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Solution to Understanding Analysis Exercise 1.2


Exercise 1.2.1

See Understanding Analysis Instructors’ Solution Manual Exercise 1.2.1


Exercise 1.2.2

We prove it by contradiction.

Suppose there exists a rational number r such that 2r=3. Let r=p/q, where p and q are integers. Then2p/q=32p=3q.Note that q0.

If q>0, then p>0. Hence we have 3|2p, which is impossible.

If q<0, then p<0. We have 2p=3q. Similarly, we have 3|2p, which is impossible. Thus completing the proof.


Exercise 1.2.3

See Understanding Analysis Instructors’ Solution Manual Exercise 1.2.2


Exercise 1.2.4

Rearrange N as follows,

Take Ai as the set of numbers in i-th column. Then it is clear that AiAj= for ij and i=1Ai=N.


Exercise 1.2.5

(a) If x(AB)c, hence xAB. Therefore, we have xA or xB which implies that xAc or xBc. Thus xAcBc and hence (AB)cAcBc.

(b) If xAcBc, then xAc or xBc. Hence xA or xB, namely xAB. Thus x(AB)c. Therefore, we have AcBc(AB)c.

(c) If x(AB)c, then xAB. Hence xA and xB, namely xAc and xB2. Thus xAcBc. We conclude that (AB)cAcBc.

If xAcBc, Then xAc and xBc, namely xA and xB. Therefore xAB and thus x(AB)c. We conclude that AcBc(AB)c.

As a consequence, (AB)c=AcBc.


Exercise 1.2.6

(a) If a and b are both non-negative, then |a+b|=a+b, |a|=a, and |b|=b. Hence|a+b|=a+b=|a|+|b|.If a and b are both non-positive, then |a+b|=(a+b), |a|=a, and |b|=b. Hence|a+b|=(a+b)=ab=|a|+|b|.

(b) Note that (a+b)2=a2+2ab+b2 and (|a|+|b|)2=|a|2+2|a||b|+|b|2=a2+2|a||b|+b2. Since ab|ab|, we have (a+b)2(|a|+|b|)2.Since x2y2 implies |x||y|, we have |a+b||a|+|b|.

(c) We applies the triangle inequality several times,  |ac|+|cd|+|db| |(ac)+(cd)|+|db|= |ad|+|db||(ad)+(db)|= |ab|.

(d) Note that ||a||b||2=(|a||b|)2=a22|a||b|+b2 and |ab|2=(ab)2=a22ab+b2.Since ab|a||b|, we have 2|a||b|2ab. Hence||a||b||2= (|a||b|)2=a22|a||b|+b2 a22ab+b2=|ab|2.Therefore ||a||b|||ab|.


Exercise 1.2.7

See Understanding Analysis Instructors’ Solution Manual Exercise 1.2.6


Exercise 1.2.8

(a) f(n)=n+1 for nN.

(b) f(1)=1 and f(n)=n1 for n2.

(c) f(1)=0, f(2n)=n and f(2n+1)=n for all nN.


Exercise 1.2.9

See Understanding Analysis Instructors’ Solution Manual Exercise 1.2.7


Exercise 1.2.10

(a) & (b) False. Take a=b=0 as a counterexample.

(c) True. Let us prove it. Consider the “if” part. Suppose a<b, then we are done. Suppose ab, then|ab|=ab.By assumption, |ab|=ab<ε for every ε>0. By Theorem 1.2.6, we have a=b. Hence we always have ab.

“Only if” part. If ab, then it is clear thata<b+εfor every ε>0.


Exercise 1.2.11

(a) True. When n is very large, then 1n is very small and close to zero.

(b) False. See above. We cannot find a positive number which is small than any positive number.

(c) True. See Theorem 1.4.3.


Exercise 1.2.12

(a) It is clear y1=6>6. Now we suppose that yn>6. Now we show that yn+1>6. Since yn+1=2yn63, we haveyn+1= 2yn63> 2×(6)63=183=6.By induction, we have yn>6 for all nN.

(b)It is equivalent to show that yn>yn+1 for all nN. Note that y2=2663=2, hence we have y1>y2.Now we suppose that yn>yn+1, namely ynyn+1<0, we would like to show that yn+1>yn+2. Since yn+1=2yn63, we haveyn+2yn+1= 2yn+1632yn63= 2(yn+1yn)3<0.Hence yn+2<yn+1. By induction, we have yn>yn+1 for all nN, which implies that the sequence (y1,y2,) is decreasing.


Exercise 1.2.13

See Understanding Analysis Instructors’ Solution Manual Exercise 1.2.12

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.
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