι(xy) by defiitio of ι Remar. The map ι : Z Q from the rig Z to the rig of fractios Q satisfies ι(x) ι(y) ι(x + y), ι(x) ι(y) ι(xy), Maps R S betwee rigs with these properties (we say that they respect the rig structure ) are called rig homomorphisms if they map the uit elemet of R to the uit elemet of S. I particular, our idetificatio map ι is a rig homomorphism. Usig these defiitios, we ca prove associativity, commutativity, distributivity, thereby verifyig that Q is a rig. I fact, Q is eve a field! A field F is a commutative rig i which, iformally speaig. we ca divide by ozero elemets: thus F is a field if F satsifies the rig axioms (i particular we have 0), ad if i additio F For every r F \ {0} there is a s F such that rs. Observe that F holds if ad oly if F F \ {0}. This is a strog axiom: together with some other rig axioms it implies that fields are itegral domais: 4
3 Propositio 3.. If F is a field ad if xy 0 for x, y F, the x 0 or y 0. Proof. I fact, assume that xy 0 ad y 0. Sice the ozero elemets of F form a group, y has a iverse, that is, there is a z F such that yz. But ow 0 xy implies 0 0z (xy)z x(yz) x x; here we have used associativity of multiplicatio. We have proved Theorem 3.. The set Q of ratioal umbers forms a field with respect to additio ad multiplicatio. We ca also defie powers of ratioal umbers: if a Q is ozero, we put a 0 ad a + a a. This defies a for all N; if is egative, we put a /a. We ow ca prove the well ow set of rules a a m a +m, a m (a m ), a b (ab) etc. Biomial Theorem The ext result is called the Biomial Theorem. Before we ca state it, we have to itroduce the biomial coefficiets. These are defied i terms of factorials, so we have to defie these first. To this ed, we put 0! ad (+)!! (+) for N. Now we set ( )!!( )! for 0 ad ( ) 0 if < 0 or >. Lemma 3.3. The biomial coefficiets are itegers. I fact, we have ( ) + ( ) ( + + +) for 0 ad. Proof. This is a simple computatio: ( ) ( )! { + +!( )! + } +! +!( )! ( )( + ) +. + This calculatio is valid for 0; for, we have ( ) ( 0, +) ( + +), ad the claim holds. Now we have Theorem 3.4 (Biomial Theorem). For a, b Q ad N, we have (a + b) 0 a b. 5
4 Proof. This is doe by iductio o. For, we have to prove (a + b) ( 0 ) a b ( ) 0 a b 0 + ( ) a 0 b, which is true sice ( ( 0) ). Now assume that the claim holds for some iteger ; the ( (a + b) + (a + b) (a + b) a b ) (a + b) 0 a + b + a b ( ) + a + b + a + l b l l 0 l ( ( ) ( ) a + ) + + a + b ( ) ( ) + + a + + a + b a + b, 0 which is exactly what we wated to prove. 3. Q as a ordered field b + b + Observe that every ratioal umber ca be writte as [r, s] with s (if s, recall that [r, s] [ r, s]). From ow o, we will assume that all our ratioal umbers are preseted lie this. We defie a order relatio < o Q by puttig [r, s] < [t, u] ru < st (recall that s, u N). This is well defied: if [r, s] [r, s ] ad [t, u] [t, u ], the rs r s ad tu t u. Now [r, s] < [t, u] ru < st by defiitio rus u < sts u sice s u > 0 r suu < ss t u sice rs r s ad tu t u r u < s t sice su > 0 [r, s ] < [t, u ] by defiitio Now we have Theorem 3.5. Q is a ordered domai (eve field). Proof. Sice exactly oe of the relatios ru < st, ru st or ru > st is true by the trichotomy law for itegers, exactly oe of x < y, x y or x > y is true for x [r, s] ad y [t, u]. 6
5 Next assume that x < y ad y < z, where z [v, w]. The ru < st ad tw < uv, hece ruw < stw ad stw < suv sice w, s > 0; trasitivity for the itegers gives ruw < suv, ad sice u > 0, this is equivalet to rw < sv, i.e., x < z. This shows that Q is simply ordered. The rest of the proof that Q is a ordered domai is left as a exercise. Thus everythig proved for geeral ordered domais holds for the ratioals; i particular, x 0 for all x Q, ad x + y x + y for x, y Q. Now let us collect a few simple results that will tur out to be useful. Lemma 3.6. We have x < y if ad oly if x < y for some N. Proof. Exercise. Propositio 3.7. Let x, y Q ad assume that for every ratioal ε > 0 we have x y < ε; the x y. Proof. Assume that this is false, i.e. that x y 0. The ε x y is a positive ratioal umber, so by assumptio we have x y < ε. This implies ε < ε, which is a cotradictio. Propositio 3.8. Let 0 < x < y be ratioal umbers. The there is a N such that x > y. Proof. Write x r s ad y s t with r, s, t, u N (here we have used that x, y > 0). The x < y is equivalet to ru < st; by the Archimedea property of the atural umbers there is a N such that (ru) > st. But the last iequality is equivalet to x > y. Divisio with remaider i Z allows us to itroduce the floor fuctio i Q: for ratioal umbers x a b with b > 0, we put x q if a bq+r with q, r Z ad 0 r < b. Note that this is well defied: if x c d with d > 0, c dq + r ad 0 r < d, the ad bc, hece ad bdq +rd, bc bdq +br, ad therefore 0 bd(q q ) + rd r b. We may assume without loss of geerality that q q ; if q q, the q q +, hece bd > r b bd(q q ) + rd bd + rd bd: cotradictio. Propositio 3.9. For x Q, the iteger x is the uique iteger satisfyig x < x x. Proof. First, there is exactly oe iteger m satisfyig x < m x because m < for itegers implies m 0, hece m. It is therefore sufficiet to prove that x < x x. To this ed, recall that q x is defied for x a b by 0 a bq < b. Dividig through by b ad addig x we get x < q x as claimed. For ay ratioal umber x, we call x x x the fractioal part of x. Note that 0 x < for all ratioal umbers x Q. 7
6 3.3 Irratioal Numbers The irratioality of, at least i its geometric form (the side ad the diagoal of a square are icommesurable) seems to have bee discovered by the Pythagoreas. Although by the time of Euclid it was ow that square roots of osquares are irratioal, Euclid s elemets oly cotai the proof that is ot ratioal; i this case, a proof ca be give depedig oly o the theory of the odd ad the eve, as the Grees called the most elemetary parts of umber theory. Theorem 3.0. If N is ot the square of a iteger, the it is ot the square of a ratioal umber. Proof. I fact, if is ot a square of a iteger, the it lies betwee two squares, that is, we ca fid a iteger a such that a < < (a + ). Assume that p q with q > 0 miimal. The p q, hece p(p aq) p apq q apq q(q ap), so p q ap q p aq. But a < p q < a + implies 0 < p aq < q: this cotradicts the miimality of the deomiator q. Proof. Assume that A/B with B > 0 miimal; the A/B B/A, hece both of these fractios have the same fractioal part, say b/b A/B B/A a/a with 0 < a < A ad 0 < b < B (ote that e.g. a 0 would imply that is a iteger). But the A/B a/b, ad 0 < b < B cotradicts the miimality of B > 0. Proof 3. Sice is ot a square, at least oe prime p divides to a odd power. If we had b a, squarig ad clearig deomiators would give b a, ad p would divide a to a odd power, cotradictig uique factorizatio. 3.4 Historical Remars Log before maid discovered 0 ad the egative umbers, they started worig with positive ratioal umbers. The Babyloias had sexagesimal fractios, ad the Egyptias essetially wored with pure fractios, that is, those that ca be writte i the form. Every fractio was represeted by a sum of differet pure fractios: for example, they would have writte 5 ot as 5 + 5, but as Fractios were ot regarded as umbers i Euclid s elemets; Eudoxos theory of proportios dealt with magitudes, that is, legths ad areas etc. Archimedes ad Diophatus, o the other had, wored freely with positive ratioal umbers. Due to Joh Coway. 8
7 Exercises 3. For a, b Q, we have a a + b b. The ratioal umber a+b is called the arithmetic mea of a ad b. 3. For a, b Q, we have The ratioal umber a + b 3.3 Prove that for all a, b Q, we have a a + b b. is called the harmoic mea of a ad b. a + b a + b. This is called the iequality betwee harmoic ad arithmetic mea. Show that equality holds if ad oly if a b. 3.4 Which of the proofs of the irratioality of for osquares geeralizes to mth roots of itegers? 9
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