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NULO Even Numbered solutions MAnuAl
Abstract Algebra: An
Interactive Approach, 2nd Edition William Paulsen by 1 / 4
Answ ers to Even-Numbered Problems Section 0.1 2)q= 15,r= 12 4)q=21,r= 17 6)q= 87,r= 67 8)q=1,r= 215 10) 1 +n <1 + (n 1) 2 =n 2
- 2(1n)< n
- 3(n 1) + 4 = 2k, thenn
- 3n+ 4 = 2(k +n+ 1).
- = 3k, then 4
- = 3(4k + 1).
- +nx
- (2n1) =n
2 12) If (n 1) 2
2
14) If 4 n1
n
16) (1+x) n = (1+x)(1+x) n1 (1+x)(1+(n 1)x) = 1+nx+x 2 (n1)
18) (n 1) 2
2 .20) (n 1) 2 ((n1) + 1) 2 =4 +n 3 =n 2 (n+ 1) 2 =4.22) (n 1)=((n1) + 1) + 1=( n(n+ 1)) =n=(n+ 1).
24) 4 100 + (11) 36 = 4.
26) (6) 464 + 5 560 = 16.
28) (2) 465 + 9 105 = 15.
30) (54) (487) + (221) (119) = 1.
32) Letc= gcd(a; b ). Thencis the smallest positive element of the setA= all integers of the formau+bv. If we multiply all element ofAbyd, we get the set of all integers of the formdau+dbv, and the smallest positive element of this set would bedc. Thus, gcd(da; db) = dc.34) Since bothx=gcd(x; y) andy=gcd(x; y) are both integers, we see that (x y)=gcd(x; y) is a multiple of bothxandy. If lcm(x; y ) =ax=byis smaller then (x y)=gcd(x; y), then (x y)=lcm(x; y) would be greater than gcd(x; y). Yet (x y)=lcm(x; y) =y=a=x=bwould be a divisor of bothxand y.
36) 2 3 23 29.
38) 7 29 31.
40) 3 13
2 101.42)u=222222223,v= 1777777788.44) 3 4 37 2
333667
2 .Section 0.2 2) Ifa=b=c=d, so thatad=bc, thenab(c 2 +d 2
- =abc
- Thus,ab=(a
- =cd=(c
2 +abd 2 =a 2 cd+b 2 cd= cd(a 2 +b 2
2 +b 2
2 +d 2 ).4) a) One-to-one, 3x+ 5 = 3y+ 5)x=y. b) Onto,f((y5)=3) =y.6) a) One-to-one,x=32=5 =y=32=5)x=y. b) Onto,f(3y+ 6=5) =y.1Note: Odd Numbered Questions Answers are the end of Text Book. 2 / 4
2Answers to Even-Numbered Problems 8) a) One-to-one, ifx >0,y <0 theny= 3x >0. b) Onto, ify]0, f(y=3) =y. Ify <0,f(y) =y.10) a) Not one-to-onef(1) =f(2) = 1. b) Onto,f(2y(1) =y.12) a) One-to-one, ifxeven,yodd, theny= 2x+ 2 is even. b) Not onto, f(x)̸= 3.14) a) Not one-to-onef(5) =f(8) = 24. b) Not onto,f(x)̸= 1.16) Supposefwere one-to-one, and let ~ B=f(A), so that ~
f:A!
~ Bwould be a bijection. By lemma 0.5,jAj=j ~ Bj, butj ~
Bj [ jBj
18) Supposefwere not one-to-one. Then there is a case wheref(a1) =f(a2), and we can consider the set ~ A=A( fa1g, and the function ~
f:
~ A!Bwould still be onto. Butj ~ Aj 2 .22)x 3 (3x+ 2.24)f(x) = { 3x+ 14 ifxis even, 6x+ 2 ifxis odd.26) Iff(g(x)) =f(g(y)), then sincefis one-to-one,g(x) =g(y). Sincegis onto,x=y.28) There is somec2Csuch thatf(y)̸=cfor ally2B. Thenf(g(x))̸=c sinceg(x)2B.30) Ifxeven andyodd,f(x) =f(y) meansy=x+ 8 is even. Onto is proven (1 (x) = { x+ 3 ifxis even, x(5 ifxis odd.32) Associative, (xy)z=x(yz) =x+y+z(2.34) Not associative, (xy)z=x(y(z,x(yz) =x(y+z.36) Yes.38) Yes.40) Yes.42)f(x) is both one-to-one and onto.Section 0.3 2) 55 4) 25 6) 36 8) 7 26) First nd 0[q[u)vsuch thatq)x(modu) andq)y(modv). Then ndkso thatk)q(modu)v) andk)z(modw). Answers to Even-Numbered Problems3 30) 4 Section 0.4 2) Since 1+2⌊an⌋is an integer, 1+2⌊an⌋anwill have the same denominator asan. Thus, the numeratoran+1is the denominator ofan. Note that the fractions will already be in lowest terms.4) Since the sequence beginsb0= 0,b1= 1,b2= 1,b3= 2,: : :we see that the equations are true forn= 1. Assume both equations are true for the previousn, that is,b2n2=bn1andb2n1=bn1+bn. Then by the recursion formula,b2n=bn1+ (bn1+bn)2(bn1mod(bn1+bn)).But (bn1mod(bn1+bn)) =bn1, sincebn1+bn> bn1. Sob2n=bn.Then we can computeb2n+1=bn1+bn+bn2(bn1+bnmodbn). But (bn1+bnmodbn) = (bn1modbn), andbn+1=bn1+bn2(bn1modbn).Thus,b2n+1=bn+bn+1.6)a2n+1=b2n+1=b2n+2= (bn+bn+1)=bn+1= (bn=bn+1) + 1 =an+ 1.8) Ifai=ajfori > j, then becausean+1is determined solely onan,a2ij= ai. In fact, the sequence will repeat forever, so there would be only a nite of rational numbers in the sequence. But this contradicts that every rational is in the sequence, which is an innite set.10) In computing the long division ofp=q, the remainders at each stage is given by the sequence in Problem 9. Since this sequence eventually repeats, the digits produced by the long division algorithm will eventually repeat.12) Ifp 3 =q 3 = 2 withpandqcoprime, then 2jp, but replacingp= 2rshows 2jqtoo.14) Ifp 2 =q 2 = 5 withpandqcoprime, then 5jp, but replacingp= 5rshows 5jqtoo.16) Ifp 3 =q 3 = 3 withpandqcoprime, then 3jp, but replacingp= 3rshows 3jqtoo.18) If 1=awere rational, thena= 1=a 1 would be rational.20) Givenxandy, choose any irrationalz, and nd a rationalqbetweenxz andyz. Thenq+zis irrational by Problem 19.22)x 2 p 6, and p 2 is too. Ifxwere rational, then x 2 would be rational.24) 2 p p 26)a6= p p
by nding the inverse:f
10) 10
12) 91
14) 43
16) 223
18) 73
20) 1498
22) 3617
24) 3875
28) 12 3 / 4
32) 35
34) 17
36) 30
38) 51
40) 3684623194282304903214
42) 21827156424272739145155343596495185185220332
44) 1334817563332517248
= 5 + 2
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