MATIENM SPECTRUM DISCRETE MAT SETS, RELATION AND FUNCTION result is true for n= k+1 132 133 SECTION-V But the result is true for n = 1 hy the method ot induction, the result is true for allnEN. if the result is true for n= k, then it is also true for n = k+ 1. PRINCIPLES OF MATHEMATICS INDUCTIO that 52n* -24 n-25 is a multiple of 576. 1.62. Statement of Principle of Mathematical Induction Example 2. Prove that s2n+2 Let P (n) be a statement such that Sol. Let Pn)= 5èn*2 24 n- 25 (a) P(1) is true P(1)= 5-24-25 625 - 24-25 = 576, which is a multiple of 576. () P(r+ ) is true whenever P () is true where n =r. result is true forn= 1. Then P (n) is true for all natural numbers n. Assume that result is true for n =k. P() 52k*2 -24 k -25 is a multiple of 576. The Well-Ordering Principle Every nonempty set of non-negative integers has a least element. Recursive Definition: Sometimes we find it dificult to define an object explicitly. However, it ma easy to define this object in terms of itself. This process is called recursion. We can use recursion to define sequence, functions, and sets. In previous discussions, we specified terms of a sequences using an explicit formula. For example: We define recursively defined functions: We use two steps to define a function with the of nonnegative integers as its domain: Basic step: Specify the value of the function at zero. may 52k+2 24k-25 576 1, where l is an integer Let 52k+2 5761+24 k +25 ..(1) Now P(k + 1) = 54*)*-24(k +1)-25 5.5k*2 -24 k-24-25 = 25(576 1+24 k + 25) -24 k-24-25 = 25.576 l+25-24 k+ 625 24 k - 24 -25 576-25l+ 576 k+ 576 = 576(25 Il+ k+ 1), which is a multiple of 576. ( of (1)] Recursive step: Give a rule for finding its value at an integer from its value at smaller integers. ILLUSTRATIVEEXAMPLES result is true for n = k+1 if the result is true for n = k, then it is also true for n =k+ 1. But the result is true for n = 1. by the method of induction, the result is true for all n EN. Example 1. Prove that 9" -8" - 1 is divisible byy8. Sol Let Pin) = 9" -8" -1 Example 3. Use the principle of mathematical induction to prove that P(1)=9-8-1 =0. which is divisible by 8 result is true for n= 1 P+2+23+. - "7*9 VnEN. 4 Assume that result is true for n = k Sol. We have to prove that P(k)= 9 -8 - 1 is divisible by 8. Let 98 - 1 = 8 I, where l is an integer P+23 ..+ n? == " 4 ...(1) 9 81+8* +1 For n=1, Now P(k 1) = 9**-8*- - 1 = 9.9 -8k*1-1 L.H.S. 1 = 1 9(81+8 +1)-8**l -1 of ( RH.S. 0D - - 72+ 9.8* +9-8-8 -1 - 72/+8* +8 L.H.S. = R.H.S. = 8(91+8*-1+1), which is divisible by 8. result (1) is true for n =1. SPECTRUM DISCRETE MA AND FUNCTION 135 134 SETS RELATI (k+D (2+7k +6) - 2+3k+4k+6] = k*l (K2 k+ 3) + 22 k- 3)) Assume that result (1) is true for n =k 22 .+k - k*k+1)2 & [k +2) (2 k+ 3)] = *D&+2)2k+3) 6 6 Adding (k+1)° to both sides of (2), we get 2+22+3+. + (k+1 = *Dk+2)(2k+3) P+2+2+ .+ A+ (k+1)3-6+), (k+1? result is true for n =k+1 But the resut is true for n = 1. u by method of induction, the result is true for all n E N. have added (k+1* to both sides of (2). This is obtained by changing n to k+ 1 in the last if the result is true for n =k, then it is also true for n = k+1. P+2+2+ (k+n3 = &+*(k+2 Note. result is true for n =k+1 em of LH.S. of(). if the result is true for n =k, then it is also true for n =k+ 1. Erample 5. Use the principle of mathematical induction to prove that nEN. But the result is true for n = 1. 1-3+2.4+3.5+..+n(n + 2) = "0*)(2n+7) 6 by method of induction, the result is true for all n E N. nn +1) (2 n+1) Sol. We have to prove that Example 4. By mathematical induction, prove that1f+2 +3* +. VnEN nn +1)(2 n+7) 6 1-3+2.4+3:5+ ..+ n{n +2) ..0) 6 Sol. We are to prove that For n=1, n(n+1) (2 n* VnEN. 1+2+32 LH.S. = 1-3=3 = 6 11+1)2+7) -Ix2x9- 3 For n= l1, R.H.S= 6 6 LH.S.= 1=1 L.H.S. = R.H.S. RHS 0-D2+) - 2x3 - =1 result (1) is true for n = 1. 6 6 Assume that result (1) is true for n =k. L.H.S. R.H.S. 13+2.4+3.5+ k(k + 2) =*(K +)(2k+7) ...(2) result (1) is true for n =1 Adding (k+ 1) k+3) to both sides of (2), we get 1-3+2-4+3.5+.. + k(k+2) +(k+ 1)(k+ 3) Assume that result (1) is true for n =k *2" +3+. k2 = R(k+)(2k+1) k(k +1) (2 k+7) 6 (6+ D&+3)=+ D 2,k+3 6 Adding (k+1) to both sides of (2), we get 27)+6k+18-+ 2k+13k+18&+ nk+2)2k+9)| +2+3+ 2+ (k+ )2 = k\k+D2k+)+4-n2= k(k +)(2k+)+6h + 6 6 6 6 6 6 6 SPECTRUM DISCRETE ATIWEM .RELATION AND FUNCTION 137 136 EXERCISE 1.10 result is true for m =k+ of mathematical induction to prove that if the result is true for n =k, then it is also true for n =k + 1. Use the principle o. 1 23-1 is divisible by 7VnEN 2m+2-8n-9 is divisible by 64 Yn EN dhe orinciple of mathematical induction to prove that But the result is true for n = l. (in 102-+1 is divisible by 11 Vn EN by method of induction, the result is true for allnEN. Example 6. Use the principle of mathematical induction to prove that .L. 13 35 5-7 n-D2n+) 2ne Vn EN. 32-1 is divisible by 8 Vn¬N. (in 10"+3.4*2+5 is divisible by 9 VnEN. Sol. We have to prove that ih62+7-n* is divisible by 43 Vn EEN. 1'35'57 * (2n-1)(2n+1) 2n+l 3. Use the principle of mathema induction to prove that x+4x+7x+ .. ... +(3n-2)x={3n-1)x Vn EN. For n 1 LHS 4 By mathematical induction, prove that 1+32+52 .+(2n-12 = 47=9n+) ynEN 3 R.H.S. 2n VnEN. 5 Prove by induction that 1+ 1+2+3 1+2+3+.+n L.H.S. R.H.S. n+l result (1) is true for n =1. 1.63. Division Algorithm Statement. For any two integers a andb, with a > 0, there exist unique integers q and r such that Assume that result (1) is true for n =k. k T (2 k-1) 2k+1) 2k+l baqtr, 0sr<|al Proof. When b=0 to both sides of (2), we get Adding Taking q 0, r=0, we have [2 (k+1)-1J[2 (k +1)+1] (2k+1)(2k+3) 0af0)+0, 0 =r<|a| or b=aq+r,0 Sr<|a| Hence the result. 13 35 5.7 (2k-1)(2 k + 1) (2k+1) (2 k +3) When b 0 Consider the infinite sequence of multiples of a i.e. ...,-2a,- a, 0, a, 2 a, ... Let aq be the greatest multiple of a such that 1 2k+1 (2k+1)2k+3) 2k+112 b2aq and b<(q+ 1)a 23k+ (k+1)(2k+1) 2k+3 2k+1 2k+3 aqsb0] if the result is true for n=k, then it is also true for n = k +1. lf possible, let there exist another set of integers q and such that But the result is true for n = 1. by method ofinduction, the result is true for all n EN. ag+ 0s <|al (2) SPECTRUM DISCREIE MA. 139 138 SETS, RELATION AND FUNCTION s|al, then 2 From () and (2), we get aqtr=dt a-al i-|a|<|a|-|a| 2 a dividesr-7 -a|<0 I 0 Sr 9, -9 =0 as a0 a b = aq1 +r+| a|, - Sr<0 from (1), = aq t| al+r for any tO integers a and b, w ith a> 0, there exist unique integers q and r such that a (q t 1)+r [: lal = a] b =aq-r. 0 Sr0 |a b =aqtr, srla if a0, then |aj>0 there exists unique integers q and r such that As 41, are unique so0 4, r are also unique. b 92 a +r. 0 Sr<|a|= 92(-a) +7, 0Sr<|a| Hence the result. -92) a +r, 0sr<|al Note:) In b=aq+r, 0 sr 0. then there exists unique integers q and satisty ing a=bq-r, where 2 b sr< 3b b 5k+r, rs-2,- 1,0, 1,2 b 5k-2, 5 k- 1,5 k, 5 k+ 1,5k+2. kE Z Sol. Given a and b are integers with b >0. By division algorithm, we have every integer is of form 5 k-2, 5 k- 1, 5 k, 5 k+ 1,5 k+2. () Putting a=6 and q =k in (1), we have a= bq +r whereq and r' are integerS b 6k+r, sr and 0r' 1. 12 (q (q+ 1)+2g+2q+2)) 9.9I are consecutive integers= q(q+ 1) is even integer ANSWERS qq )+ 2(q+2q+2) is also even integer. .164 2 qq 1) 24g+2q+2) 12.2| 12 (g (q+ 1)+ 2(g+2q+2)) 24 m23 (: of (2)) Hence the result L64. Common Divisor, Greatest Common Divisor Common Divisor: Ifc | a and c | b, then c is called a common divisor of a and b. Note: As there is only a finite number of divisors of any non zero integer so there is only a finite number of common divisors of a and b except when a = b=0. If atleast one of a and b is non zero, then the greatest among their common divisors is known as greatest common divisor of a andb. Greatest Common Divisor Let a and b be two integers such that atleast one of them is non zero. lhen positive integer d is the Greatest Common Divisor of a and b if Example 5. Show that 2is an integer for al n EN. Sol. By division algorithm, ifn is divided by 3, then n =3 q+r, 0sr<3, ,rEZ n =3q+r, r=0, 1,2 n=3 q. 3 q+ 1,3q+2 0 da, d b ie. dis a common divisor of a and b. cla,c| b cs dfor any positive integer c. abon:(0 The g.c.d. of a andb is denoted by g.c.d. (a, b) or simply (4, D). nn+2) 3g9g *2) - q942 +2) An integer When n=3 q, Then 4,6) is defined for each pair of integers a and b except when a =0, b=0. Gi) (a, b) 2 . When n 3 q+ 1, Then n +2) _ Gq+D(Gq+1)2 +2) 3 Tor example : To find g.c.d. of- 12 and 48. he positive divisors of-12 are 1,2,3,4, 6, 12 g-)9g-6q+3) (3q+1)3(G4+24+) The positive divisors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 3 3 = (3 g+1)(3q+2g+1) the common divisors are 1, 2,3, 4, 6, 12 C.d.of- 12 and 48 is 12 i.e. -12, 48)= 1 An integer SPECTRUM DiSCRETE ETE MAT SETS, RELATION AND FUNCT andlv, we want to show g.c.d. (a, b) is unique CTION 1.65. Ifa and & are any two integers, not both zern, then prove that g.c.d ofa and b Proof: Firsty, we shall prove that gcd. (a, 6) exists. WIog assume that & and & both are positive and b 2a exists and is unig 145 144 Second tf possible, let d as well as d be g.c.d. (a, b). [gcd. (a, b) is not affected by si ted by signs of a d, and d2 are also common divisors of a and b. d 2 d2 d, -gcd. (a, b) and d, is divisorof a, b] by division algorithm a9+ where 0s n 0; x,y EI} Putting value of a in (), we have 792)9 + 9291+1) As a,b are not both zero W.lo.g. let a*0 * |a| >0 Thus a and b Ial =a.x+b.0 where x- if a <0 and a a.x+b.0 where x= if a>0 Take p a and p b. Then p |b- aq alEA p A pS by well ordering principle, A has a least element say d. Hence gc.d. (a. b) = : of (3)a d-ax+by for some integers x, y. Thus so that g.c.d. (a, b) exists. We claim d=g.c.d. (a b) If r0, we repeat the above process, which ends after n steps (say) and we get the remainde after th step So, we get a sequence of integers r, (1 sj s n) such that Divide a by d, so by division algorithm 3 q,r EI such that aqdtr, 0 sr0 then rES which is a contradiction to the choice of d. 0 so that a = q d and 'n-2n-19,t n23 9n+ dla Similarly on dividing b by d, we get d\b disa common divisor of a and b. nn-1n'n-2 a and , lb Further, if p is a common divisor of a and b then p b,pla plb-a9 P\ : COnsider c be any other positive common divisor of a and b ela and elb clax+ by cld»c|S|d| csd Hence d= g.c.d. (a, b). P2.P so that g.c.d. (a, b) = r Hence g.c.d. (a, b) exists. sPETRUNM DISCREH 146 , RELATIONAND FUNCTIC 147 L67. Other Definition of G.C.D 67 s 4, b not both zero, the g.cd.(a, b) is defined as the smalest positive integer of type ax + by NOw we will prove thal d is the least positive valuc of r a+ yb Det. For integers a, b not both EZ. Let u - nbbe any other member of the form x a + yb. Sunce ( ) a.b not both zero, the g.c.d.(a, b) = d where d is positive integer satisfying a andd dn6) dS (ma +n b) Del. 0 dla, d1b (in If cla, c|b Then c | d. dis the least positive value of ra *yb. Cor. 1. If u and h are tho integers not both zero, then prove that a positive integer Result: If alk and b |k. Then (a, b)|k dgcda. A) iff () da. db ) if cla and c|b then c|d mof: Let d= a, b) > d|a, d|b Given a & b]k so that dk (a, 6) |. Proof: Fiursthy. let d=gcd (a. 6) da and d|b. And as c| a and ci b so that e If d, a2 1are m integers, not all zero, then g.c.d. of a, a2, .am is positive integer apc and b=q c where p. qEZ Let d = (a, b) so 3 x.y EZ such that satistying d a, for Si sm (i) If cl a, for I siSm, thenc|d. dar- by =pcx+gcy Note: G.C.D. of aj, 02., m is denoted as d=(a,. as m L68, Prove that (a, b) = (a, b * ax) =(a + by, b) Vx,y ¬ Z. c(pr*qy) =cr wherer = px+qy E Z d =cr cjd * c|(a, b). Comversely: Let dbe positive integer satisfying conditions (i) and (i) Proof: Let (a, b) =d and (a, b+ax)= d, By condition (i) if c is a common divisor of a and b, then (a, b) = d dla and d|b * d|ax and d|b Vr EZ d| (b +ax) d| (a, b+ ax) d d dcsd d2 c dis g.c.d. of a and b. dla and d| (b + ax) Hence the result. Cor. 2. Ifa and b are given integers, not both zero, then prove that the set (1) B ax by|x.yE Z} consists ofall multiplies of g.c.d. (a. b). Proof: Let d=gcd (a. b) And (a, bt ax) = d dla and d, |(b + ax) d |ax and d, |(b+ax) dia and d b d ax + by for all x, y E Z d (b+ ax) - a x) > d, |b dis a divisor of each element of B d la and d, |b d, |(a, b) ie Each element of B is a multiple of d dld As dis the smallest positive integer of the form ax+ by. (2) From (1) and (2), we get 3 integers x. such that d= ax +by d d For any nteger m, (a, b) = (a, b+ ax) Vx¬Z md =m(axy +by,) = a(mxj)+ b (m y,) = ax, +by, (Say) urther to prove (a, b) = (a+ by, b) VyEZ Le (atby, b)= d2 md EB where 2.y2 EZ each multiple of d is an element of B (4, b)=d » d|a and d|b » dla and d|by Vy EZ Combining (1) and (2), we see that the set B consists of all multiples of d. datby d|a+ by and d|b d| (a+ by, b) Hence the result. d d ..3) SPECTRUM DISCRETE MAn MATN 148 RELATION AND FUNCTION 149 For a, b any two integers, not both zero and m any positive integer, prove that Proof: Let d= (a, b) and d (m a, m b) (a by. b) = d SETS, R and d, la+ by and d b d, la+ by and d, |by (m a, m b}= mla, b) d ja+by -by and d|a da and d b d (a. b) d, Id We want to prove d md. From (3) and (4), we get = d dla and d|b As d=(a, b) mdjm a and m d|mb (a. 6) = (a +by. b) V y¬Z. md (m a, m b) (a, 6)-(a. b+ar) = (a + by, b) Vx,yEZ mdi d .(1) 1.69. For a. b any two integers, not both zero, prove that (a. b)=(-a. 6) = (a. - b) =-a,-b). Proof: Let d= (a. b) and d =(-a, b) Eurther as d= (a, 6) so there exist ntegers x, y such that d ax+ by As d (a. 6) da and d|b md =m (ax+by) d-a and d|b md (m a)x+ (m b) y (2) d|(-a, b) d| d d (m a, m b) Also Further, as d -a, b) dma and d |mb d-a and d d (m a) x+ (m b)y dla and d md .(3) d (a, b) 41d (of (2)] From (1) and (2). d d (a, b)=-a, b) From (1) and (3), we get d md Similarly Further to show (a, b) = -a,- b) Let d-(a, b) and d, =-a,- b) (a, b) = (a, - b). m a, m b) = m (a, b) Hence the result. As d (a, b) dla and d|b COr. 1. For any non zero integer m, prove that (m a, m b) =|m|(a, b). Proof: If m>0, we have d-a and d-b d(-a, - b) d d2 m a, m b) =m (a, b) (m a, m b) =|m |(a, b) (m> 0 m= |m|] d -a, -b) If m<0, then And m>0 d2 -a and d21-b m a, m b) =(-ma, m b) d2 a and d2 1b - m (a, b) d2 1(a, b) d2ld =|m |(a, b) :m<0 -m =|m|] for m 0, we have From (3) and (4) d= d2 (a, b) = (-a, - b) (a, b) = (-a, b) = (a, -b) = (-a,-b). m a, m b) = | m|(a, b). Hence SPECTRUM DISCRETE M. -uETE MATMEM SETS, RELATION AND FUNCTI1O d =e 1ON 151 1SU From (1) and (2) (a, b, c) =(a, b), c) Cor. 2. If dla, d|b and d> 0. Prove (a, b, c) (a, (6, c)=((a, c), b) Similarly Hence the result. Further more if d, = (a. 6). Prove 1. 1LLUSTRATIVE EXAMPLES Proof: Since (m a, m b}= m{a, b) for positive integer m. Example .Ifa = bq + r, then sho that (a, b) = (b, r). Sol. Let (a, b) = d, (6, r) = d2 (a, b) = d (a. b 4/a. 4/6 d(a-b q) » d,Ir Now db and d r - d is a common divisor of b andr d,/ d, as d2 is the g.c.d. of b andr. Hence the result. Similarly d, /d Further given d (a, b) d >0. Now d,/ d2. dz l d and d, d2 are natural numbers. Putting d= d in (1), we get ) - 4) (a. b)=(b, ). Note. If (a, b)= 1, then (6, r) = 1. Example 2. (a, m n) = 1 if and only if (a, m) = 1 and (a, n) =1. Sol. () Assume that (a, m) = 1, (a, n) = 1 (a,m)=1 there exist integers x and y such that Hence the result. 1.71. If a, b, c are any integers, no two of which are zero and d= (a, b, c). ax+my=1| (1) (a, n)=1 Prove d (a, b, c) = ((a, b), c) = (a, (b, c) = ((a, c),. b). there exist integers z and t such that Proof: Let d=(a, b, c) and e= ((a, b), c) aztnt =1 dla, djb, djc aztn(l1) =1 d (a, b) and d|e az+nax + my) =1| d1 ((a, b), c) dle of (1)] aztantx +nm ty =l| Further as e ((a, b), c) a2+ntx) t m n{ty) = 1| e 1(a, b) and e c ar+ (m n) s = 1 where r=z +ntx, S =ty e la, el b and elc (a, m n) =1| e l (a, b, c) eld SPECTRUM DISCRETE MAT ETS, RELATION AND FUNCTIONn dla 155 152 and db Assume that (a. m n) = 1| d (a, b) d11 d-:1 (ii) But d is positive so that d-1 (c, b) =e so that there exist integersu and v such that (c, a)=1. au (m n) v =1| aut m( v) =1 (a, m)= Further let e lc and elb ela+b Given Similarly (a n) =1 and elb e la+b Note. Let (a m)=1 e la+b-b ela we have (m, a)= 1, (m, a) = 1 e l(a, b) el e-|1 (m.a)= 1 (a,m) =1 so thar Bute is positive Now (m, a) = 1. (m, a) = 1 so thatel > (c, b) = 1. (c, )=(c, b)= 1. (m. a) =1 and so on. Hence the result. Example 5. If(¢, a)=1, then prove that (a, b c) =(a, b). Sol. Let (a, b) = d and (a, b c) =e Proceeding in this way, we get (m, a) =l or (a, m) =I d la, db and ela, elbc ~ dla, d|bc if (a, m)= 1, then d I(a, b c) (a, m) = 1. (a, m) =1,.. (a, m) = 1. e ..(1) Example 3. Ifal b, c/d and (b, d) = 1, then prove that (a, c) = 1. ela and (c, a) =1 Sol. Sincealb,cld. (e, c)=1. there exist integers m, n such that e lbc elb b =am, d = cn ela and e|b el(a, b) eld ...2) (b. d)=1 From (1) and (2), we get e=d there exist integers x and y such that (a, be) = (a, b) bx + dy 1 Hence the result. Example 6. Prove that there are infinitely many pairs x, y satisfying a mx+ cCny =l| a(m x) +cny) = 1| x+y= 100 and (r, y) = 5. Sol. Given (a, c)=1 x+y 100 ...(1) Example 4. 1f (a, b) = I and c| (a + b), then prove that (c, a) = (c, b) = 1. Sol. Let (c, a)=d Let x= 5p where p is odd integer such that (p, 5)=1. From (1), y= 100-x = 100 -5p 520-p). We claim (x, y)=5 Let (x,y)= d dz5 f possible, suppose d> 5 d|e and d|a Given cla+b (: 5|x and 5 |y] dla+b and dla d a+b-a d|b AND FUNCTION SETS, RELATION ANDDF - (m+1)-m m-m (m Also my(m-45 dr' (m-m) (m' -4) + 5 (m dl 100 15 ) m (m -ly = (m -2) (m - 1) m (m+ )(m+2)+ 5 m d-10 20, 25, S0. 100. -4)+5-) .-2. m-1, m, +1,m+2 are 5 consecutive integer so that m) AS 5 (m-2) (m -1) m (m +1) (m+2) must hr odd 5 5 (m- m) Also 25 5p 5 I(m-2) (m - 1) m (m+ 1)(m +2)+5 (m. 5 5iP -m) which is impossible as (p. 5) =1 5 m° -m As there are infinitely manyp such that (p. 5)=1 there are infinitely many r and so infinitely many pairs x, y satisfyino ur supposition is wrong so that d= 5 )=5. : of2)) From(1) and (3), we get 6x 5 | m -m as (6, 5) = 1 [: a\n, bin abn where (a. b)= 1] + 100 and (x, y) = 5. 30 1(m-m). (a. be) d and (a, (a, b) c)=e dja and d|bc d|ac and d|bc Eumple 7. For any integers a, b, c, prove that (a, b)= (a, (a, b) c). Hence the result. Sol Let d |(ac, bc) d|(a, b)c dla and d| (a, b) c d|(a, (a, b) ) Example 9. If (a b)-1, prove that (a, a b, b) =1. Sol (a. b)= I (a",ab) =a .abb)- (a, b) dle (a. (a, b) ©) =e ela, el(a, b) c e|a, el(ac, bc) ela, e|bc AS = 1 :.ab)- ea, be) eld Example 10. For any integer p, show that : (a b)-1 (a, b) =1) From (1) and (2). d=e (a, bc)= (a, (a, b) c). 0 2p+ 1, 9p+4)=1 () (Sp+2, 7p+3)=1 (i) (Gp.3 p+2)=1 for odd p. Sol. (9 Let (2p+1, 9p+4) =d Erample8. Prove for any integer m, 30| (m° -m). d|2p+1 and d|9p+4 Sol Here m-m m (m" -1) d |2 9p+ 4)-9(2p+ 1) m (m-1) (m +1) d |8-9 (m-m)m+1). d1-1 As 6 (m' - m) m-m = (m- 1) m (m + 1) and m- 1, m, m + 1 are three cons But d is positive integers and their product is divisibley 6| (m -m) (m* +1) 6|(m° -m) d =1. (2p+1,9p +4) = 1. SPECTRUM DIscr RETE MATHEM SETS, RELA77ON AND FUNCTI For all posi h f(a b) = 1, then prove that ( (atb, a-b) = I or2 157 ositive integers p and q where p is odd, show that (2P -1,2" +1) = 1. wLet Sp:2. 7p+ 3)-d d 5p+2 and d|7p+3 in (a+b, ab) = 1 d 5(p+3)-76p+2) a2 +b) = I or 2 (iv) (2a+ b, a + 2b) =1 or 3 () (atb, a I or 3 17 Ifla. b)=d then 1 d|l d=tl (i) (a tb, a2+8 But d is positive d-1 (p +2,7p + 3) =1. (tu) Grven p be odd so let p=2k+ 1, kEI 3p 3(2 1)+ I =6k+4 = 1, then (a, b) = c. 18 ifcla c/b 19. Prove that every two consecutive Prove that one of any three consecutive integers is divisible by 3, p+2 3(2k+1)+2=6k+5 3p.3p+2)d integers are coprime. Let ie (64, 6+ 5) =d 20. Ltand be fraction 1fa., an, are integers wncn are teiavey prine in pairs, then prove that (a, ay, ..., )= 1. d6k+4 and d|6k+ 5 be fractions such that (a, b) = (¢, ad) = 1. Prove it + is an integer, then |b|= |d|. d (6k+5)-(6k+4) d1 d=t1 But dis positive. Is converse true ? 23. Prove ( If g.c.d. (a, b) = g.c.d. (a, c) = 1, then g.c.d. (a, b c)= |1. (G) Ifg.c.d. (a, b) = I and c|a then g.c.d. (b, c) = 1 (Git) If g.c.d. (a, b) = 1, then g.c.d. (a c, b) = g.c.d. (c, b). () If g.c.d. (a, b) = 1, d|acand d| be then d| c. d=l p. 3p+2)= 1. EXERCISE 1.12 1. Ifa b) = l and c / a, then (c, b) = 1. 2 Ifa and bare relatively prime, then any common divisor of a c and b is a divisor of c. 3 Ifa b)= 1, then (a c, b) = (c. b) 24. If a and b are non zero integers, prove that gc.d.(2a-3 b, 4 a-5 b) divides b and hence g.c.d. (2 a+ 3, 4 a + 5) = I. 4.Ifla e)=d alband e/b, then show that a c/ bd. ANSWERS S Show that (a b) = (a + b, b). 6. Ifa b)- 1, then (a. b+ ka) = 1. 7. If (a m)=1,then (m- a, m)= 1. 22. Not true. 1.72. Euclidean Algorithm 8. If (a, 4)-2 and (b, 4)=2, then prove that (a + b, 4) =4. For any two positive integers a and b, on applying the division algorithm repeatedly to obtain a set of nainers 'z defined successively by the following relations 9. Ifa>I and n is positive integer such that b| (a" -1), then prove that (a, b) = I. 10. Prove that there are no pair of integers x, y satisfying x +y= 100 and (, y) =3. I1. Prove that (a,b) = (a, b)*. b=aq+ 0sa 0sr a=4+2 12. Ifr and y are prime to 3, then prove that r+y can not be a perfect square. 43+3 13. Let dand g be two positive integers. Prove that there are integers x and y satisfying x T (a. y)=d iff dlg. 14. Let d and p be two positive integers. Prove that there are integers x and y satisfying X .0) **********************" **************** *** ******************* ***************** a, )=d iff 4 1p. n-2n-19, * 0Sr, he n the last non zero remainder in the above process is g.c.d. of a ando. SPECTRUM DISCRETE M EMATHEMAN 15 SETS, RELATION AND FUNCTION Proof: 'e have for any integers i. y (a. 6) =(a. b + ax) = (a + by, b) 2a-792 Using equations () 6aq (a. 6) = (ab-aq,) (a, b) ' -2-197 -(a.r)(a-792.7) Now -2 3 -29n-)9, - (1+9,-9,)n-2 *+ -4,),-3 Continuing like this, we get - (1+9n-9,a-4 a-34n-2)+-9,),-3 (1) = (1+9-19,)4 *9, 4n-2 4n-29-19,) ,-3 2) equation () gives , as linear combination of_ and Equation (2)gives , as linear combination of r and n-4 Continuing like this, we go on eliminating the remainders--2-3 the last non zero remainder is g.c.d. of a, b. 321 untilr, is expressed as a linear combination of a and b. 1.73. Common Multiple, Least Common Multiple on Multiple. Ifa and b are non zero integers such that a | n, b |n. Then n is called a common Cor. 1. Express G.c.D. of a and b as a linear combination ofa and b. Proof: From Euclidean Algorithm, we have b a9 0 a multiple ofa and b. Teast Com mon Multiple. Let a and b be non zero integers. Then a positive integer I is the least a 192+2 common multiple (L.C.M.) ofa and b if or 2 2433 al, b|l ie. lisa common multiple ofa and b. (in al n, b|n ISnfor any positive integer n. Notation: The L.C.M. ofa and b is denoted by LCM. (a, b) or [a, b]. **************** o n-2 n4 -39n-2*n-2 Important Results : n-3 -29-*'- On-1 'n-2 1. Ifaln, b|n, then [a, b]|n. 2. Prove [ka, kb} =k [a, 6] for positive integer k. 'a-2-19,* 0S n n-i 3. Let a, b be two positive integers. Prove that (g.c.d. (a, b)) (1.c.m. [a, b]) = a b. Rewrite above equations as Or n-2n-19n (a, 6) la, b]=ab. n-3 a-24n-1 1.74. Ifa b - c" and (a. b) = 1, then each of a and b is an exact rth power. n-39n-2 Proof. We are given that ...(1) **** ******************************** ab = c" and (a, b)=1 ******* ****************************** OcREIE M NE MATHED 160 TS, RELATION AND FUNCTION Let (a c) a (d c) =D we can take D/d and D/c aaß. c=ay 161 D/d and (a, b) =d where .) D/dand d/ a, dib from (1), we get D/a, D/b aßb= a"y" we have D/ a, D/b, D/c D is a common divisor of a b, c. Let D' be any other common divisor of a, b, c. (6.7)=1 D'l a, D'/b D/ (a b) D'1d (.")=1 Now D'/d, Die ® D 7(4 c) » D'/D :1b and ( (a. b, c)= D from (3), "/b D is g.c.d. of a, b, c. 176. Prove that (m a, m b, m c) = m(a. b, c). Prof. Let (a b) = d and (d c)=D (a. b, c) = D we take b= y'6 From (3) and (4), we get, B'6=a Also (m a, m b) = md ..0) and (m d, mc) =mD (m a, mb, m c)=mD Bo a (a, b)=1 and a/a, d/b From (1) and (2), we get ..2) (a. 6)=1 (a,) =1 m a, mb, mc)= m(a, b, ) from (5), a-B Cor. fmla, m/b, m/c, tnenmmm (a,b, ) m we can take B= al Proof is quite simple. From (5) and (6), we get, ILLUSTRATIVE EXAMPLES 10=1 i=1,ð=1 Example 1. Find g.c.d. of 24 and 138 and express it as linear combination of these numbers from (6), a as = 1 Sol. Here 138 24 (5)+ 18 From (2), a=aß=a a = a" 24 18 (1) +6 From (4), b= as d=1 18 6 (3) +0 24) 138 120 each of a and b is an exact nth power. 6 g.c.d. (24, 138) 18 )24(1 1.75. Explain the method of finding g.c.d. of three numbers a, b, c. And 6 24 18 18 Proof. Let (a b)= 4 6) 18(3 18 24-(138-24 (5)) and (d c)=D 6 x 24- 138 =6 x 24 +(-1) x 138 We shall prove that (a, b, c)= D. = 24x+ 138y where x =6, y=- 1. TE MATNO SPECTRUM DISCRETE M RELATIONAND FUNCTION Emple 2. Find imtegers 1 andy so that 12378 r + 3054 y=6 12216 4. Findx, y such that 71x-50y = I. (71, 50) SETS REL 63 4 Eample 4. F irstly, we shall prove so)71 Sol Firsth show gcd (/2375, 3054) =6. 12378 3054 (4)+ 162 71 50 (1)+21 50 21 (2)+8 21 -8(2)+ 8-5(1)+33 3054) 12378 Sol Here 50 162) 3054 (18 21) 50 (2 Here 3054 162(18)- 138 42 162 162 = 138 (1) +24 5 =3 (1) +2 1434 8)21 (2 I38 24(5)- 18 1296 3 2(1)+ 16 24 18(1)-6 138) 162 (1 5)8(1 2 I (2)+0 18-6(3)+0 138 gcd(71, 50) = 1 5 gcd (12378, 3054) =6 6 24-18 24) 138 ( : (a, b)= (a,-b)] 3)5 ( g.c.d. (71, - S0) = 1 and 120 I =3-2-3-(5-3)=2 x3-5 3 -24-(138-24 (5) Now 18 )24T -2 x (8-5)-5 =-3 x 5+2 x8 2)3( =6 (24) - 138 6 (162-138)-138 --7x 138+6 X 162 =-3 x (21 -8(2)) +2 x8 --3 x 21+8 x8=-3 x 21 +8 x (50-21 (2)) =8x 50-19 x 21 =8 x 50- 19 x (71-50) =- 19 x 71 +27 x 50- 71 (-19)-50 (-27) 71 x- 50 y where x =- 19, y=-27. 18 2 6) 1)2 (2 =-7 x (3054- 162 X 18) +6 X 162 = 132 x 162-7x 3054 = 132 (12378-(3054) 4)-7x 3054 =-535 x 3054+ 132 X12378 = 12378 r+3054 y where r=132 andy=-535 71x-50y= I tor -19, y=- 27. Example 5. Evaluate g.c.d. (198, 288, 512) and express it as a linear combination of r, y, : (integers). Sol Here gc.d. = (198, 288, 512) = g.c.d. (g.c.d(198, 288), 512) 288 198 (1) +90 198 90 (2)+ 18 12378 r+3054y =6 forr= 132 and y=- 535. Eumple 3. Use the Euclidean Algorithm to find integers x and y such that gcd (1769, 2378) = 1769 r + 2378y 2378 1769 (1)+ 609 Here 198) 288 Sol Here 90 18 (5) +0 198 1769 609 (2)+551 1769) 2378 (198, 288) = 18. 90) 198 (2 609 551 (1)+58 1769 609) 1769 (2 = 198 90 (2) 180 551 =58 (9)+29 58 29 (2)+0 18) 905 = 198-(288 198) (2) 1218 gc.d. (1769, 2378) =29 551 ) 609 (1 3 (198)-2 (288) 90 .1) Now g.c.d. (198, 288, 512) =g.c.d. (18, 512) 29 551-58 (9) 551-(609-551) (9) and 551 28 Here 58)51 S12 18 (28) +8 18) 512 10 x 551 -609 x 9 52 18 8 (2)+2 36 152 10 x (1769609 x 2)-609 x 9 9 8 2 (4)+0 gcd. (18, 512) = 2. = 10 x 1769-29 x 609 144 =10 x 1769-29 (2378- 1769) Hence g.c.d. (198, 288, 512) - 2. 8) 18 (2 39 x 1769-29 x 2378 16 = 1769 x + 2378 y where x=39, y =-29. 2 18 8(2) = 18-(512 - 18 (28)) 2 SPECTRUM DISCRE MAM STTS REL4I1OVA V ANSWERS 2. 11, 11 = 726r + 275 y 164 165 (3 (/98) -2288)) 57- 2 * 512 2xS12- 114 x 288 171 x 198 - 18 x $7 2x S12 252 mt 395 m 5. 14, 14 826 r + 1890y 3. 13,13 35%1+ 325y - SI2-2)+288 (- 14)+ 198 (171) - 512r +288 y+ 198 : where r=- 2,y5- 114 and: = 1 71 & 47y=-22 n) r = 94, y=-9 11 = l 109 r + 4999 -22 (i)* =41, y--9 (i)x - 7. y = - 3 (v) 22., y = - 15 8. (a. b, c) = 13. 9. x = 11,y = - 11,z =3 10. 3838 gcd -2- 512r + 288y+ 1982. 1 - 36, y= 71 Sol Firsty find g cd (657, 306) By Euchdean algorithm Example 6& Evaluate le.m. [306, 657) reater than I is called a prime number if it has no proper divisor 17. Prime Number sitive integer greater Or 657 306 (2) +45 >l is called prime number if it has only two divisors 1 and itself 306 =45 (6) + 36 Any positive integer>I is called prim 17, 19, .. are prime numbers. ny positive integer greater than 1, which is not prime, is called a composite 45 36 (1) +9 23,5,7, 11, 13, 1: 36 9(4) +0 lumber. Any positive Composite Num gcd (657, 306) =9. number. 46,8,9, 10, .. are composite numbers. Now using (a, b) [a, b] =ab ab [a.b(a.b) divide natural numbers into three classes we have Unity 306 x657 34 x 657 =22338. (i) Prime Numbers [306, 657) =. 9 (i) Composite Numbers EXERCISE 1.13 Note l. 1 is neither prime nor composite. I. Find the gcd. of 595 and 252 and express d in the form 252 m+ 595 n. 2. Find the g.cd. of 726 and 275 and express it in the form 726 x+ 276y. 3. Find the gcd. of 858 and 325, and express it in the form 858 x + 325y. 4. Find the gcd. of | 109 and 4999, and express it in the form 1109r+ 4999 y. 5. Find the g.cd. of 826 and 1890, and express it in the form 826x + 1890 y. 6. Find x,y (integers) satisfying Note2 2 is the only even number which is prime. Note 3. Any number, which 1s not prinme, iS not necessarily composite. Two prime numbers are called twin-primes ir there is only one composite number between them. eg, for any odd integer p; p and p +2 are Twin Primes. eg, 3,5; 5,7; 11, 13 ; 17, 19 are twin primes. 1.78. Prove that the least divisor (other than 1) ofa composite number is a prime. ( 6409x +42823 y 17 (i) 256 x + 1166y = 2 i) 119x +272y = 17 Prof: Let us take n be any given composite number. (iv) 68 x+710y =2 ()657x+963 y = 9 there is atleast one divisor of n other than I so n has a least divisor q such that 1 1 has a prime factor. sPECTRUM DISCRETE MA AND FUNCTION I.79.Ifp is any prime and m any integer, then prove thaf either (p, m) = 1 or pln Proof: Let us take gc.d. (p. m) = d dlp and d|m Now dlp d=l or p ELATION AND FUNCTI m. SETS A : Given PlPiP2 16 ********** Pn cor 1, PlP for some , I stSn only divisorsa are 1, P Prool Then by cor 67 If d=1, then g.c.d. (p. m) = If d=p. then djm pim As P, is pime so its on/v. = P Butp> I so that p = ience either gcd. (p. m) =| or p|m. Hence proved 1, Euclid's Theorem ve that number. 1.80. Ifp is a prime and plab, then prove that pa or p|b. Proof: Given that p|a b er of primes is infinite Pia, then theorem is proved. If pta then gcd (p, a) = 1 pr+ay = 1 for some integers x,y Multiplying both sides by b, we get bpr+bay =b end to the sequence of primes 2, 3, 5, 7, 11, 13, 17, .. ible, suppose that numbe of primes is finite. Let p be the greatest prime. ie there is noe roon. If possible, suppose nsider an integer N defined as ********************* * N = (2 x 3 *5 * ...... Xp) +1 Given plab plaby plbpx fN is prime, then N>p. composite, then it has Nhas at least one prime factorgreater thannp. PIP dhen it has atleast one prime factor. But none of primes from 2 to p divides N. Also plbpr+a by) p|b Henceeither pla or p|b. Cor. 1. If p| a,a.. , then prove pla, for some 1, 1 Si Sn ie p divides atleasta cases, 3 a prime which is greater than p, which is impossible. in both Proof: We shall prove this result by induction on n. our supposition is wrong. number of primes is infinite. Step 1. If n=1, then p|a and there is nothing to prove. Ifn=2, then p| a,a, pla or pl02 Note 1. The above theorem can be set as Out side any given set of primes, there is another one". duct of two numbers of the form 4 n+ 1 is again of the same form. result is true for n=1,2 Step 2. Suppose result is true for n = k Note 2 (4m+ 1)(4 n' + 1)= 16nn' +4n+4 n'+1 Lepl a,a Pla, for some , 1 SIsk. = 4[4 nn' +ntn']+1 Step 3. Suppose p| a,a2. ... +1 = 4k+ 1, where k = 4nn' +n +n' Plba+1 where b= a@2.. Plb or p| ak+1 Note 3. Any number of type 4 k-l is of type 4 k +3 Pl442 ag or Plk+1 p 4k-1 = 4k-4 +4-1 =(4k-4) +4-1 Pla, or pla+1 for some , 1 sIsk 4(k-1)+3 ( ofs Pla, for some , 1 S t k+1 result is true for n=k+1 Hence by mathematical induction, result is true for all n. = 4k+3 181. Prove that primes of the form 4 k +3 are infinite. Prol. f possible, suppose that number of primes of the form 4 k+3 is finite. Remark: The above result holds only if p is prime e.g, 6 | 12 i.e. 6 |24 x3. But 6{3 and b {4. Letp be the greatest prime of this type. Cor. 2. Given p, P1. P2. P, are all prime numbers and pl pP2 P then prov for some t, 1 SiSn. Consider a number N defined as N 2(3x 5x7x ..... p)-1| Clearly N is of the type 4 k- I and hence of 4 k +3 type. that sPECTRUM DISCRET MATHD Ir Nis aprime number, then certainlyN>p and N is of type 4k+3. TN is composite, then it has a prime factor of 4 k+ 3 type. N AND FUNCTION SETS, RELATIONA Now ither A is even or A+l is even A +1) is even 168 ain of the sa k+ 1 is ag 169 product of two numbers of the form 4k+1 But none of the primes 2, 3, 5, .pdivides N. this prime divisor of N is greater than p. Let a+ )=2k greater than p, 4(2 k) +1 = 8k+1. nd is that there are infinit initely many prime numbers of the form 8 +1. Proof: Let, ifpossible, set of Let pbe the largest possible 3.32.52-72.11 132..+ in both the cases, we have got a prime which is of form 4 k + 3 and 1.84. Prove that e ssible, set of primes of the eform 8 k+ 1, kEZbe finite. mpossible. prime of the form 8 +1, kEZ. Consider N be any integer our pposition is Wrong. Cor. Prove that number of primes of the form 4 k- 1 are infinite. Proo. Primes of the form 4 k-1 are of the form 4 k+ 3. 1.83. Prove that primes of the form 6 k+ 5 are infinite. Proof. If possible, suppose that, primes of the form 6 k *S are finite, and let p be t ype Let N=2 x3 x5 x... P-T number of primes of the form 4 k + 3 is infinite. Since square ofev ofevery 8k+1. odd integer is of the form 8k+ 1 and product of integers of the form 8k+ 1 is again pbe the greates an integer of the förm 8 Prine 2.52.72 114 13 ..p' is of the form 8 k+ I and let it be k'. 12.132 p) is ofthe form8 Clearly N is of 6 k- 1 type and hence of 6 k + 5 type If Nis prime, then certainly N is of type 6 k + 5 and>p. If Nis composite, then it has at least one prime factor of6 k + 5 type. Nis of the form 8 K +1 Nowe either N is a composite number ber or N is a prime number. omnosite number, then clearly none of the odd primes 3, 5, 7, 11, 13, .p divides N. Also as N But none of the primes 2, 3, 5,. p divides N. Nhas got a prime factor of 6 k+ 5 type and which is >p. in both the cases, there is a prime of6 k * 5 type and > p, which is impossible. IfN is a composite s an odd number, not divisible by 2. N has odd prime factor>p of the form 8 k+ 1. Further if N is a prime number, again we get a prime number >p. in both the cases, we get a contradiction ie. our supposition is wrong Our supposition is WTong. number of primes of the type 6 k+ 5 is infinite. Cor. Primes of the form 6 k-1 are infinite. Proof. Primes of the form 6k-1 are clearly of the form 6 k + 5. Note 1. Ifa and b have no common divisor> 1, then every odd prime factor of a +b* is of 4n+l185. Prove that primes ofthe form 8k+5 are infnits Hence the set of primes of the form 8 k + 1,kEZ is not finite. there are infinite many primes of the form 8 k+1. Example. Take a = 5, b = 12 Proo. f possible, suppose that number of primes of type 8k+5 is finite and let p be the greatest prime of a andb have no common factor s type. a +b 25 + 144 169 = 13 x 13 (4K + 1)4k+ 1) = 4n+1 Consider N= 3.5.72.. p+ 22 Note 2. Prove that square of every odd +ve number is of 8 k +1 type. square of every odd number is of 8 k + 1 type. Proof. Let n be any odd+ ve integer. n2A +1| each

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