Solution Manual for Concise Introduction To Pure Mathematics, 4th Edition

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solutions MAnuAlFoRA Concise Introductionto Pure Mathematics,Fourth EditionMartinliebeckbyK25658_SM_Cover.indd117/12/151:17 pm

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Chapter 1Solutions1.TTFTFFTFT2.(a) Notice thatB={x|xreal,−2<x<2}andD={0}. So the only pair,neither of which is contained in the other, isD,E.(b)Xcannot beD, but could beB,CorE.3.(a) Not valid.(b) Valid. LetCbe the statement “I eat chocolate” andDthe statement “Iam depressed”. We are givenC⇒D. Therefore ¯D⇒¯C. We are told that ¯Distrue, so I am not eating chocolate.(Actually I am, it’s delicious.)(c) Valid.LetEbe the statement “movie was made in England”,Wbe“movie is worth seeing”,Ibe “movie reviewed by Ivor”. Then we are given¯W⇒¯EandW⇒I. Hence ¯I⇒¯W⇒¯E. Therefore the movie was not made inEngland.4.The implications between these statements are (a)⇔(d)⇔(e).5.(a) True(b) False(c) False(d) False: for example ifa=b=2 thenabis a square buta,bare not.(e) True: ifa,bare squares thena=m2,b=n2withm,nintegers, soab=m2n2= (mn)2, a square.6.(a) By contradiction: suppose√6− √2≤1.Squaring both sides gives6+2−2√12≤1, hence 7≤2√12, hence 49≤48, a contradiction. Therefore√6− √2>1.(b) Supposen2is even. Ifnis odd thenn2is odd, by Example 1.2. Hencen1K25658_SM_Cover.indd517/12/151:17 pm

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2SOLUTIONS TO CHAPTER 1cannot be odd, sonis even.(c) Observe thatm3−m=m(m2−1) = (m−1)m(m+1), a product of threeconsecutive integers. Of three consecutive integers, exactly one must be divis-ible by 3, and at least one must be even. Hence their product is divisible by3×2=6.7.We produce counterexamples to each of these statements:(a) 24−2=14 is not divisible by 4.Hence the statement is false whenn=2,k=4.(b) 7 is not the sum of three squares.8.We are given that 8881 is not a prime, so there are integersa,bgreater than1 such that 8881=ab. Choose such a factorization withaas small as possible.Thenais prime (otherwise it would have a smaller factor, contradicting thechoice ofabeing as small as possible). Alsoa≤b, so 8881=ab≥a2, andsoa≤ √8881. Since 972>8881, this implies thata<97, and since the nextprime down from 97 is 89, this means thata≤89.9.(a) Negation:∃n∈Zsuch thatnis a prime number andnis even. Thisnegation is true, asn=2 is an even prime.(b) Negation:∃n∈Zsuch that∀a,b,c,d,e,f,g,h,∈Z,n=a3+b3+c3+d3+e3+f3+g3+h3.The original statement is true — i.e. every integer is the sum of 8 cubes. Thisis tricky to prove. Here is a proof.Letnbe any integer. By Q6(c),n3−nis a multiple of 6. Writen3−n=6xwithx∈Z. Observe that(x+1)3+ (x−1)3=2x3+6x, and hencen=n3−6x=n3−(x+1)3−(x−1)3+2x3.Thusnis in fact the sum of 5 cubes.(Notice that it is crucial in this proof to allow negative cubes. If we insist onall of the cubesa3, . . . ,h3beingnon-negative(i.e.a≥0, . . . ,h≥0), then thenegation is true — for example, 23 is not the sum of 8 non-negative cubes.)(c) Negation:∀x∈Z,∃n∈Zsuch thatx=n2+2. This negation is false,for examplex=4 is not of the formn2+2. So the original statement is true.(d) Negation:∀x∈Z,∃n∈Zsuch thatx=n+2. This negation is true: taken=x−2.(e) Negation:∃y∈ {x|x∈Z,x≥1}such that 5y2+5y+1 is not prime. Thisnegation is true: e.g. ify=12 then 5y2+5y+1=781, which is not prime asit is a multiple of 11.K25658_SM_Cover.indd617/12/151:17 pm

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SOLUTIONS TO CHAPTER 13(f) Negation:∃y∈ {x|x∈Z,x2<0}such that 5y2+5y+1 is not prime.This negation is false, as the set in question is the empty set. So the originalstatement is true.10.Letxbe a real number which is less than every positive real number.Suppose for a contradiction thatxis positive, i.e.x>0. Thenx>12x>0.But this means thatxisnotless than the positive real number12x, which is acontradiction. Thereforexcannot be positive.11.(a) Letx=a1. Thenax=3. Sinceax>ax−1> ... >a1≥1, it follows thatx≤3. Easily see that onlyx=2 is possible. Thusa1=2.(b) Sincea1=2, havea2=aa1=3, and thena3=aa2=6. Thereforea6=aa3=9. Asa3<a4<a5<a6, follows thata4=7,a5=8. Thena7=aa4=12,a8=15,a9=18.(c) Work as in (b) to see thata10,a11, ....is19,20,21,22,23,24,25,26,27,30,33,36,39,42,45,48,51,54,55,56, ......,81,84,87, ....,162,163,164, ....,243,246,249, .....and so on. In particular,a100=181.K25658_SM_Cover.indd717/12/151:17 pm

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Chapter 2Solutions1.(a) By contradiction. Suppose√3 is rational, so√3=mnwherem,nareintegers andmnis in lowest terms.Squaring, we getm2=3n2.Thusm2isa multiple of 3, and so by Example 1.3,mis a multiple of 3.This meansm=3kfor some integerk. Then 3n2=m2=9k2, son2=3k2. Thereforen2is a multiple of 3, hence so isn. We have now shown that bothmandnaremultiples of 3. Butmnis in lowest terms, so this is a contradiction. Therefore√3 is irrational.(b) By contradiction again. Suppose√3=r+s√2 withr,srational. Squar-ing, we get 3=r2+2s2+2rs√2. Ifrs=0 this gives√2=3−r2−2s22rs. Sincer,sare rational this implies that√2 is rational, which is a contradiction. Hencers=0. Ifs=0 thenr2=3, sor=√3, contradicting the fact that√3 is irra-tional by (a). Thereforer=0 and 3=2s2. Writings=mnin lowest terms, wehave 3n2=2m2. Now the proof of Proposition 2.3 shows thatmandnmustboth be even, which is a contradiction.2.(a) Supposex=√2+√3/2 is rational. Thenx2=2+32+2√3, hence√3=12(x2−72). Asxis rational this implies√3 is rational, a contradiction byQ1(a). Hence√2+√3/2 is irrational.(b) By (a), 1+√2+√3/2 is the sum of a rational and an irrational, henceis irrational by Proposition 2.4(i).(c) We have 2√18−3√8+√4=6√2−6√2+2=2, which is rational.(d) Letx=√2+√3+√5. Then(x− √2)2= (√3+√5)2, which givesx2−6−2√15=2x√2. Squaring again,(x2−6)2+60−4(x2−6)√15=8x2,hence√15= ((x2−6)2−8x2+60)/4(x2−6). Therefore ifxis rational, thenso is√15. But√15 is irrational by the hint in the question. Hence so isx.(e) This is sneaky one. Observe that(√2+√3)2=5+2√6, so√5+2√6=√2+√3, and hence√2+√3−√5+2√6=0, which is rational.5K25658_SM_Cover.indd917/12/151:17 pm

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6SOLUTIONS TO CHAPTER 23.(a) True: ifx=m/nandy=p/qare rational, so isxy=mp/nq.(b) False: for a counterexample take the irrationals√2 and−√2.Theirproduct is−2, which is rational.(c) False: the product of the two irrationals√2 and 1+√2 is√2+2, whichis irrational.(d) True: we prove it by contradiction. Suppose there is a rationala=0 andan irrationalbsuch thatc=abis rational. Thenb=ca, and sinceaandcarerational, this implies thatbis rational, a contradiction.4.(a) Letc=x+ax+band supposecis rational. Thenx+a=c(x+b), whichgivesx(c−1) =a−bc. Ifc=1 thenx=a−bcc−1, which is rational sincea,b,care rational. Asxis irrational, this implies thatc=1, hencea=b.(b) Letc=x2+x+√2y2+y+√2and supposecis rational. Multiplying up gives√2(c−1) =x2+x−cy2−cy. Ifc=1 this gives√2=x2+x−cy2−cyc−1, which is rational.Hencec=1, which impliesx2+x=y2+y. This yields(x−y)(x+y+1) =0,hence eitherx=yorx+y=−1.5.By contradiction. Letα=√2+√n, and supposeαis rational. Thenα−√2=√n. Squaring both sides,α2+2−2α√2=n, so 2α√2=α2+2−n.Since clearlyα=0, we can divide through by 2αto get√2= (α2+2−n)/2α.Asαis rational, this implies that√2 is rational, which is a contradiction.Henceαis irrational.6.Letaandbbe two different real numbers withb>a. Choose a positiveintegernsuch thatb−a>1n.Then there is a rational of the formmnlyingbetweenaandb.Also, choose a positive integermsuch thatb−a>√2m.Then there is anumber of the formk√2mlying betweenaandb, wherekis an integer; byProposition 2.4(ii),k√2mis irrational unlessk=0, in which casea<0 andb>0 and we apply the above argument replacingaby 0.7.Letan=√n−2+√n+2. Thena2n= (n−2)+(n+2)+2√(n−2)(n+2) =2n+2√n2−4. We are given thatanis an integer. This implies√n2−4 is ra-tional. By the hint given, this means thatn2−4 must be a perfect square, i.e.n2−4=m2for some integerm. Thenn2−m2=4. Staring at the list of squares0,1,4,9,16, . . ., we see that the only way the two squaresn2,m2can differ by4 is to haven2=4,m2=0. Hencen=2, soan=2.8.Letr,b,gbe the numbers of red, blue and green salamanders at some pointin time. If a red and a green meet, these numbers change tor−1,b+2,g−1, soK25658_SM_Cover.indd1017/12/151:17 pm

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SOLUTIONS TO CHAPTER 27the difference betweenbandris increased by 3; if two reds meet, the numberschange tor−2,b+1,g+1, and again the difference betweenbandrincreasesby 3; and so on — you can easily check that whenever two salamanders meet,the difference betweenbandreither stays the same, or increases or decreasesby 3.Initially,r−bis 15−7=8.This cannot be changed by adding andsubtracting multiples of 3 to 30−0=30. Hence it is not possible for all thesalamanders to be red at some point.K25658_SM_Cover.indd1117/12/151:17 pm

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Chapter 3Solutions1.1812999.2.We know that√2 is irrational. By Proposition 3.5, periodic decimal expres-sions are rational, hence irrationals have non-periodic decimal expressions.Therefore√2 is non-periodic.3.(a) This number is 0.12340123401234....which is periodic, hence rational.(b) Suppose the given number is rational.Then the decimal must be pe-riodic, so for somenthere is a sequencea1a2...anofndigits which repeatsitself. But if we go far enough along the decimal, there will be a sequence ofat least 2nconsecutive zeroes. The sequencea1a2...anmust appear within thissequence of zeroes, hence must be just 000...0. This means the decimal ends0000....(going on for ever), which is plainly false. Hence (by contradiction)the number is irrational.(c) This is 1.1001000010000001..... There are ever-increasing sequences ofzeroes, so the argument for (b) shows that this number is irrational.4.Letαbe the cube root of 2.Since(1.25)3= (54)3=12564<2, we knowα>1.25.Also(1.3)3>2, soα<1.3.Thereforeαis 1.3, correct to 1decimal place.5.Letx=1009899. It’s easy to check that 100+x+100x=10000x. Writing thisequation out in full, using the fact thatx=0.010102. . ., and spacing it out abit to make the point, we get100+0.01 01 02 03 05...+1.01 02 03 05 08...=101.02 03 05 08 13...9K25658_SM_Cover.indd1317/12/151:17 pm

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10SOLUTIONS TO CHAPTER 3You can see that, at least near the beginning of the decimal expansion ofx,the numbers formed by looking at successive pairs of digits are being forcedto obey the rule defining the Fibonacci sequence. Eventually one has to startcarrying digits and the whole thing goes wrong, but it does work for a while(in fact up to the term 55 in the Fibonacci sequence).If you try for examplex=1000/998999, then it works for even longer.6.Consider the proof of Proposition 3.4. If one of the remainders is 0, then thedecimal expression ends with repeated zeroes and the period is 1. If not, theremainders are all among 1,2, . . . ,n−1, and so the period is at mostn−1.The first fewnfor which the period isn−1 aren=2,7,17,19,23,29. Noticethat they are prime numbers! You will find an explanation for this in Chapter26 (see Proposition 26.8 and Exercise 14).7.Supposex=mnis a rational which has decimal expression ending in repeat-ing zeroes. Sayx=a0.a1a2. . .ak000. . .. Thenx=A10k, whereAis an integer(in factA=ak+10ak−1+. . .+10ka0). Sox=A2k5k. Cancelling out commonfactors ofAand 2k5k, we see that the denominatornis of the form 2a5b.Conversely, ifn=2a5b, thenmnis equal top10kfor some integersp,k, sothe decimal expression formnhas repeating zeroes from thek+1thposition atleast.8.Suppose1nhas decimal expansion of period 1, and letrbe the repeatingdigit in the decimal, repeating from thek+1thdecimal place.Then as inthe previous solution,1n=A10k+r910−k−1. So the denominatornmust divide9·10k+1. Hencen=2a5b3cwherea,bare arbitrary andc≤2. Conversely, forsuch numbersnthe period of1nis 1.K25658_SM_Cover.indd1417/12/151:17 pm

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Chapter 4Solutions1.(50)3/4(5√2)−1/2= (2.52)3/4(5.2−1/2)−1/2=23/453/25−1/221/4=2151=10.2.This is again 10.3.2617; 3(3332).4.21/231/3= (23)1/6(32)1/6= (2332)1/6. So taken=72,t=1/6.5.Note that 10000100= (1002)100=100200. So 10010000is bigger.Suppose 21/2≥31/3. Taking 6th powers, this implies 23≥32, which is false.Hence 21/2<31/3.6.Squaring both sides givesx+(2−2x)−2x1/2(2−2x)1/2=1, hence 2x1/2(2−2x)1/2=−x+1; squaring again gives 4x(2−2x) =x2−2x+1, hence 9x2−10x+1=0, which has solutionsx=1 andx=19. Of these, onlyx=1 satisfiesthe original equation.7.x+y≥2√xy⇔(x+y)2≥4xy⇔x2+y2+2xy≥4xy⇔(x−y)2≥0, whichis true. Equality holds whenx=y.8.Takea=b=2,c=3 andd=2,e=1,f=any.8.Taking sixth powers of both sides of the equationy4/3=x5/6givesy8=x5.Hencey=25,x=28is a solution. It is the smallest solution, since if 1<y<25andyis an integer, thenx=y8/5is not an integer.So the bill was£512.32. Pretty expensive meal.11K25658_SM_Cover.indd1517/12/151:17 pm

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Chapter 5Solutions1.Ify<0 then−y>0, sox.(−y)>0, hence−xy>0, hencexy<0.Ifa>b>0 thena−b>0,1a>0 and1b>0 (using Example 4.4). Hence theproduct(a−b).1a.1b>0. This meansa−bab>0, so1b−1a>0, so1b>1a.2.The inequality isx2+x+1−x−12x−1≥0, which isx2(2x+1)2x−1≥0. This is trueprovidedx≤ −12orx>12orx=0.3.This translates to 3x2−4x+1<0, which is(3x−1)(x−1)<0, which istrue provided13<x<1.4.(a) The inequality is the same as 2x+1x−3<0, i.e.2x2−3x+1x<0. Factoring,this is(2x−1)(x−1)x<0. So for the inequality to hold, either one or three of thenumbers 2x−1,x−1,xmust be negative. This holds precisely whenx<0 or12<x<1.(b) By the formula for the roots of a quadratic, there are two distinct realsolutions if and only if “b2−4ac>0”, that is,t2−12>0. This occurs whent>√12 ort<−√12.5.Inequality translates to(u−1)(v−1)1+uv>0, which is true if 0<u,v<1. Theother ranges for which it is true are:u,v>1; oru,v<1 anduv>−1; oru<1,v>1 anduv<−1; oru>1,v<1 anduv<−1.6.Ifx,y≥0 then|x|=x,|y|=yand|xy|=xy=|x| |y|. Ifx≥0,y<0 then|x|=x,|y|=−yand|xy|=−xy=|x| |y|. The other cases are similar.7.(i) The inequality says eitherx+5≥1 orx+5≤ −1, i.e. eitherx≥ −4 orx≤ −6.(ii) The inequality is true if and only if(x+5)2>(x−2)2, i.e.x2+10x+13K25658_SM_Cover.indd1717/12/151:17 pm

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14SOLUTIONS TO CHAPTER 525>x2−4x+4, i.e.x>−32.(iii) Note thatx2+2x+3= (x+1)2+2, which is always positive, so theright hand side of the inequality is alwaysx2+2x+3.Whenx≥ −5, theinequality isx2+2x+3>x+5, i.e.(x+2)(x−1)>0, which is true providedx>1 orx<−2. And whenx<−5 the inequality isx2+2x+3>−x−5, i.e.x2+3x+8>0, which is always true. Hence the range for which the inequalityholds isx>1 orx<−2.8.Letx,ybe the sides of a rectangle of areaA=xy. The perimeterPis 2(x+y).By Example 5.14,P≥4√xy=4√A. The perimeter of a square of side√Aisequal to 4√A, so this is smallest possible perimeter.9.First apply Example 5.14 takingx= (a1a2)1/2,y= (a3a4)1/2, to get(a1a2a3a4)1/4≤12((a1a2)1/2+ (a3a4)1/2).By Example 5.14,(a1a2)1/2≤12(a1+a2)and(a3a4)1/2≤12(a3+a4). Apply-ing these to the above inequality gives the result.To prove the Arithmetic-Geometric Mean Inequality forn=8, just repeatthe above argument:(a1a2. . .a8)1/8≤12((a1. . .a4)1/4+ (a5. . .a8)1/4)by Example 5.14≤12(14(a1+· · ·+a4) +14(a5+· · ·+a8))=18(a1+· · ·+a8).Repeating this argument we can prove the Arithmetic-Geometric Mean In-equality for any case wherenis a power of 2.10.(i) In Question 9, takea1=x4,a2=a3=a4=y4.(ii) As in (i) we havex3y≤34x4+14y4. Add this to the inequality in (i).11.Squaring both sides ofx+y+z=0, we getx2+y2+z2+2(xy+yz+xz) =0. Hencexy+yz+xz=−12(x2+y2+z2)≤0.12.Letnbe a 1000-digit number. Thenn≥10999. But the sum of the 1000thpowers of the digits ofnis at most 1000×91000, which is much smaller than10999. Soncannot be a Smallbrain number.Remark:Would this have worked for 100-digit numbers ? Where does thetrick stop working ?K25658_SM_Cover.indd1817/12/151:17 pm

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