1.1 Numbers 101: The Very Basics

i i i i i i i i 1 Exercises

1.1 Numbers 101: The Very Basics

• (a) The claim makes sense and is true.

(b) The claim makes no sense; p 8isn't a subset.(c) The claim makes sense and is true.(d) The claim makes sense but is false; considera= 0andb= p 2.(e) The claim makes sense and is true.

(f) The claim makes sense but is false: considera= 0.

• (a) The claim is false; leta=

p 2.(b) The expressionQ 2 doesn't make sense.(c) The claim is true. Sincea 2 >0, some largenwill work.(d) The claim is true; see Theorem??.(e) Ifa2Qanda6= 0, thena p 2=2Q.

• The number1=ais an integer only ifa=1. The number1=ais rational

for all nonzero integersa. The equation1=a=aholds only ifa=1.

• (a)12S1but1=2S1

(b)22S2but1=2=2S2 (c) p 22S3but1= p

• =

p

2=2=2S3

• (a)12S1but1=2S1

(b)22S2but1=2=2S2 (c) p 22S3but1= p

• =

p

2=2=2S3

(d)2S4but 2 =2S4

• Theorem??is useful. BecauseQis closed under addition, multiplica-

tion, and division (if denominators aren't zero), expressions like those in (a) and (b) are rational—unless, for (b),p=q= 0. Expressions in- volving square roots are different. Ifp=q= 1, for instance, then p p 2 +q 2 = p 2is irrational; the same expression is rational ifp= 3andq= 4. The quantity p p 2

• 2pq+q

2 is always rational, since p p 2

• 2pq+q

2 = p (p+q) 2 =(p+q). (Note that p (p+q) 2 =p+q may be false.)

• All ofxy,x+y,xyandx=ycan be either rational or irrational. Examples

are easy to nd.Solutions Manual for Understanding Real Analysis, 2e by Paul Zorn (All Chapters) 1 / 4

i i i i i i i i 2 8.contradiction. (E.g., ifx=rwere rational, then we could multiply by the rational numberr. Then the productxis also rational, a contradiction.) p r can go either way.

9.p

• =a=bfor integersaandb, where

a=bis in reduced form. Then squaring both sides gives3b 2 =a 2 . This implies (essentially as in the proof of Theorem??) that3divides botha andb, which contradicts the assumption thata=bis in reduced form.

• x

2 =2Q. Ifx2Q, then (by Theorem??)x 2 is rational, too, which contradicts our assumption.(b)x= p

• +

p 3is rational, then x 2

= 5 + 2

p 6is rational, too. This implies, in turn, that p 6is rational, which is absurd.(c)x= p

• +

p

• +

p 5, then we havex p

• =

p 2+ p 3, and supposexis rational. Squaring both sides of the last equation and simplifying gives x 2 2x p

• = 2

p 6; which is progress, since onlytwosquare roots remain. Squaring again gives x 4 4x 3 p

• + 20x

2 = 24; which is even better, as onlyonesquare root is left. The last equation implies that p

• =

x 4

• 20x

2 24 4x 3

:

Becausexis rational, so is the right-hand side above, and thus so is the left. This absurdity completes the proof.

11.1< a=b <2. For part (iii), note that a 02 b 02 = (2ba) 2 (ab) 2 = 4b 2 4ab+a 2 a 2 2ab+b 2 ; and substitutinga 2 = 2b 2 shows that the last fraction is2.This all shows that if p

• =a=bholds foranypositive integersaandb,

then we can nd a new fractiona

=b

with p

• =a

=b

andb

< b, which is absurd.

12. Z2is not closed under addition:1 + 1 = 2=2Z2.

(b)Z2satises all the requirements in Theorem??. 2 / 4

i i i i i i i i 3 13.M22is commutative, but multiplication is not; exam- ples are easy to nd. Every matrixAinM22has an additive inverseA, but multiplicative inverses exist only for some nonzero matrices (those with nonzero determinant); again, examples are easy to nd. Distributivity does indeed hold inM22.

• aandbare rational, then

1 a+b p 2 = ab p 2 a 2

• 2b

2

:

This shows that elements ofFhave multiplicative inverses inF. The rest is easier.

• ln ln lnntends to innity, it must exceed two for largen. The well-

ordering property guarantees that a smallest suchn0exists. (Using a calcu- lator we can see thatn0has about 703 decimal digits.) 16.f1;1=2;1=3; : : :gis a subset ofQ, but has no least element.(b) R=f1;10;100;1000; : : :gdoes have the well-ordering property; every nonempty subset includes asmallestpower of 10.(c) T=f 3;2;1; : : : ;41;42g(like all nite sets of real numbers)doeshave the well-ordering property, since every nonempty subset ofTis also nite, and hence has a least element.(d) result no longer holds forN, sinceNitself has no greatest element.The propertydoeshold for the nite setT, and also forZnN= f: : : ;3;2;1;0g.

1.2 Sets 101: Getting Started

1. DI;D2C.

(b)B=fm2Ajmhas 31 daysg.(c)ADis the set of ordered pairs(January;2),(February;2), . . . , (December;2),(January;3),(February;3), . . . ,(December;3). There are 24 such pairs.(d)AnB=fFebruary, April, June, September, Novemberg;BnA=;; A\C=fNovemberg;B\A=B;D\I=D;D[I=I.

• S=f0;1g;Tis the interval of numbers between(1

p 21)=2

2:791and(1 +

p

21)=21:791. 3 / 4

i i i i i i i i 4 (b)

explain:SNis false because1=2N;STis true;T\Q6=;

is true, since02T\Q;2:82QnTis true.

(c)U=

x2Rjx 2 +x <0

=

(1;0).

• RnA= (1;1)[(3;1)

(c)RnA= (1;1])[[2;3][[4;1) (e)RnA=f0g

• Rn[a; b] = (1; a)[(b;1)

(b)I= (1;1)has empty complement;I= (1;17)has closed complement[17;1);I= (0;17)has complement(1;0][[17;1).(c)RnZ= (0;1)[(1;0)[(1;2)[(2;1)[: : :.

• ais inRn(RnA)means thataisnotinRnA; this means, in

turn thata2A.

6.A=RandB=;. Then Rn(A[B) =;but(RnA)[(RnB) =R.(b) A=B, thenA[B=A\B=A, and both claims are clearly true.(c)x2Rn(A[B). Thusx =2A[B, sox =2Aandx =2B; in other words,x2RnAandx2RnB. This is just another way of saying thatx2(RnA)\(RnB). The “vice versa” implication is similar.

• RnA1= (1;1][[3;1)andRnA2= (1;2][[5;1). Also,

Rn(A1\A2) = (1;2][[3;1)andRn(A1[A2) = (1;1][[5;1).It's easy to see that, as claimed,Rn(A1\A2) = (1;2][[3;1) = (1;1][[3;1)[(1;2][[5;1) = (RnA1)[(RnA2). Similarly,Rn (A1[A2) = (1;1][[5;1) = ((1;1][[3;1))\((1;2][[5;1) = (RnA1)[(RnA2)).

• A1[A2= (0;1)[(2;1)andA1\A2=;. ThusRnA1=

(1;0][[1;1)andRnA2= (1;2]. This implies that(RnA1)[ (RnA2) = (1;1)and(RnA1)\(RnA2) = (1;0)[[1;2].Consistent with De Morgan,Rn(A1\A2) = (1; infty);Rn(A1[A2) =

(1;0][[1;2].

• x2T

thenx =2T. SinceTS, we havex =2S, which meansx2S

, as desired.

• / 4