6 Integers Modulo n. integer k can be written as k = qn + r, with q,r, 0 r b. So any integer.

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1 6 Itegers Modulo I Example 2.3(e), we have defied the cogruece of two itegers a,b with respect to a modulus. Let us recall that a b (mod ) meas a b. We have proved that cogruece is a equivalece relatio o. The equivalece classes are called the cogruece classes or residue classes (modulo ). The cogruece class of a will be deoted by a. Notice that there is ambiguity i this otatio, for there is o referece to the modulus. Thus 1 represets the residue class of 1 with respect to the modulus 1, also with respect to the modulus 2, also with respect to the modulus 3, i fact with respect to ay modulus. However, the modulus will be usually fixed throughout a particular discussio ad a will represet the residue class of a with respect to that fixed modulus. The ambiguity is therefore harmless. By the divisio algorithm (Theorem 5.3), ay. iteger k ca be writte as k = q + r, with q,r, 0 r b. So ay iteger. k is cogruet (mod ) to oe of the umbers 0,1,2,..., 1. Furthermore,. o two distict of the umbers 0,1,2,..., 1 are cogruet (mod ), for if r 1, r 2 {0,1,2,..., 1} ad r 1 r 2 (mod ), the r 1 r 2, so r 1 r 2 by Lemma 5.2(11), ad so ( 1) 0, which is impossible. Thus ay iteger is cogruet to oe of the umbers 0,1,2,..., 1, ad these umbers are pairwise ico-gruet. This meas that 0,1,2,..., 1 are the represetatives of all the residue classes. Hece there are exactly residue classes (mod ), amely 0 = {x : x 0 (mod )} = {z : z } =: 1 = {x : (mod )} = {z + 1 : z } =: = {x : x 2 (mod )} = {z + 2 : z } =: = {x : x 1 (mod )} = {z + ( 1) : z } =: +( 1). The set {0,1,2,..., 1 } of residue classes (mod ) will be deoted by. A elemet of, thas is, a residue class (mod ) is called a iteger modulo, or a iteger mod. A iteger mod is ot a iteger, ot 54

2 a elemet of ; it is a subset of. A iteger mod is ot a iteger with a property "mod ". It is a object whose ame cosists of the three words "iteger", "mod(ulo)", "". 6.1 Lemma: Let, a,a 1,b,b 1. If a a 1 (mod ) ad b b 1 (mod ), the a + b a 1 (mod ) ad ab a 1 b 1 (mod ). Proof: If a a 1 (mod ) ad b b 1 (mod ), the a a 1 ad b b 1. Hece (a a 1 ) + (b b 1 ) by Lemma 5.2(5), which gives (a + b) (a 1 ), so a + b a 1 (mod ). Also, b(a a 1 ) + a 1 (b b 1 ) by Lemma 5.2(7), which gives ba a 1 b 1, so ab a 1 b 1 (mod ). We wat to defie a kid of additio ad a kid of multiplicatio o. We put a b = a + b ( * ) for all a,b (for all a,b additio ad multiplicatio o a b = ab ( ** ) ). This is a very atural way of itroducig. ( * ) ad ( ** ) seem quite. iocet, but we must check that ad are really biary operatios o. The reader might say at this poit that ad are clearly defied o ad that there is othig to check. But yes, there is. Let us remember that a biary operatio o is a fuctio from ito (Defiitio 3.18). As such, to each pair (a,b) i, there must correspod a sigle elemet a b ad a b if. ad are to be biary operatios o (Defiitio 3.1) We must check that the rules ( * ) ad ( ** ) produce elemets of that are uiquely determied by a ad b. The rules ( * ) ad ( ** ) above covey the wrog. impressio that a ad a b are uiquely determied by a ad b. I. order to peetrate ito the matter, let us try to evaluate X Y, where X,Y give directly as the residue classes of itegers a,b the discussio applies equally well to.) How do we. fid X. b are ot. (We discuss ; Y? Sice X,Y, there are itegers a,b with a = X, b = Y. Now add a ad b i to get a+b, the take the residue class of a+b. The result is X Y. 55

3 The result? The questio is whether we have oly oe result to justify the article "the". We summarize telegrammatically. To fid X Y, 1) choose a from X, 2) choose b from Y, 3) fid a + b i, 4) take the residue class of a + b. This souds a perfectly good recipe for fidig X use some auxiliary objects, amely a ad b, to fid X Y,. but otice that we Y,. which must be determied by X ad Y aloe.. Ideed, the result a + b depeds explicitly o the auxiliary objects a ad b.. We ca use our recipe with differet auxiliary objects. Let us do it. 1) I choose a from X ad you choose a 1 from X. 2) I choose b from Y ad you choose b 1 from Y. 3) I compute a + b ad you compute a 1. I geeral, a + b a 1. Hece our recipe gives, geerally speakig, distict elemets a + b ad a 1. So far, both of us followed the same recipe.. I caot claim that my computatio is correct ad yours is false.. Nor ca you claim the cotrary. Now we carry out the fourth step.. I fid the residue class of a + b as X Y, ad you fid the residue class of a 1 as X Y. Sice a + b a 1 i, it ca very well happe that a + b a 1 i. O the other had, if is. to be a biary operatio o, we must have a + b = a 1 + b. This is 1 the cetral issue. I order that be a biary operatio o, there must work a mechaism which esures a + b = a 1 + a 1 wheever a = a 1, b 1 = b, eve if a + b a 1. If there is such a mechaism, we say is a well defied operatio o. This meas operatio o : X is really a geuie Y is uiquely determied by X ad Y aloe. Ay depedece of X Y o auxiliary itegers a. X ad b Y is oly apparet. We will prove that ad are well defied operatios o, but before that, we discuss more geerally well defiitio of fuctios.. A fuctio f: A B is essetially a rule by which each elemet a of A is associated with a uique elemet of f(a) = b of B. The importat poit is that the rule produces a elemet f(a) that depeds oly o a. Sometimes we cosider rules havig the followig form. To fid f(a), 1) do this ad that 2) take a x related to a i such ad such maer 3) do this ad that to x 4) the result is f(a). 56

4 A rule of this type uses a auxiliary object x. The result the depeds o a ad x. At least, it seems so. This is due to the ambiguity i the secod step. This step states that we choose a x with such ad such property, but there may be may objects x,y,z,... related to a i the prescribed maer. The auxiliary objects x,y,z,... will, i geeral, produce differet results, so we should perhaps that the result is f(a,x) (or f(a,y), f(a,z),... ). I order the above rule to be a fuctio, it must produce the same result. Hece we must have f(a,x) = f(a,y) = f(a,z) =.... The rule must be so costructed that the same result will obtai eve if we use differet auxiliary objects. If this be the case, the fuctio is said to be well defied. This termiology is somewhat ufortuate. It souds as though there are two types of fuctios, well defied fuctios ad ot well defied fuctios (or badly defied fuctios). This is defiitely ot the case. A well defied fuctio is simply a fuctio. Badly defied fuctios do ot exist. Beig well defied is ot a property, such as cotiuity, boudedess, differetiability, itegrability etc. that a fuctio might or might ot possess. That a fuctio f: A B is well defied meas: 1) the rule of evaluatig f(a) for a A makes use of auxiliary, foreig objects, 2) there are may choices of these foreig objects, hece 3) we have reaso to suspect that applyig the rule with differet choices may produce differet results, which would imply that our rule does ot determie f(a) uiquely ad f is ot a fuctio i the sese of Defiitio 3.1, but 4) our suspicio is ot justified, for there is a mechaism, hidde uder the rule, which esures that same result will obtai eve if we apply the rule with differet auxiliary objects. The questio as to whether a "fuctio" is well defied arises oly if that "fuctio" uses objects ot uiquely determied by the elemet a i its "domai" i order to evaluate f(a). We wrote "fuctio" i quotatio marks, for such a thig may ot be a fuctio i the sese of Defiitio 3.1. Give such a "fuctio", which we wat to be a fuctio i the sese of Defiitio 3.1, we check whether f(a) is uiquely determied by a, that is, we check whether f(a) is idepedet of the auxiliary objects that we use for evaluatig f(a). If this be the case, our supposed "fuctio" f is ideed a fuctio i the sese of Defiitio 3.1. We say the that f is well defied, or f is a well defied fuctio. This meas f is a fuctio. I fact, it is more accurate to say that a fuctio is defied istead of sayig that a fuctio is well defied. 57

5 6.2 Examples: (a). Let L be the set of all straight lies i the Euclidea plae, o which we have a cartesia coordiate system.. We cosider the "fuctio" s: L { },. which assigs the slope of the lie l to l. How do we fid s(l)? As follows: 1) choose a poit, say (, ), o l; 2) choose aother poit, say (x 2,y 2 ), o l; 3) evaluate x 2 ad y 2 ; 4) put s(l) = (y 2 )/(x 2 ) if x 2 ad s(l) = if = x 2. Clearly we ca choose the poits i may ways. For example, we might choose (, ) (, ) as the first poit, (x 2,y 2 ) (x 2,y 2 ) as the secod poit. The we have, i geeral, x 2 x 2 ad y 2 y 2, so we might suspect that (y 2 )/(x 2 ) (y 2 )/(x 2 ). It is kow from aalytic geometry that these two quotiets are equal, hece s(l) depeds oly o l, ad ot o the poits we choose. Thus s is a well defied fuctio. Ultimately, this is due to the fact that there passes oe ad oly oe straight lie through two distict poits. The ext example shows that well defiitio breaks dow if we modify the domai a little. (b) Let C be the set of all curves i the Euclidea plae.. We cosider the "fuctio" s: C { },. which assigs the "slope" of the curve c to c. How do we fid s(c)? As follows: 1) choose a poit, say (, ), o l; 2) choose aother poit, say (x 2,y 2 ), o l; 3) evaluate x 2 ad y 2 ; 4) put s(l) = (y 2 )/(x 2 ) if x 2 ad s(l) = if = x 2. This is the same rule as the rule i Example 6.2(a). Let us fid the "slope" of the curve y = x 2. 1) Choose a poit o this curve, for example (0,0). If you prefer, you might choose ( 1,1). 2) Choose aother poit o this curve, for example (1,1). If you prefer, you might choose (2,4) of course. 3) Evaluate the differeces of coordiates. We fid 1 0 ad 1 0. You fid 2 ( 1) ad 4 1. Hece 4) the slope is 2/1. You fid it to be 3/3. So s(c) = 2 ad s(c) = 1. This is osese. We see that differet choices of the poits o the curve (differet choices of the auxiliary objects) give rise to differet results. So the above rule is ot a fuctio. We do ot say "s is ot a well defied fuctio". s is simply ot a fuctio at all. s is ot defied. (c) Let F be the set of all cotiuous fuctios o a closed iterval [a,b]. We wat to "defie" a itegral "fuctio" I: F, which assiges the real umber a b f(x)dx to f F. So I(f) = a b f(x)dx. I is a "fuctio" whose "domai" is a set of fuctios. How do we fid I(f)?. As follows. 1) Choose a idefiite itegral of f,. that is, choose a fuctio F o [a,b] such that F (x) = f(x) for all x [a,b]. (we take oe-sided derivatives at a 58

6 ad b). 2) Evaluate F(a) ad F(b). 3) Put I(f) = F(b) F(a).. There are may fuctios F with F (x) = f(x) for all x [a,b].. For two differet choices F 1 ad F 2, we have F 1 (b) F 2 (b) ad F 1 (a) F 2 (a) i geeral. So we may suspect that F 1 (b) F 1 (a) F 2 (b) F 2 (a). I order to show that I is a well defied fuctio, we must prove F 1 (b) F 1 (a) = F 2 (b) F 2 (a) wheever F 1 ad F 2 are fuctios o [a,b] such that F 1 (x) = f(x) = F 2 (x) for all x [a,b]. We kow from the calculus that, whe F 1 ad F 2 have this property, there is a costat c such that F 1 (x) = F 2 (x) + c for all x [a,b]. So F 1 (b) F 1 (a) = (F 2 (b) + c) (F 2 (a) + c) = F 2 (b) F 2 (a). Therefore, I is well defied. After this legthy digressio, we retur to the itegers mod ad to the "operatios" ad. 6.3 Lemma: ad are well defied operatios o. Proof:. We are to prove a b = a b ad a b = a b wheever a = a ad b = b i (differet ames for idetical residue classes should ot yield differeret results). This follows from Lemma 6.1. Ideed, if a = a ad b = b, the a a (mod ) ad b b (mod ) by defiitio, so we obtai a + b a + b (mod ) ad ab a b (mod ) by Lemma 6.1, hece a + b = a + b ad ab = a b, which gives a b = a + b = a + b = a b ad a b = ab = a b = a b. Havig proved that ad are well defied operatios o, we proceed to show that ad possess may (but ot all) properties of the usual additio ad multiplicatio of itegers. First we simplify our otatio. From ow o, we write + ad. istead of ad. I fact, we shall eve drop. ad use simply juxtapositio to deote a product of two itegers mod. Thus. we will have a + b = a + b ad a. b = ab or simply a b = ab. The. reader should ote that the same sig "+" is used to deote two very distict. operatios: i the old otatio ad the usual additio of itegers.. If aythig, they are defied o distict sets ad. The same remarks apply to multiplicatio. 59

7 6.4 Lemma: For all a,b,c (1) a + b ; (2) (a + b) + c = a + (b + c ); (3) a + 0 = a ; (4) a + a = 0; (5) a + b = b + a ; (6) a. b ; (7) (a. b). c = a. (b. c ); (8) a. 1= a ; (9) if (a,) = 1, the there is a x (10) a. b = b. a ;, the followig hold. such that a. x = 1; (11) a. (b + c) = a. b + a. c ad (b + c). a = b. a + c. a ; (12) a. 0 = 0. Proof: (1) is obvious. (2) follows from the correspodig property of additio i. We ideed have (a + b) + c = ( a + b ) + c = (a + b) + c = a + (b + c) = a + (b + c) = a + (b + c). The remaiig assertios are proved i the same way by drawig bars over itegers i the correspodig equatios i. We prove oly (9), which is ot as straightforward as the other claims. If (a,) = 1, The there are itegers x,y with ax y = 1 (Lemma 5.10). Usig (3) ad (12), we get 1 = ax y= ax y = a. x. y = a. x 0. y = a. x 0 = a. x. Exercises 1. Determie whether the "fuctio" g: 13 is well defied, if g is defied as follows. (a) g(a ) = (a,13); (b) g(a ) = (a,26); (c) g(a ) = (a,169); (d) g(a ) = (a 2,13); 60

8 (e) g(a ) = (a 3,169); (f) g(a ) = (a,6); (g) g(a ) = (a 2,65); where a 2) Let f: 13 ad a be such that (a,b) f a 2 +ab+b 2. Is f well defied? 3) For a iteger a, we deote by a the residue class of a (mod 12), by ã the residue class of a (mod 6), ad by â the residue class of a (mod 5), so that a 12, ã 6 ad â. Determie whether the followig 5 "fuctios" are well defied. (a) 12 6, a ã; (b) 6 12, â a ; (c) 12 5, a â; (d) 5 6, â ã; (e) 5 6, â a+1 ~. 61