SPECTRUM DISCRETE MAT 132 MATIENM SETS, RELATION AND FUNCTION 133 result is true for n= k+1 SECTIONV result is true for n= k, then it is also if the true for n = k+ 1. true for = 1 INDUCTIO is n result PRINCIPLES OF MATHEMATICS But the hy the method ot induction, the result is true for allnEN. P r o v e that 52n* that s2n+2 24 n25 is a multiple of 576. 1.62. Statement of Principle of Mathematical Induction Example2. Let P (n) be a statement such that P n ) =5èn*2 24 n 25 Let Sol. (a) P(1) is true P(1)= 52425 625 2425 = 576, which is a multiple of 576. () P(r+ ) is true whenever P()istrue where n =r. result is true forn= 1. Then P (n) is true for all natural numbers n. result is true for n =k. The WellOrdering Principle Assume that Every nonempty set of nonnegative integers has a least element. P() 52k*2 24 k 25 is a multiple of 576. Recursive Definition: Sometimes we find it dificult to define an object explicitly. However, may itma 52k+2 24k25 576 1, where l is an integer easy to define this object in terms of itself. This process is called recursion. Let We can use recursion to define sequence, functions, and sets. In previous discussions, we specified 52k+2 5761+24 k +25 terms of a sequences using an explicit formula. . ..(1) For example: We define recursively defined functions: We use two steps to define a function withthe Now P(k + 1) = 54*)*24(k +1)25 5.5k*2 24 k2425 ofnonnegative integers its domain: 25(576 1+24 k as = + 25) 24 k2425 ( of (1)] Basic step: Specify the value of the function at zero. = 25.576 l+2524 k+ 625 24 Recursive step: Give a rule for finding its value at an integer from its value at smaller integers. k  24 25 57625l+ 576 k+ 576 = 576(25 Il+ k+ 1), which is a multiple of 576. result is true for n = k+1 ILLUSTRATIVEEXAMPLES i f the result is true for n = k, then it is also true for n =k+ 1. Example 1. Prove that 9" 8"  1 is divisible byy8. Butthe result is true for n = 1. by the method of induction, the result is true for all n EN. Sol Let Pin) = 9" 8" 1 Example 3. Use the principle of mathematical induction to prove that P(1)=981 =0. which is divisible by 8 result is true for n= 1 P+2+23+.  "7*9 VnEN. Assume that result is true for n = k 4 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? == " ...(1) 4 9 81+8* +1 For n=1, Now P(k 1) = 9**8*  1 = 9.9 8k*11 L.H.S. 1 = 1 9(81+8 +1)8**l 1 of( 7 2 +9.8* +988 1  72/+8* +8 RH.S. 0D   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 134 SETS RELATI 135 =k Assume that result (1) is true for n (k+D (2+7k +6)  2+3k+4k+6] k*l (K2 k+ 3) = + 22 k 3)) .+k  k*k+1)2 22 *D&+2)2k+3) &6 [k +2) (2 k+ 3)] = 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)36+), (k+1? =k+1 result is true for n ifthe result is true for n =k, then it is also true for n = k+1. true for n = 1. resut is But the uby method of induction, the result is true for all n E N. (k+n3 &+*(k+2 P+2+2+. added (k+1* to both sides = Note. have of (2). This is obtained by changing n to k+ 1 in the last of(). result is true for n =k+1 em ofLH.S. principle of mathematical induction to prove that =k+ 1. 5. Use the ifthe result is true for n =k, then it is also true for n Erample "0*)(2n+7) But the result is true for n = 1. 13+2.4+3.5+..+n(n +2) = nEN. 6 E N. by method of induction, the result is true for all n that have to prove Sol. We nn +1) (2 n+1) Example 4. By mathematical induction, prove that1f+2 +3* +. 6 VnEN 13+2.4+3:5+ . . + n{n +2) nn +1)(2 n+7) 6 ..0) Sol. We are to prove that For n=1, 1+2+32 = n(n+1) (2 n* VnEN. LH.S. = 13=3 6 For n= l1, R.H.S= 11+1)2+7) Ix2x9 3 6 6 L.H.S. = R.H.S. LH.S.=1=1 result (1) is true for n 1. = RHS 0D2+) 6 2x3 6   =1 Assumethat result (1) is true for n =k. L.H.S. R.H.S. 13+2.4+3.5+ k(k + 2) =*(K +)(2k+7) result (1) is true for n =1 ...(2) Assume that result (1) is true for n =k Adding(k+ 1) k+3) to both sides of (2), we get R(k+)(2k+1) 13+24+3.5+.. + k(k+2) +(k+ 1)(k+ 3) *2" +3+. k2 = 6 k(k +1) (2 k+7) 6 (6+ D&+3)=+ D 2,k+3 Adding (k+1) to both sides of(2), we get 2+ (k+)2 k\k+D2k+)+4n2= k(k +)(2k+)+6h + 27)+6k+18+ 2k+13k+18&+ nk+2)2k+9) +2+3+ 6 = 6 6 6 6 6 6 SPECTRUMDISCRETE ATIWEM .RELATION AND FUNCTION 137 136 result is true form=k+ it is also true for n =k + 1. EXERCISE 1.10 for n =k, then principle of mathematical induction to prove that o. result is true Use the ifthe 1 231 isdivisible by 7VnEN for n = l. (in 102+1 is divisible by 11 Vn EN is true But the result the result is true for allnEN. bymethod of induction, mathematical induction to prove that 2m+28n9 is divisible by 64 Yn EN of Use the principle Example 6. 2ne Vn EN. dhe orincipleofmathematical induction to prove that .L. nD2n+) by 8Vn¬N. is divisible (in 10"+3.4*2+5 is divisible by 9 VnEN. 1 3 35 57 321 Sol. We have to prove that ih62+7n* is divisible by 43 Vn EEN. * (2n1)(2n+1) 2n+l principle of mathema induction to prove that the 1'35'57 3. Use For n 1 x+4x+7x+.. ...+(3n2)x={3n1)x Vn EN. LHS 4 Bymathematical induction, prove that 1+32+52 .+(2n12 = 47=9n+) ynEN 3 R.H.S. Prove by induction that 1+ 1+2+3 2n VnEN. L.H.S. R.H.S. 5 1+2+3+.+n n+l =1. result (1) is true for n 1.63. Division Algorithm for n =k. Assume that result (1) is true Statement. For any two integers a andb, with a > 0, there exist unique integers q and r such that T k baqtr, 0sr<al 2k+l ( 2 k1) 2k+1) Proof. When b=0 to both sides of (2), we get have Taking q 0, r=0, we Adding [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 (2k1)(2 k +1) (2k+1) (2 k +3) When b 0 1 Consider the infinite sequence of multiples of a i.e. ...,2a, a, 0, a, 2 a, ... 2k+1 (2k+1)2k+3) 2k+112 Let aq be the greatest multiple of a such that b2aq and b<(q+ 1)a 23k+ (k+1)(2k+1) aqsb<alq +1) 0sbaq<a 2k+ 2k+3 2k+1 2k+3 Put ba q=r 0sr<a 13 3.5 5.7 (2k+1)(2k+1) (2k+1)(2k +3) 2k+3 baq+r, 0 sr<al ..(1) a>0] result is true for n = k + 1  i fthe result is true for n=k, then it is also true for n = k +1. Uniqueness lf possible, let there exist another set such that But the result is true for n = 1. ofintegers q and by method ofinduction, the result is true for all n EN. ag+ 0s <al (2) SPECTRUM DISCREIE MA. FUNCTION RELATIONAND 139 138 SETS, From ( ) and (2), we get s  a l , then 2 aqtr=dt aal ia<aa 2 a dividesr7 I 0 a<0 But r Id Sr<a, 0s , r 0 where lal or =r+a. =0 as a0 g)=0 > 9, 9 a trom().dl from (1), b = aq1 + r +  a,  Sr<0 . = aq t al+r and b, w ith a> 0, there exist unique integers q and r such that for any tO integers a a (q t 1)+r [: lal = a] b =aqr. 0 Sr<ja Cor 1. If and b are any integers, with a = 0, then there exist unique integers q and r such that =aqtr (say),  sr<0 a .(3) b = aq+r. 0 Sr<ja have Combining (2) and (3), we Proof. We have proved the result for a>0 a if a0, then aj>0 and such that b =aqtr, srla there exists unique integers q r 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<jal EL. qa+r, 0 Sr<a where q= 92 bh,a,4,rare called dividend, divisor, quotient, remainder respectively. b =aq+r. 0Sr<a (i) The remainder is zero iff a divides b. Hence the result. Cor. 2.If a and b are integers with a 0, then there exists unique integers q and r such that ILLUSTRATIVE EXAMPLES Lample 1. Prove that fourth power of any integer is either of the form 5 k or 5 k + 1. b=aqrr, Sr SoL Let b be any integer, divide b by a Proof: By Division algorithm, for any integers a 0 and b, 3 unique integers , Sucn tnat By division algorithm, we have b = ag + . 0snal. aqtr, 0sr<a, for q,rE Z Put a= 5 Now 0s <la a b 5q +r, 0 Sr<5 If 0s a on taking 4 4, =rin (1), we have b=5q,5 q+ 1,5q+2, 5 q+ 3,5q+4 Everyinteger is of form 5 q, 5 q+ 1, 5 q+2,5qt3, 5 q*4 If b baqtr, 0 sr< henb 625 q =5(125 qt)=5k; k= 125q* EEZ. SETS, RELATIONAND FUNCTION SPECTRIMDISCRETE MA 140 integer b b e any Let Ifb 5q I then 6  (5y+)° sol. Wehave b=aqtr, s r . rez and qk in (1), we have c,5q* *C39)+1 P u t t i n g a 3 sc c,s° () SAI (sa). & E Z. b3ktr sr If b= 5q 2 then  5q+ 2) b =3 k+r, r= 1,0,1 b 3k1, 3k,3k+1, kEZ every integer is of form 3k1, 3k, 3 k+ 1. scs C25)2q')+ °C,(20)q+ "C, 8q+ 3) +1 =5k+1 (say), kEz and 4 *kin (), we have (m Putting a=4 If b5q 3 then b* =(5q+3) b 4k+r,  b4k+r, r=2,1,0, 1  s(tc,s. tc,s2g'3). tc,5q 0)+ "C3 q27)+16)+1 5k+1 (say). kEz b 4 k2, 4 k 1, 4 k, 4 k+ 1. k EZ If b 5 q4 then b" = (5q +4) e v e r y integer is ot form 4 k2,4 k 1,4 k, 4 k + 1 c5 c,sg'(4)+c,(5q)?42 C,(5q)4*+*c,4* (Gn Putting a=5 and q =kin (I), we have s c s  'c,s*g (4) fc,5q°0)C, a(64)+51) +1 5k+ 1 (say). kez b Sktn,sr< is either form 5 kor 5 k+ 1. Hence fourth power of any integer of the b 5k+r, rs2, 1,0, 1,2 with b> 0. then there exists unique integers q and Example 2. Prove that if a and b are integers, satistying a=bqr, where 2 b sr< 3b b 5k2,5 k 1,5 k, 5 k+ 1,5k+2. kE Z Sol. Given a and b are integers with b >0. every integer is of form 5 k2, 5 k 1, 5 k, 5 k+ 1,5 k+2. By division algorithm, we have ()Putting a=6 and q =k in (1), we have a= bq +r whereq and r' are integerS and 0r'<b b 6k+r, s r a bg2+2)r b =6k+r, r=3, 2,  1,0, 1,2 b q 2 ) 2 b+r' =bq +r where q=q2 andr =2b+r are integers b 6k3,6 k2,6 k1,6k,6 k+ 1,6 k+2, kEZ. a n dg . r are unique. every integer is of form 6k3,6 k2,6 k 1,6 k, 6 k + 1,6k+2. Also 0 r <b i e 4.If m is an integer not divisible by 2 or 3 show that 24  :* +23. 2 b sr2bs3h 2b sr<3b Ely, divide m by 6 and let q be there exists unique and r such thai so that quotient and r be remainder. a= bq*r, where 2 bsr<3 b. m 6q+r, 0 sr<6 But r 0 , 2,3,4 Example 3. Show every integer is of form [:m is not divisible by 2 or 3] 3 k1,3 k, 3 k1 (i) 4k2, 4k1, 4k,4k+ 1, 1,5 (iin) 5k2, 5 k1, 5k,5k 1,5k+2 S0 that m 6 k2,6 k 1.6 k, 6 k+ 1,6k+ 2, where kE Z =6q +l or 6qts (iv) 6k 3, SPECTRUNM DISCRETEMA RELATIONAND FUNCTION 142 14 When m 69 1. Then en n=3q+2,Then nn+2) (3q+2) ((3q+2)* +2) 3 When m23(6y+)* 23 369 129 1 23 6 q* +12q+24 12 (3q+ q+2) 12(9+q2 3q+2)9 q+12q+6) 34+2)3 3g4q+2) 3 12 i91)2e*+)] = (3 q+ 2) 3qt+4q+ 2) 9.91 arr oonsecutiveintegers. An integer q(q+ 1)+ 2(q* +1) is also even integer EZ] q + 1) is even integer n i s an integer for all n E N. 29g1)+2q+1 Hence 24 m' +23 122 When m 6q+5. Then 12(g(g +1)+2(9*+)) form: EXERCISE 1.11 that every integer of is 2 3 (6q+5)* +23 1.Show 36q+60q+25+23 =36q +60q+48 (0 29,2g+1 (n 3q.3 q+1,3 q+2 (ii 4g, 4q +1,4q+2, 4q+3 (v) 5q.5 q+1,5q+2.5q+3,5q+4 +q+2gt +4q+4) 64,6q+ 1, 6q+2.6q+3,6q+4,6q+5 where q¬Z 12(3q5q+4) = 12 (g . Show that b leaves the remainder 0 or I when divided by 4 for any integer b. 12(q (q+ 1)+2g+2q+2)) Find remainder r, when 1059, 1417, 23 12 are divided byp> 1. 9.9I are consecutive integers= q(q+ 1) is even integer qq ) + 2(q+2q+2) is also even integer. ANSWERS 2 qq 1) 24g+2q+2) 12.2 12 (g (q+ 1)+ 2(g+2q+2)) .164 L64. Common Divisor, Greatest Common Divisor 24 m23 (: of (2)) Common Divisor: Ifc  a and c  b, then c is called a common divisor of a and b. Hence the result Note: As there is only a finite number of divisors of any non zero integer so there is only a finite number Example 5. Show that 2is an integer for al n EN. ofcommon divisors of a and b except when a b=0. If atleast one of a and b is non zero, then the greatest = amongtheir common divisors is known as greatest common divisor of a andb. Sol. By division algorithm, ifn is divided by 3, then Greatest Common Divisor Let a and b be two lhen positive integers such that atleast one of them is non zero. n =3 q+r, 0sr<3, ,rEZ integer d is the Greatest Common Divisor of a and b if n =3q+r, r=0, 1,2 n = 3 q. 3 q+ 1,3q+2 0 da, d b ie. dis divisor a common of a and b. nn+2) 3g9g *2) q942 +2) An integer cla,c b c s dfor any positive integer c. When n=3 q, Then abon:(0 The g.c.d. of a andb is denoted by g.c.d. (a, b) or simply (4, D). 4,6) is defined for each pair of integers a and b except when a =0, b=0. n +2) Gq+D(Gq+1)2 +2) Gi) (a, b) 2 . n _ When 3 q+ 1, Then 3 Torexample : To find g.c.d. of 12 and 48. g)9g6q+3) (3q+1)3(G4+24+) 3 3 he positive divisors of12 are 1,2,3,4, 6, 12 The positive divisors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 =(3g+1)(3q+2g+1) thecommon divisors are 1, 2,3, 4, 6, 12 C.d.of 12 and 48 is 12 i.e. An integer 12, 48)= 1 SPECTRUM DiSCRETE ETE MAT AND FUNCT CTION 145 144 that g.c.d ofa and b exists RELATION two integers, not both zern, then prove and is unig SETS, andlv, we want to show g.c.d. (a, b) is unique 1.65. Ifa and & are any 6) exists. Second that gcd. (a, shall prove well d be g.c.d. (a, b). Proof: Firsty, we and & both are positive and b 2a tf possible, let d as as that & WIog assume ted by signs of [ g c d . (a, b) is not affected by si a d, and d2 are also common divisors of a and b. by division algorithm d 2 d2 d , gcd. (a, b) and d, is divisorofa, b] where 0 s n <a a9+ and d d : d, gcd (a, h) and d, is divisor of a. b] and g.c.d. (a, 6) = a ab If 0then a d, d2 Hence g c d (a, 6) exists g.c.d. (a, b) is unique. It 0. again by division algorithm a 92+2. 0s < L66. 1fd= (a, 5) where a, bnotboth zero, then prove that 3 integers and ysuch that value of xa+y b. x d=ax+ by and d least positive isthe f 0. then a = 92 a Consider the set Proof: Putting value of a in (), we have A {ax+ by ax+ by> 0; x,y EI} 792)9 + 9291+1) As a,b are not both zero W.lo.g. let a*0 * a >0 i f a>0 Thus a and b where x= and aIal a=a.x+b.0 . 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 Thus dax+by for some integers x, y. 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 Divide a by d, so by division algorithm 3 q,r EI such that after th step aqdtr, 0 sr<d So, we get a sequence ofintegers r, (1 sj s n) such that a  q d = a  q (ax+ by) = a (1 4x)+ b  4 ) n0S <2 <<a ax +by where 1qx= ,  4y= y and 'n2n19,t n23 fr>0then rES which is a contradiction to the choice of d. 0 so that a = q d 9n+ dla nn1n'n2 aand , lb Similarly on dividing b by d, we get d\b Further, if p is a common divisor of a and b then : disa common divisor of a and b. p b,pla plba9 P\ COnsider c be any other positive common divisor of a and b P2.P ela and elb sothat g.c.d. (a, b) = r clax+ by cld»cSd csd Hence d= Hence g.c.d. (a, b) exists. g.c.d. (a, b). sPETRUNM DISCREH , RELATIONANDFUNCTIC 146 of r a+ yb 147 valuc positive of G.C.D NOw will prove thal d is the least Definition we form x a + yb. Other other member ofthe 67 L67. s o t both not bb n both zero, the g.cd.(a, b) is Let  nbbe any 4,, a defined as the smalest u integers Det.For positive integer of type ax + by Sunce ( ) EZ. dn6) d S (ma +n b) a.b not both zero, the g.c.d.(a, b) d where dis positive = a andd Del. 0 integer satisfying *yb. dla, d1b (in If cla, cb Then  d. dis the least positive value of ra c both zero, then prove that a positive integer If alk and b k. Then(a, b)k Cor. 1. If u and h are tho integers not Result: da, db d=a,b) > db ) if claand cbthen cd mof: Let dgcda. A) iff () da. b]k Given a & da and db. Proof: Fiursthy. let d=gcd (a. 6) ( a , 6)  . sothat dk 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, . a m is positive integer a p c and b=qc where p.qEZ satistying b) 3 x.y EZ such that d d a, for S i sm so = Let (a, (i) If cl a, for I siSm, thencd. dar by =pcx+gcy Note: G.C.D. of aj, 02., m is denoted as d=(a,. as m E Z c(pr*qy) =cr wherer =px+qy L68, Prove that (a, b) = (a, b * ax) =(a + by, b) Vx,y ¬ Z. d =cr cjd * c(a, b). Proof: Let (a, b) =d and (a, b+ax)= d, and (i) Comversely: Let dbe positive integer satisfying conditions (i) (a, b) = d dla and db * dax and db Vr EZ By condition (i) if c is a common divisor of a and b, then d (b +ax) d l a and d (b + ax) dcsd d2 c dis g.c.d. of a and b. d  (a, b+ ax) Hence the result. Cor. 2. Ifa and b are given integers, not both zero, then prove that the set d d (1) And (a, bt ax) d = B ax byx.yE Z} consists ofall multiplies of g.c.d. (a. b). Proof: Let d=gcd (a. b) dla and d, (b + ax) d ax and d, (b+ax) dia and d b d ax + by for allx, 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 (2) As dis the smallest positive integer ofthe form ax+ by. 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 md EB where 2.y2 EZ (atby, b)= d2 each multiple of d is an element of B (4,b)=d » da and db » dla and dby Vy EZ Combining (1) and (2), we see that the set B consists of all multiples of d. datby da+ by and db d (a+ by, b) Hence the result. d d ..3) SPECTRUM DISCRETE MAn 148 MATN RELATIONAND FUNCTION 149 d SETS,R by. b) = and (a For a, b any two integers, not both zero and m any positive integer, prove that b d, la+ by and d, by d, la+ by and d m b}= mla, b) (m a, and da (m a, b) d ja+by by d and m Let d= (a, b) d (a. b) d, Id Proof: and da d b prove d md. We wantto (3) and (4), we get From As d=(a, b) dla and db = d V y¬Z. mdjm a and m dmb (a. 6) =(a +by. b) md (m a, m b) (a, 6)(a. b+ar) = (a + by, b) Vx,yEZ prove that mdi d 1.69. For a. b any two integers, not both zero, .(1) b) =a,b). Eurther as d= (a, 6) so there exist ntegers x, y such that (a. b)=(a. 6) (a. =  Proof: Let d= (a. b) and d =(a, b) d ax+ by d a and db md =m (ax+by) As d (a. 6) d  a and db md (m a)x+ (m b) y (2) d(a, b) d d Also d (m a, m b) Further, as d a,b) d m a and d mb da 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 Similarly (a, b) = (a,  b). d md Further to show (a, b) =  a ,  b) m a, m b) = m (a, b) Let d(a, b) and d, =a,b) Hence the result. As d (a, b) dla and db COr. 1. For any non zero integer m, prove that (m a, m b) =m(a, b). d  a and db Proof: If m>0, we have d(a, b) m a, m b) (a, b) =m d d2 (m a, m b) =m (a, b) If m<0, then (m> 0 m= m] And d a, b) m>0 d2  a and d21b m a, m b) =(ma, m b) d2a and d2 1b  m (a, b) d2 1(a, b) d2ld =m (a, b) :m<0 m =m] for m From (3) and (4) d= d2 (a, b) = (  a ,  b) 0, we have (a, b) = (  a , b) = (a,  b ) = (  a ,  b ) . m a, m b) Hence =  m(a, b). uETE M. SPECTRUM DISCRETE MATMEM AND FUNCTI1O 1ON 151 RELATION SETS, d =e 1SU From (1) and (2) 0. Prove (a, b, c) =(a, b), c) and d> Cor. 2. If dla, db (a, b, c) (a, (6, c)=((a, c), b) Similarly Hence t h e r e s u l t . 1. Prove 6). Furthermoreif d, =(a. Since (m a, m b}= m{a, b) for positive integer m . 1LLUSTRATIVE EXAMPLES Proof: Example .Ifa =bq + r, then sho that (a, b) = (b, r). d, (6, r) = d2 Let(a, b) = Sol. (a, b) = d 4/6 d ( a  b q) d,Ir » (a. b 4/a. r Now db and d  d is a d,/ d, as d2 common divisor is the g.c.d. of b andr of b andr. Hence the result. d >0. Similarly d, /d Further given d (a, b) Now d,/ d2. dz l d and d, d2 are natural numbers. d= d in (1), we get Putting . ) 4) (a. b)=(b, ). Note. If(a, b)= 1, then (6, r) = 1. Example 2. (a, m n) = 1 if and only i f (a, m) = 1 and (a, n) =1. Sol. () Assume that (a, m) = 1, (a, n) = 1 (a,m)=1 Hence the result. there exist integers x and y such that and d= (a, b, c). ax+my=1 1.71. If a, b, c are any integers, no two ofwhich are zero (1) Prove d (a, b, c) = ((a, b), c) = (a, (b, c) = ((a, c),. b). (a, n)=1 there exist Proof: Let d=(a, b, c) and e= ((a, b), c) integers z and t such that dla, djb, djc aztnt=1 aztn(l1) =1 d (a, b) and de d1 ((a, b), c) dle az+nax +my) =1 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 e la, el b and elc n) s 1 where r=z +ntx, S =ty = (a, m n) =1 e l (a, b, c) eld SPECTRUM DISCRETE MAT FUNCTIONn 155 RELATION AND ETS, 152 m n) = 1 dla and db that (a. that d (a, b) d11 d:1 Assume such (ii) integersu and v so that exist there d1 au (mn)v =1 But d is positive so that (c, a)=1. =1 Further let (c, b) =e aut m( v) (a,m)= e lc and elb ela+b n) =1 Given Similarly(a m)=1 e la+b and elb Note. Let (a 1 a)= 1, (m, a) e la+bb ela = we have (m, (m.a)=1 (a,m) =1 so thar e l(a, b) el e1 1 Bute is positive 1. (m, a) = (m, a) = Now 1. > (c, b) = sothatel (m.a) =1 (c, )=(c, b)= 1. and so on. Hence the result. get Example 5. If(¢, a)=1, then prove that (a, b c) =(a, b). we Proceeding in this way, =l or ( a ,m) =I Sol. Let(a, b) = d and (a, b c) = e (m,a) 1, then d la, db and ela, elbc ~ dla, dbc if(a, m)= d I(a, b c) 1. (a, m) =1,.. (a, m) = 1. (a, m) = e ..(1) c/d and (b, d) = 1, then prove that (a, c) = 1. Example 3. Ifal b, e l a and (c, a) =1 Sol. Sincealb,cld. there exist integers m, n such that (e, c)=1. e lbc elb b =am, d = cn ela and eb 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. a mx+ cCny =l Example 6. Prove that there are infinitely many a ( mx) + c n y ) = 1  pairs x, y satisfying x+y= 100 and (r, y) = 5. (a,c)=1 Sol. Given x+y 100 ...(1) (c, b) 1. Example 4. 1f(a, b) I and c (a + b), then prove that (c, a) = = = Let x= 5p where p is odd integer such that (p, 5)=1. Sol. Let (c, a)=d From(1), y=100x = 100 5p 520p). de and da We claim (x, y)=5 Given cla+b Let f (x,y)= d dz5 (: 5x and 5 y] dla+b and dla possible, suppose d> 5 d a+ba db S E T S ,R E L A T I O N AND FUNCTION ANDDF mm (m  (m+1)m my(m45 Also d r ' d l 100 (mm) (m' 4) 5(m + 15 100. ) S0. m (m d10 20, 25, = (m 2) (m  1) m(m+ ly .2. m1, m, +1,m+2 are 5 consecutive )(m+2)+ 5m m) 4)+5) AS 5 (m2) integer so that (m 1) m(m +1) (m+2) must h r o d d 5 5 (m m) Also 2 5 5p 5 I(m2) (m  1) m (m+ 1)(m 5 5iP +2)+5 (m.m) 5) = 1 )=5. (p. impossible as d= 5 5 m° m which is is wrong sothat 5)=1 : of2)) (p. (3), we get supposition ur that satisfyino and pairs x, y infinitely manypsuch infinitely many From(1) thereare so and 6x 5 As many r m m as (6, 5) = 1 there are infinitely [: a\n, bin (a, (a, b) c). 5. 30 1(mm). abn where = y) + 100and(x, prove that (a, b)= (a. b)= 1] a, b, c, Eumple 7. For any integers and (a, (a,b) c)=e Hence the result. d dbc dac and (a. be) Sol Let dja and dbc Example 9. If(a b)1, prove that (a, a b, b) =1. d(a, b)c (ac, bc) d d(a, (a, b) ) (a",ab) =a dla and d(a, b) c Sol (a. b)= I dle .abb) (a,b) =e = 1 :.ab) AS (a. (a, b) ©) ela, ebc ea, el(ac, bc) : (a b)1 ela, el(a, b) c (a, b) =1) e a ,be) eld Example 10. For any integer p, show that 0 2p+ 1, 9p+4)=1 () (Sp+2, 7p+3)=1 (a, bc)=(a, (a, b) c). (i) (Gp.3 p+2)=1 for From (1) and (2). d=e Sol.(9 Let (2p+1, 9p+4) =d odd p. Erample8. Prove for any integerm, 30 (m° m). d2p+1 and d9p+4 mm m (m" 1) Sol Here d 2 9p+ 4)9(2p+ 1) m (m1)(m +1) d 89 (mm)m+1). d11 As 6 (m' mm =(m 1) m (m + 1) and m 1, m, m + 1 are three cons m) But d is positive integers and their product is divisibley d =1. 6  (m m) (m* +1) 6(m° m) (2p+1,9p +4) = 1. FUNCTI AND 157 SPECTRUM DIscrRETE MATHEM RELA77ON SETS, integers p and q where p is odd, show that (2P 1,2" +1) = 1. ositive F o rallposi prove that then b) = 1, 3)d h f(a ab) = I or2 in (a+b, ab) = 1 Sp:2. 7p+ d7p+3 (atb, wLet d 5p+2 and ( tb, a2 +b) a2+8 = I or 2 (iv) (2a+ b, a + 2b) =1 or 3 d 5(p+3)76p+2) (a (i) (atb, a I or 3 d  l d = t l () b)=d then 1 But d is positive (p+2,7p+ 3)=1. 17 Ifla. d1 kEI = 1, then (a, b) = c. p=2k+ 1, let (tu) Grven pbeodd so =6k+4 18 ifclac/b 1)+ I 3p 3(2 consecutive integers are coprime. p+2 3(2k+1)+2=6k+5 Prove that every two consecutive integers is divisible by 3, 19. one of any three Prove that 20. 3p.3p+2)d 1.Prove it + is an integer, then b= d. = (a, b) = (¢, ad) fractions such that Let (64, 6+ 5) =d Ltand befraction be ie and d6k+5 . in pairs, then prove that (a, ay,...,) = 1. d6k+4 are integers w n c n are teiavey prine d (6k+5)(6k+4) 1fa., an, converse true ? d1 d=t1 Is 23. Prove then g.c.d. (a, b c)= 1. g.c.d. (a, c) 1, = = But dis positive. If g.c.d. (a, b) d=l p. 3p+2)=1. ( I and ca then g.c.d. (b, c) = 1 Ifg.c.d. (a, b) = (G) (c, b). 1, then g.c.d. (a c, b) g.c.d. = EXERCISE 1.12 If g.c.d. (a, b) = (Git) be then d c. If g.c.d. (a, b) 1, dacand d = () that then (c, b) = 1. zero integers, prove 1. Ifa b) = l and c / a, is divisor of c. 24. If a and b are non 3, 4 a +5) =I. then any common divisor of a c and b a a5 b) divides b and hence g.c.d. (2 a+ 2 Ifa and bare relatively prime, gc.d.(2a3 b, 4 3 Ifa b)= 1, then (a c, b) (c. b) = ANSWERS 4.Ifla e)=d alband e/b, then showthat a c/bd. S Show that (a b) = (a + b, b). 22. Not true. of 6. Ifa b) 1, then (a. b+ ka) = 1. 1.72. Euclidean Algorithm repeatedly to obtain a set applying the division algorithm positive integers a and b, on 7. If (a m)=1,then (m a, m)= 1. For any two the following relations 8. If (a, 4)2and (b, 4)=2, then prove that (a + b, 4) =4. defined successively by nainers 'z (a, b) I. is positiveintegersuch that b (a" 1), then prove that = 9. Ifa>I and n b=aq+ 0sa 10. Prove that there are no pairofintegersx,y satisfying x +y= 100 and (, y) =3. a=4+2 0sr (a,b) .0) I1. Prove that = (a,b)*. 43+3 **************** **********************" 12. Ifr andy are prime to 3, then prove that r+y can not be a perfect square. ********************** ***************** 13. Let dand g be two positive integers. Prove that there are integers x and y satisfying x T (a.y)=d iff dlg. n2n19, * 0Sr, 14. Let d and p be two positive integers. Prove that there are integers x and y satisfying X is g.c.d. of a ando. a,)=d iff 4 1p. he the last non zero remainder in the above process n SPECTRUM DISCRETE M 15 EMATHEMAN SETS, RELATION AND FUNCTION ' e have for any integers i. y Proof: + ax) = (a + by, b) b (a. 6) =(a. 2a792 Using equations () 6aq (a. 6) = (abaq,) (a.r)(a792.7) Now (a,b) ' 2197 2 3 29n)9, Continuing like this, we get  (1+9,9,)n2 4,),3 *+  (1+9n9,a4 a34n2)+9,),3 (1) = (1+919,)4 *9, 4n2 4n2919,) ,3 equation ( ) gives , as linear combination of_ and 2) Equation (2)gives , as linear combination of r and n4 like this, we go on eliminating the the last non zero remainder is g.c.d. of a, b. Continuing linear combination o f a and b. remainders23 321 untilr, is linear combination o f a and b. expressed a s a G.c.D. of a and b as a Least Common Cor. 1. Express 1.73. Common Multiple, Multiple Algorithm, we have Proof: From Euclidean on Multiple. Ifa and b are non zero integers such that a  n, b n. Then b a9 0 a and b. n is called a common multiple ofa a 192+2 Teast Com mon Multiple. Let a and b be non zero integers. Then a positive integer I is the least o f a and b if common multiple (L.C.M.) 2433 or 2 a l , bl ie. lisa common multiple ofaand b. (in al n, bn ISnfor any positive integer n. **************** Notation: The L.C.M. ofa and b is denoted by LCM. (a, b) or[a, b]. o n2 n4 39n2*n2 Important Results : 1. Ifaln, bn, then [a, b]n. n3 29*' On1 'n2 2. Prove [ka, kb} =k [a, 6] for positive integer k. 'a219,* 0S n ni 3. Let a, b be two positive integers. Prove t h a t ( g . c . d . ( a , b ) ) (1.c.m. [ a , b ] ) = a b. Rewrite above equations as Or n2n19n (a, 6) la, b]=ab. 1.74. Ifa b  c" and (a. b) = 1, then each of a and b is an exact rth power. n3 a24n1 Proof. We are given that n39n2 ...(1) ab = c" **** ******************************** and (a, b)=1 ******* ****************************** NE M OcREIE MATHED 160 (a c) a TS, RELATION AND FUNCTION Let we can take (d c)=D D/d and D/c aaß.c=ay D/d and (a, b) =d 161 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'la, D'/b D/ (a b) D'1d Now D'/d, Die ® D 7(4 c) » D'/D (.")=1 (a. b, c)= D from (3), "/b :1band ( D is g.c.d. of a, b, c. wetake b= y'6 176. Prove that (m a, m b, m c) = m(a. b, c). (a b) d and (d c)=D = From (3) and (4), we get, Prof. Let (a. b, c) = D B'6=a Also (m a, m b) = md and (m d, mc) =mD ..0) Bo a (m a, mb, m c)=mD (a, b)=1 and a/a, d/b From (1) and (2), we get (a,) =1 ..2) (a.6)=1 m a, mb, mc)= m(a, b, ) from (5), aB Cor. fmla, m/b, m/c, tnenmmm (a,b,) m we can take B= a l Proof is quite simple. From (5) and (6), we get, 10=1 i=1,ð=1 ILLUSTRATIVEEXAMPLES Example 1. Find g.c.d. of 24 and 138 and express it linear combination from(6), a as = 1 as of these numbers 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) 1.75. Explain the method of finding And 18 )24(1 g.c.d. of three numbers a, b, c. 6 24 18 18 Proof. Let (a b)= 4 6) 18(3 and (d c)=D 24(13824(5)) 18 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 MMATNO SPECTRUM DISCRETE R E L A T I O N A N DF U N C T I O N 63 that 71x50y= I. SETSREL such FF i n d x , y + 3054 y = 6 E a m p l e4 . 4. prove (71, 50) so)71 r we shall so that 12378 4 irstly, 71 50 (1)+21 50 Sol andy 1 =6. 3054)12378 Here 50 21 (2)+8 2. Find imtegers (/2375, 3054) 21)50 (2 Emple gcd 162 12216 21 8(2)+ 42 Sol Firsth show 3054 12378 3054 162(18)138 (4)+ 162) 3054(18 85(1)+33 8)21 (2 Here 162 5 =3 (1) +2 +24 16 138 (1) 1434 = 3 2(1)+ 162 I38 24(5) 18 1296 2 I (2)+0 5)8(1 24 1 8 ( 1 )  6 138) 162 (1 5 138 gcd(71, 50)=1 3)5( 186(3)+0 S0) = 1 : (a, b)= (a,b)] 24) 138 ( 3  3054) =6 g.c.d. (71, I =323(53)=2 x35 2)3( (12378, gcd 120 Now (85)5 =3 6 2418 x 5+2 x8 24(13824(5) 2 x 2 and 18 )24T =3 x (21 8(2)) +2 x8 =6 (24)  138 18 1)2 (2 3 x 21+8 x8=3 x 21 +8 x (5021 (2)) 6 (162138)138 138+6 X 162 6) =8x 5019 x 21 =8 x 50 19 x (7150) 7x X 162 18) +6 =7 x (3054 162 X = 19 x 71 +27 x 50 71 (19)50(27) 1627x 3054 71 x 50y where x = 19, y=27. = 132 x 3054 (12378(3054) 4)7x 71x50y= I tor 19, y=27. = 132 X12378 132 Evaluate g.c.d. (198, 288, 512) and express it as a linear combination of r, y,: (integers). =535 x3054+ where r=132 andy=535 Example5. y g.c.d. (g.c.d(198, 288), 512) gc.d. (198, 288, 512) = 12378 r+3054 = Sol Here = 132 and y=535. r+3054y =6 forr= and y such that Here 288 198 (1) +90 12378 Euclidean Algorithm to find integers x 198 90 (2)+ 18 198)288 3. Use the 198 Eumple 2378y 1769r + 90 18 (5) +0 = (1769, 2378) gcd 609 90)198 (2 2378 1769 (1)+ 18. (198, 288) = Sol Here 180 1769) 2378 1769 609(2)+551 609 551 (1)+58 1769 = 198 90 (2) 18) 905 609) 1769 (2 = 198(288 198) (2) 90 551 =58(9)+29 1218 3 (198)2 (288) 58 29(2)+0 .1) 28 551 ) 609 (1 gc.d. (1769, 2378) =29 Now g.c.d. (198, 288, 512) =g.c.d. (18, 512) 551 18) 512 and 29 55158(9) Here S12 18(28) +8 58)51 36 551(609551)(9) 52 18 8 (2)+2 152 10 x 551 609 x 9 9 8 2 (4)+0 144 10 x (1769609 x 2)609 x 9 = 10 x 176929 x 609 g c d . (18, 512) = 2. 8) 18(2 16 Hence g.c.d. (198, 288, 512)  2. =10 x 176929 (2378 1769) 2 18 8(2) 39 x 176929 x 2378 = = 1769 x 2378 y where x=39, y =29. + 18(512  18 (28)) 2 SPECTRUM DISCRE STTS REL4I1OVAV MAM 165 2. 11, 11 ANSWERS = 726r 2x S12 252 mt 3 9 5m 275 y + 3. 13,13 35%1+ 164 18 x $7 2 * 512 4999 5. 14, 14 826 r 1890y + 325y  57 198 109r + (/98) 2288)) 171 x & = l 47y=22 22 (i)*=41, y9 (3 x 288 1 1 (i)x  7. y =  3 114 (171) (v) 22.,y =  15 2xS12 14)+ 198 114 and: = 1 71 y =  9 ( 2,y5 r = 94, SI22)+288 where r= n) 71 8. (a. b, c) =13. 9. x = 11,y  = 198: y=  11,z =3 10. 3838  512r +288y+ 1982.  36, + 288y+ 1 1 7 .P r i m eN u m b e r 512r 2 gcd reater than I is called sitiveinteger greater a prime number if it has le.m. [306, 657) no proper divisor Evaluate Example 6& Or 306) (657, find g c d Sol Firsty By Euchdean algorithm integer>I >l is called prim is called prime number if it has only two divisors 1 and itself +45 positive 657 306(2) Any 17, 19, 1: .. . are prime numbers. + 36 11, 13, = 4 5 (6) positive 23,5,7, 306 lumber. Anyny positive integer greater than 1, which is not prime, is called 45 36(1) +9 Composite N u m a composite 36 9(4) +0 number. numbers. are composite 10, . . =9. 46,8,9, 306) numbers into three classes gcd (657, divide natural =ab Now using (a, b) [a, b] ab Unity wehave [a.b(a.b) (i) (i) Prime N u m b e r s Composite N u m b e r s 306x657 34 x 657 =22338. [ 3 0 6 ,6 5 7 ) = . 9 composite. 1 is neither prime nor Note l. EXERCISE 1.13 Note2 2 is the only even number which is prime. m+ 595 n. 1s not prinme, iS not necessarily composite. d in the form 252 3. Any number, which and 252 and express Note I. Find the gcd. of 595 it in the form 726 x+ 276y. Two prime numbers are called twinprimes ir there is only one composite number between them. and 275 and express 2. Find the g.cd. of 726 + 325y. it in the form 858 x and p +2 are Twin Primes. Find the gcd. of858 and 325, and express eg, for any odd integer p; p 3. 4999, and express it in the form 1109r+ 4999 y. eg, 3,5; 5,7; 11, 13 ; 17, 19 are twin primes. 4. Find the gcd. of  109 and + 1890 y. and 1890, and express it in the form 826x 5. Find the g.cd. of 826 1.78. Prove that the least divisor (other than 1) ofa composite number is a prime. 6. Find x,y (integers) satisfying Prof: Let us take n be any given composite number. = 2 i) 119x +272y =17 ( 6409x +42823y 17 (i) 256x + 1166y there is atleast divisor of n other than I so n has a least divisor qsuch that 1 <q<n. one (iv) 68 x+710y =2 ()657x+963y =9 lg is a composite number, then it has atleast one divisor other than I and so that I < 7. Find x,y such that 3587 x + 1819y= 17. q q 4 <q. 8. If a 780, b = 728 and c = 585, find (a, b, c). 419 and qln q n which is a contradiction to the fact that q is the least divisor of n. 9. Evaluate g.c.d. (228, 342, 420) and find x, y, z (integers) so that isnot acomposite number ie. q is ote: prime. z.c.d. (228, 342, 420) 228 x + 342y 420 + Ihuseach integer> 1 has a 10. Find I.c.m. [1819, 3587]. prime factor. ELATION AND FUNCTION A N DF U N C T I sPECTRUM DISCRETE MA 67 SETSA PlPiP2 . ********** Pn m. pln or = 1 (p, m) Prool: Given some , for I stSn 1, PlP either thaf T h e nb yc o r cor prove then integer, dm so itsonly o n / v . divisorsaare 1, P and 16 m any dlp pime P,is and prime d As =P = is any m) p = I.79.Ifp gc.d. (p. that take I so orp us Butp> Let Proof: d=l m)= H e n c ep r o v e d dlp g.c.d.(p. Now then 1 ,E u c l i d ' s T h e o r e m d=1, pim If djm pm. pb. infinite of primes is then or or d=p. = pa If m) eerr . v et h a tn u m b gcd.(p. that ience either plab, then prove end to the sequence ofprimes 2, 3, 5, 7, 11, 13, 17,. . ********************* * isa prime and there isnoe ible, suppose that numbe of primes is finite. Let p be the greatest prime. 1.80. Ifp b ie suppose pa that proved. Given is possible, defined as roon.If Proof: theorem then integer N P i a , an = 1 nsider (p, a) 3 *5 * ...... Xp) +1 ptathen gcd integersx,y N =(2 x If p r + a y = 1 forsome we get then N>p. sides by b, Multiplying both fN isprime, dhen it has then it has atleast one prime factor. But none of primes from 2 to p divides N. bpr+bay =b composite, least one prime factorgreater thannp. Given plab plaby Nhas at plbpx cases, 3 a prime which is greaterthan p, which is impossible. Also PIP plbpr+a by) pb in both ie divides atleasta is wrong. Si Sn p supposition pb. 1 our some1, or pla Henceeither pla, for is infinite. , then prove numberof primes 1. If pa,a.. on n. Out side any Cor. this result by induction prove. above theorem can given set of primes, there is another one". be set as We shall prove there is nothing to Note 1. The Proof: then pa and duct of two numbers of the form 4 n+ 1 is again of the same form. If n=1, Step 1. pl02 Note2 or 16nn' +4n+4 n'+1 Ifn=2, then p a,a, pla (4m+1)(4 n' + 1)= result is true for n=1,2 = k = 4[4 nn' +ntn']+1 is true for n Step 2. Suppose result 1 SIsk. = 4k+ 1, where k =4nn' +n +n' Pla, for some, a,a 4 kl is of type 4 k +3 Note3. Any number oftype Lepl +1 ... Step 3. Suppose p a,a2. p 4k1 =4k4 +41 =(4k4) +41 Plb or pak+1 where b= a@2.. 4(k1)+3 Plba+1 Pl442 ag or Plk+1 ( ofs = 4k+3 1 sIsk Pla, or pla+1 for some, 181. Prove that primes ofthe form 4 k +3 are infinite. k+1 Prol. f possible, suppose that number of primes of the form 4 k+3 is finite. Pla, forsome , 1 S t Letp be the greatest prime of this type. result istrue for n=k+1 for all n. Consider a number N defined as Hence by mathematical induction, result is true {4. But 6{3 and b 6  12 i.e. 6 24 x3. The above result holds only pis prime e.g, if Remark: that N 2(3x 5x7x ..... p)1 Cor. 2. Given p, P1. P2. P, are all prime numbers and pl pP2 P then prov Clearly N is of the type 4 k I and hence of 4 k +3 type. forsome t, 1 S i S n . sPECTRUMDISCRET N AND FUNCTION SETS,RELATIONA 169 4 k + 3 . MATHD ither A is even or A + l is even type Now N isof is even +1) certainlyN>p and type. k+ 1 is again of A k+ 3 4k+1 the sa Let a + ) = 2 k of4 form ofthe factor then numbers 8k+1. k)+1 = number, prime 168 oftwo has a Nis aprime then it product 4(2 Ir N. composite, .pdivides . initely many prime numbers ofthe form 8 infinit +1. greater than p, are TN is + 3 nd is and that e there 4 k that primes 2, 3, 5, greater than p. which is of form 1.84. Prove ssible, ifpossible, set set ofprimes of of theeform 8 k+ 1, kEZbe finite. the N is noneof prime ofthe form 8 +1, kEZ. Consider N be any integer of prime Let, a possible divisor But got Proof: thelargest have prime this we pbe 3.32.5272.11 132..+ cases, the Let both in infinite. 3 is odd integer is of the form 8k+ 1 and product of integers of the form 8k+ 1 is again pposition is Wrong. + mpossible. form 4 k infinite. are oo Since s q u a r e ff e ev ve r y our of the k 1 form 4 primes 88k+1. number of ofthe k+ 3. förm primes 4 ofthe Prove that number of k1 are ofthe form infinite. let ppbe be tthe greates an integer 2.52.72 114 13. . p 'is ofthe form 8 k+ I and let it be k'. Cor. ofthe form 4 6 k+ 5 are are finite, and Prine Proo. Primes primes ofthe form ofthe form 6 k *S 12.132 p ) isoftheform8 that primes Prove 1.83. that, 8K +1 suppose possible, form Proof. If PT Nis ofthe ber or number N is a prime number. x... k + 5 type composite ype N=2 x3 x5 hence of6 either N is a Let and Nowe 1 type 6k + 5 and>p. then clearly none of the odd primes 3, 5, 7, 11, 13, . p divides N. Also as N omnosite number, k N is of6 is oftype + 5 type. composite Clearly then certainly N prime factor of6 k IfN is a not d i v i s i b l e by 2. Nisprime, one number, at least an odd If it has k+ 1. factor>p ofthe form 8 then divides N. s N has odd prime composite, If Nis p primes 2, 3, 5,. and which is >p. ofthe 5 type impossible. is prime number, again get prime number >p. we a 6k+ is none But factor of Further if N > which a and p, got prime a k * 5 type N h a s primeof6 ie. supposition is wrong contradiction our both the cases, there is a i n both the cases, geta we in form 8 k 1,kEZ is not finite. + Our supposition is WTong. 6 k+ 5 is infinite. Hencethe set ofprimes ofthe of the type number of primes infinite. there are infinite many primes ofthe form 8 k+1. k1 are 6 5. ofthe form infnits 6 k + form 8k+5 that primes ofthe form are Cor. Primes clearly of the Prove form 6k1 are of a +b* is of 4n+l185. Proof. Primes ofthe odd prime factor divisor> 1, then every Proo. f possible, suppose that number of primes oftype 8k+5 is finite and let pbe the greatest prime of Note 1. Ifaand bhave common no s type. = 5, b = 12 Example. Take a common factor Consider N= 3.5.72.. p+22 a andb have no 13 (4K + 1)4k+ 1) square of every odd number is of 8 k + 1 type. 144 169 = 13 x 25 a +b + =4n+1 number is of 8 k +1 type. each of 3, 54,74,.p of 8 k+ 1 type. is Note 2. Prove that square of every odd +ve Proof. Let n be any odd+ ve integer. 345.7. p is of type 8 k+ 1 n2A +1 354.7... p + 2 is of type 8 k+5. 4R+43+1 =40+1)*1 AND FUNCTION sE78,RELATIONA SPECTRUNM DISCRETE M. 173 NTWEMUA 3. ao Sh ww that for eeach that for a c h prir prime p2 5, p + 2 is a composite number. of form 3 or 3 m+ + snle it iiss o m prime so s o it 2 for some integer m. is p rime 172 As p s ( every integer is Or Torm 3 m, 3 m*l or 3 m+ 2. soh But 3 m is a composite number) +2 (3m +1)2 +2 = 1m+ 1, then p p*.P = 1 = 9m +6m+1+2 9m +6 m+3  = 3 (3 m +2 m+1) composite number. + 2 is a of primes injust way. one way. product is represented as pP +2 + 2 = (3m +2) +2 number >I 2, tthen hen natural =3m*2, every 9x5 =24 x32 x5 Ifp =16x45 = 16x = 9m 12 +4+2 + m Etample:( 720=8x90 7 x 9x 35 2*x3fx5x = 3 (3 m +4m +2) (i) s040 16 x 315 = 16 x5x 7x11 385 = 3 x5x 77=3 composite number. p + 2 is a x 1155 = 9 x3 10395 =9 x 3x5x7x 1l (i) 11 = 25 x 539 =3 X 25 x 49 X + 2 IS a composite number allp2 5. prime. 40425 =3 x 13475 =3 x Hence p (i) 2.2+2=4+2=6 is composite but for p =3, p +2=9+2=11 is prime. Another statement. Note: F primes such that pq=2 umple 4, If p,4 are normal. integer is Every positive ILLUSTRATIVE EXAMPLES is divisible by p + q showthat p +q' composite number. p +q" is a ie 6 + I or 6m +5 ;m EZ. and 3 is ofthe fom m such thatp q=2 Example 1. Show thatevery primeexcept 2 m + 3, 6m+4, 6m +5. Soal.Given P, 9 are primes 1,6m+2, 6 We know any integercan be put in the form 6m, 6 m+ Now pP+q = (p" )+(q +1) .2) Sol. For m =0, 6m+2=2 and 6m+ 3=3. Here pP1 =pP 1P =(p  IMpP='+pP2 +. +p+) 6m+4 =2 (3 m +2)) are composite numhe 6m +3 =3 (2 m+ 1), (p1)(2 /+ 1) (say) Also 6 m, 6m+22(3m+ 1), for each non zero integer ( pP+ pP2 t...+p Is even for prime p) +1 or 6 m +5. is form 6 m every prime except 2 and 3 ofthe and q+1=q +19 =(q +1) (q9q92 +q93.. ** +1) ofthe form 6 k+ 1. form 3 k + l is Example 2. Show that any prime ofthe in the form 6m, 6 m + 1, 6m +2, 6m+3, 6 m + 4, 6 m+ 5. =(q+ 1)2 m+ 1) (: q492+ .........9 IS evC Sol. We know any integercan beput from (2), pP +qi = (p  1)(21 + 1) +(q + 1)(2 m +1) And 616m. 216m+2, 316m+3, 26m+4 1 or 6 m *). 6 nm + 4 are all composite so prime is of the form = (q+ ) (21l+ 1)+ (q +1)2m+1) (q +1)(21+1+ 2m+ 1) = [: of(1)] 6 m , 6m+2, 6m +3,6 m + 2 ( q+1)((+m +1) = (2q+2)( +m + 1) If possible, suppose p= 3k+1 =6m +5 3k6m=4 3(k2 m) =4 (2q+ pq)(+ m +1) I of(1) (p+9)(0+m+ ) Now 3LH.S. but 3 4=R.H.S. (1) is absurd. p+q is divisible by p+ g 6k+ 1. le p +q is a composite number. Thus a prime p =3 k + I must of the form 6m+1 or ECTRUM DisCRETE EMATEA RELATIONN ANDFUNCTI IS prime CTION and p a7 by 24. Given P is divisible 6/+5 :0 a. a. a. a. 175 174 > 3. p r o v e pI 6/+3, 6/+ 4, p l a 4.a. p 61,6/+1,6/+2, divide at least one of the factors and anyprime 5. For nilSt her Example form composite Then each factor is a Sol Any outof which the integerisof 61,6/+2,6/+3, 6/+ 4 are 6/+5 (:616, 2]6/+2, 36/42 61+3.261 P l a () Given pis primeand pla2+b or 6/+1 (a+6)6+c2) p>3 is ofform =3614 + 12/ anyprime 1 . = (61+ 1)* pla2 then p 1, +601+24 36/2+ 0 =61+ Ifp 1 is 12/(3/+1) 251 2""1 If 2 a prime number, show that n is 1=36/" + 601+ Example 7. (a) If Drime (61+5)* nositive integer and = then p1 = 12 (3/+2)(0+1) Ifn> 1, a positi a" lis prime show a =2. Ifp6/+5,  12(3/+5/+2) sol 6) ( ) ible, suppose that n is not a prime Ifpossible n p4,where 1 p, q<n then = 2m let If l is even, +1) = 24 m (6m+1) 2P91 12(2 m) (6 m = = 2"1 p1 Now =24(3m+ 1)(2m+1) = (2P)" l= a" 1 where a =2P 12 (6m+2) (2m+1) 1 = p1 a1 al Weclaim p21 is divisible by 24. = (a)(a?+a92 al **** TI /=2m+1 Since If 1 is odd, then + 1) (6m + 4) +1)+1) =12 (2 m 12 (2 m + 1)(3(2m a1a 1 p1 2 4 (2m+ 1)3m+2) 2P12"1 or ( a 1=2" 1) 2" 1 is prime which contradicts that  a m + 5) (2 m + 2) m +1+1) =12 (6 p1 = 12 (6m +3+2)(2 . our supposition is wrong. 24 (6 m +5) (m + 1) Hence n must be a prime. pI is divisible by 24. (b) Ifa= 1 then a" l=11=0, not a prime number so that a 11 Hence the result. a 2 2 then prove that Example 6 Ifp is prime, Ifpossible suppose that, a>2 a1>1 i) Pla plaa plb Pla and pla +b for n> 1, a< a" (in pla+b2, pb+c2 pl a2 2 a1<a" 1 Sol Given p is prime and p a And a 1= (a 1) (a"l +a"+. Then Pla +....+) a1 divides a" 1 where 1<a1<a" 1 Also given pla +b aI is composite, contrary to given pla+b a plb2 so (: pBP a=2. Hence proved. Plb MATHENMA SPECTRUM DISCRETEM EXERCISE 1.14 p>3, odd. odd by 12 if 11 *91s divisible thatp prove primes, composite. nwin 4" is Ifp.gare + I. I, prove n number e. m> composite integer a 2. Forany thatn t 4 Is n>1, show number. integer composite any is a 3. For that 8" +  show n 2 1, square. perfect integer is a any 4. For pq+l that show twin primes, number. I f p and q are +n* +1 is a composite show that n> I, integer For any form n*  1 is 7. primeof the that only Show 7. 4 is 5. n MODULE2 form only primeofthe Show that () that p"a". prime and p  a", prove numbers. 8. Ifp is a written as a sum of two composite can be n>1 integer 9 Prove each a power of2. show n is equal to 10. If 2"+1 is odd prime, an 177
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