solutions MANUAL FOR

solutions MANUAL FOR Introduction to Abstract Algebra (2 nd Ed) Jonathan D.H.Smith by

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1.NUMBERS

(1) Sincez≤xandx≤y, we havez≤yby transitivity. But y≤zis given, soy=zby antisymmetry.(2) Use transitivity, and induction onr, to show thatx 0≤xrand x r≤x n. Sincex r≤x n≤x 0, transitivity givesxr≤x 0.Together withx 0≤xr, antisymmetry givesx0=xr.(3) First,x≤ymeansy−xis nonnegative. Theny≤zmeansz−y is nonnegative. Thus (z−y) + (y−x) =z−xis nonnegative, sox≤z.(4) Becausex−x= 0 is nonnegative.

(5) 11.

(6) 4.

(7) (a) In this case, if a natural numbernlies inS, thenn̸= 0,1, . . . , s−1. Thuss≤n.(b) Letm= min({0,1, . . . , s−1} ∩S). Note thatmis an element ofS. Nowmis less thans. By transitivity, it is less than each element ofSgreater thans. By definition, it is less than or equal to each element ofSless thans.(8) Use induction onn. The induction basis (n= 0) is 0 = 0. For the induction step, suppose 1 3

• 2

3

• 3

3

• . .+n

3 = ( n(n+ 1) 2 ) 2 .Then 1 3

• 2

3

• 3

3

• . .+n

3

• (n+ 1)

3 = ( n(n+ 1) 2 ) 2

• (n+ 1)

3 = 1 4 ( (n 2 +n) 2

• 4(n+ 1)

3 ) = 1 4 ( n 4

• 6n

3

• 13n

2

• 12n+ 4

) = 1 4 ( (n 2

• 2n+ 1)(n

2

• 4n+ 4)

) = ( (n+ 1)(n+ 2) 2 ) 2 .(9) (a) The induction basis (n= 0) is 0 = 0. For the induction step, suppose

• + 2 + 3 +. . .+n=

n(n+ 1) 2 .1

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2 Then 1+2 + 3 +. . .+n+ (n+ 1) = n(n+ 1) 2

• (n+ 1)

= 1 2 ( n 2 +n+ 2n+ 2 ) = 1 2 ( n 2

• 3n+ 2

) = (n+ 1)(n+ 2) 2 .(b)Note 2 (

• + 2 +. . .+ (n−1) +n

) =

• + 2 +. . .+ (n−1) +n

+n+ (n−1) +. . .+ 2 + 1 = nterms �

���

(n+ 1) + (n+ 1) +. . .+ (n+ 1) + (n+ 1) =n(n+ 1).(10)Use induction onn. The induction basis (n= 0) is 0<1. For the induction step, supposen <2 n . Thenn+ 1<2 n

+ 1≤

2 n

• 2

n = 2 n+1 .(11) (a)Supposem=nrfor some integerr. Sincemis nonzero, ris nonzero, and|r| ≥1. Then|m|=|nr|=|n| · |r| ≥ |n| ·1 =|n|.(b)Supposem=nrfor some integerr. Sincemis nonzero, ris nonzero, andr 2 ≥1. Thenm 2 = (nr) 2 =n 2 ·r 2 ≥ n 2 ·1 =n 2 .(c)For a given integerm, there are only finitely many integers nwith|n| ≤ |m|orn 2 ≤m 2 .(12)Letnbe an integer. Then 0 =n·0, so 0 is a multiple ofn.

(13)9.

(14)The proofs of (R) and (T) are the same as for Proposition 1.4.Now supposem|nandn|m, sayn=muandm=nvfor integersuandv. Ifm= 0, thenn=mu= 0u= 0, som=n in this case. Now supposem̸= 0. Sincem=nv=muv, we haveuv= 1. The only integral solutions to this equation are u=v= 1 oru=v=−1. Sincem >0 andn >0, the latter possibility is excluded. Thusu=v= 1 andm=n.(15)The divisibility relationx|ymeansy=xrfor a certain real numberr. Thusy= 0 andxis arbitrary, or elsey̸= 0, in which casexis a nonzero real number.

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3 (16) (a)NoteS={1}, so infS= 1.(b)The requirementa=dq+rbecomes 1 = 0q +r=r, so r= 1 andqis arbitrary.(c)No. The hypothesis of Proposition 1.6 states thatdmust be positive, according to (1.10). This excludes the case d= 0.(17)Use the Division Algorithm to geta=dq+r ′ with 0≤r ′ < d.If 0≤r ′ < d/2, thenais approximated as required bydq. If d/2< r ′ < d, thenais approximated as required byd(q+ 1).(18) (a)11·11 + 5 = 126.(b)106 = 11·9 + 7, so the coordinates are (9,7).(19)Use induction onito show that n=q id i +nid i−1 +ni−1d i−2

• . .+n 2d+n 1.

The desired result is the casei=k.(20)EE9, since 3817 = 14·16 2

+ 14·16 + 9.

(21)17 = 2·8 + 1.

(22)The sets are marked as shown:

divisors of 18 divisors of 24 common divisors gcd S a d q 1 S a 3 a S 2 S 6 S a a 4 a 12 a S S 8 S 24 S 9 a 18 a 36 a 72 Sa

(23){2,3},{2,5},{2,7},{2,9};

{3,4},{3,5},{3,}),{3,8};

{4,5},{4,7},{4,9};

{5,6},{5,7},{5,8},{5,9};

{6,7};

{7,8},{7,9};

{8,9}, as well as{1, m}for 1≤m <10.(24)Ford >1, we have gcd(d,2d) =d >1, sodis not relatively prime to 2d.

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