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Boca Raton London New York CRC Press is an imprint of the Taylor & Francis Group, an informa business solutionS MANUAL A Course in Real Analysis Hugo D. Junghenn by

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Chapter 1 Solutions Section 1.2 1.(a) Since( −a) +a= 0, uniqueness of the additive inverse of(−a)implies that−(−a) = a.(b) ( (ab) + (−a)b )= ( a

• (−a)

) b = 0∙b= 0, so uniqueness of the additive inverse implies−(ab) = (−a)b. A similar argument works for the second equality.(c) By (b) and (a),(−a)(−b) = −(a(−b)) =−(−(ab)) = ab.(d) By (b),(−1)a = 1(−a) = −a.(e) By commutativity and associativity of multiplication, (a/b)(bc) = a(b −1 b)c=ac=c(d −1 d)a= (c/d)(ad ), hence the first equality follows from 1.2.1(h). For the second equality, by commutativity and associativity of multiplication and 1.2.1(i), (a/b)(c/d) = ( ab −1 )(cd −1

• = (ac)(b

−1 d −1

• = (ac)(bd)

−1 = (ac)/(bd).(f) Using commutativity and associativity of multiplication, the distribu- tive law, and 1.2.1(i), a/b+c/d=ab −1 (dd −1

• +cd

−1 (bb −1

• =ad(b

−1 d −1

• +bc(b

−1 d −1 ) =ad(bd) −1 +bc(bd −1

• = (ad+bc)/(bd).

2.Letr=m/nands=p/qwherem, n, p, q∈N andnq== 0. By Exercise 1, r±s = (mq±pn )/(nq)andrs= (mp)/(nq), which are rational. Since1 /s= (pq −1 ) −1 =p −1 q=q/p,r/sis the product of rational numbers hence is rational.

3.

Ifs:=r/x∈Q , then, by Exercise 2,x=r/s∈Q , a contradiction.

Therefore,r/x∈I. The remaining parts have similar proofs.

4.(a) By commutativity and associativity of multiplication and the dis- tributive law, (x−y) n + j=1 x n−j y j−1 = n + j=1 x n−j+1 y j−1 − n + j=1 x n−j y j = n−1 + j=0 x n−j y j − n + j=1 x n−j y j =x n −y n .3

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(b) Replaceyin part (a) by−y.(c) Replacexandyin part (a) byx −1 andy −1 , respectively.

5.The left side of (a) is n−1n n−2 n

∙ ∙ ∙

1 n = n!n n .For (b), (2n)! = − 2n(2n −2)(2n −4)∙ ∙ ∙4∙2

∙−

(2n−1)(2n −3)∙ ∙ ∙3∙1 ∙ = 2 n − n(n−1)(n−2)∙ ∙ ∙2∙1

∙−

(2n−1)(2n −3)∙ ∙ ∙3∙1 ∙ .

6.

:

n k−1 9 +

:

n k 9 = n!(n−k+ 1)!(k −1)!+ n!(n−k)!k!= kn! + (n −k+ 1)n!(n−k+ 1)!k !=

:

n+ 1 k 9 7.Letandenote the difference of the two sides of the equation in (a).Combining fractions in the resulting summation leads to a n= n 8 k=1 n−2k (n+ 2)(k + 1)(n −k+ 1) .Making the index changej=n−kresults in a n= n 8 j=1 2j−n (n+ 2)(j + 1)(n −j+ 1) =−a n.Therefore,a n= 0. Part (b) is proved similarly.

8.f(k) =k 3 −(k−1) 3 = 3k 2 −3k+ 1.Section 1.3 1.(a) Ifa >0andb <0, then−(ab) = a(−b) >0henceab <0.(b) Ifa >0and1/a <0, then1 =a(1/a)<0. The converse is similar.(c) Follows froma/b−c/d= (ad−bc)/bd.

2.Multiply the given inequalities byx, using (d) of 1.3.2.

3.Part (a) follows from a double application of 1.3.2(d). Part (b) follows from (a) by noting that−y <−x and0 <−b <−a . Part (c) follows from (a).

4.If0< x < y, then multiplying the inequality by1/(xy)and using (d) of 1.3.2 shows that1 /y <1/x. Ifx < y <0, then0 <−y <−x hence, by the first part,1/(−x )<1/(−y )so1/x >1/y.4

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5.If−1< x < yorx < y <−1, then(y+ 1)(x + 1)>0hence y y+ 1 − x x+ 1 = y−x (y+ 1)(x + 1) >0.Ifx <−1< y, then(y+ 1)(x + 1)<0and the inequality is reversed.

6.(a) By Exercise 1.2.4,y n −x n = (y−x) n 1 j=1 y n−j x j−1 .Each term of the sum is positive and less thany n−j y j−1 =y n−1 . Since there arenterms, part (a) follows.(b) The inequality is equivalent to n(n+ 1)xy +ny+nx+x+ 1< n(n + 1)xy +ny+nx+y+ 1, which reduces tox < y.

7.The given inequality implies thatmx > nx−n andm < n. Therefore, n >(n−m)x > x.

8.a=ta+ (1−t)a < tb+ (1−t)b=b.

9.If the inequality holds, takex=y= 1to geta≥ −2andx= 1, y=−1 to geta≤2. Conversely, suppose that0≤a≤ 2. The inequality then holds trivially if xy≥ 0, and ifxy <0thenx 2 +y 2 +axy= ( x+y) 2

• (2−a)(−xy)≥0. A similar argument works for the case

−2≤a≤0.

10.

Ifa > bthenx:=a−b > 0anda > b+x, contradicting the hypothesis.

11.Note thatb >0. Supposea > b. Thenx:= (1 +a/b)/2>1and

bx= (a+b)/2< a, contradicting the hypothesis.

12.The inequality is equivalent toa < x 2 +xfor allx >0. Assumea >0. If

a≥1thenx= 1/2violates the condition. If0< a <1, thenx:=a/4<1

soa > x+x > x 2 +x, again, violating the condition. Therefore,a≤0.

13.(a) Follows from0≤(x−y) 2 =x 2 −2xy+y 2 .(b)0≤(x−y) 2

• (y−z)

2

• (z−x)

2 = 2(x 2 +y 2 +z 2 )−2(xy+yz+xz).(c) By expansion, the inequality is equivalent to2 xyzw≤ (yz) 2

• (xw)

2 , which follows from (a).(d) Follows from (a).

14.Expand(x−a) 2 ≥0and divide byx.

15.(a) Writex−y= (x−z) + (z −y)and apply the triangle inequality.(b)|x−L|< εiff−ε < x−L < ε.5

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