rudin solved.pdf, Study Guides, Projects, Research of Mathematical Analysis

Download rudin solved.pdf and more Study Guides, Projects, Research Mathematical Analysis in PDF only on Docsity!

lA

mo ~ 1976

)upp.

/lATH

Solutions Manual to Walter

Rudin's Principles of

Mathematical Analysis

Roger Cooke, University of Vermont

Su

Chapter 1

The Real and Complex

Number Systen1s

Exercise 1.1 If r is rational (r =f. 0) and x is irrational, prove that r + x and

rx are irrational.

Solution. If r and r + x were both rational, then x = r + x - r would also be

rational. Similarly if rx were rational, then x = 7 would also be rational.

Exercise 1.2 Prove that there is no rational number whose square is 12.

First Solution. Since v'f2 = 2.)3, we can inv~ke the previous problem and prove that .J3 is irrational. If m and n are integers having no common factor and such that m 2 ....:. 3n 2 , then m is divisible by 3 (since if m 2 is divisible by 3, so ism). Let m = 3k. Then m 2 = 9k^2 , and we have 3k^2 = n 2. It then follows that n is also divisible by 3 contradicting the assumption that m and n have no common factor.

Second Solution. Suppose m 2 = 12n 2 , where m and n have no common factor. It follows that m must be even, and therefore n must be odd. Let m = 2r. Then we have r 2 = 3n 2 , so that r is also odd. Let r = 2s + 1 and n = 2t + 1. Then

4s 2 + 4s + 1 = 3(4t 2 + 4t + 1) = 12t 2 + 12t + 3,

so that

4( s 2 + s - 3t 2 - 3t) = 2.

But this is absurd, since 2 cannot be a multiple of 4.

(c) If x is real, define B ( x) to be the set of all numbers bt, where t is rational and t ::; x. Prove that

br = supB(r)

when r is rational. Hence it makes sense to define

bx =sup B(x)

for every real x.

(d) Prove that bx+y = bx bY for all real x and y.

Solution. (a) Let k = mq == np. Since there is only one positive real number c

such that cnq = bk (Theorem 1.21), if we prove that both (bm)lfn and (bP) 1 fq

have this property, it will follow that they are equal. The proof is then a routine

computation: ((bm)Ifntq = (bm)q = bmq = bk, and similarly for (bP) 1 fq.

(b)^ Let^ r^ =^ !!!:.n^ and^ s^ =^ ..!!..w^ •^ Then^ r^ +^ s^ =^ mw+vnnw 'and

by the laws of exponents for integer exponents. By the corollary to Theorem 1.21 we then have

where the last equality follows from part (a).

(c) It will simplify things later on if we amend the definition of B(x) slightly,

by defining it as {bt : t rational, t < x }. It is then slightly more difficult to

prove that br = sup B(r) if r is rational, but the technique of Problem 7 comes

to our rescue. Here is how: It is obvious that br is an upper bound of B(r).

We need to show that it is the least upper bound. The inequality b^1 fn < t if n > (b- 1)/(t- 1) is proved just as in Problem 7 below. It follows that if 0 < x < br, there exists an integer n with b^1 fn < br jx, i.e., x < br-l/n E B(r).

Hence x is not an upper bound of B(r), and so br is the least upper bound.

(d) By definition bx+y = supB(x + y), where B(x + y) is the set of all

numbers bt with t rational and t < x + y. Now any rational number t that is

less than x + y can be written as r + s, where r and s are rational, r < x, and s < y. To do this, let r be any rational number satisfying t- y < r < x, and let s = t- r. Conversely any pair of rational numbers r, s with r < x, s < y

gives a rational sum t = r + s < x + y. Hence B(x + y) can be described as the

set of all numbers brbs with r·< x, s < y, and rands rational, i.e., B(x+y) is

the set of all products uv, where u E B(x) and v E B(y).

Since any such product is less than supB(x)supB(y), we see that the num-

ber M = sup B(x) sup B(y) is an upper bound for B(x + y). On the other hand, suppose 0 < c < supB(x)supB(y). Then cj(supB(x)) < supB(y). Let

m = (1/2)(c/(supB(x)) + supB(y)). Then c/ supB(x) < m < supB(y), and

there exist u E B(x), v E B(y) such that cjm < u and m < v. Hence we have

8 ' CHAPTER 1. THE REAL AND COMPLEX NUMBER SYSTEMS

c = (cjm)m < uv E B(x + y), and soc is not an upper bound for B(x + y). It

follows that supB(x) supB(y) is the least upper bound of B(x + y), i.e.,

as required.

Exercise 1.7 Fix b > 1, y > 0, and prove that there is a unique real x such that bx = y, by completing the following outline. (This xis called the logarithm of y to the base b.) (a) For any positive integer n, bn - 1 2: n(b - 1). (b) Hence b-12: n(b 1 fn -1). (c) If t > 1 and n > (b- 1)/(t- 1), then b^1 fn < t. (d) If w is such that bw < y, then bw+(l/n) < y for sufficiently large n; to see this apply part (c) with t = y · b-w. (e) If bw > y, then bw-(l/n) > y for sufficiently large n.

(f) Let A be the set of all w such that bw < y, and show that x = sup A

satisfies bw = y. (g) Prove that this x is unique. Solution. (a) The inequality bn - 1 2: n(b- 1) is equality if n = 1. Then, by

induction bn+l -1 = bn+^1 -b+ (b-1) = b(bn -1) + (b-1) 2: bn(b-1) + (b-1) =

(bn + 1)(b- 1) 2: (n + 1)(b- 1).

(b) Replace b by b^1 fn in part (a). (c) The inequality n > (b -1)/(t- 1) can be rewritten as n(t- 1) > (b -1), and since b- 1 2: n(b^1 fn- 1), we have n(t- 1) > n(b^1 fn- 1), which implies t > blfn. (d) The application of part (c) with t = y · b-w > 1 is immediate. (e) The application of part (c) with t = bw · (1 j y) yields the result, as in part (d) above. (f) There are only three possibilities for the number x =sup A: 1) bx < y; 2) bx > y; 3) bx = y. The first assumption, by part (d), implies that x+ (1/n) E A

for large n, contradicting the assumption that xis an upper bound for A. The

second, by part (e), implies that x- (1/n) is an upper bound for A if n is large, contradicting the assumption that x is the smallest upper bound. Hence the

only remaining possibility is that bx = y.

(g) Suppose z =/:. x, say z > x. Then bz = bx+(z-x) = bxbz-x > bx = y. Hence x is unique. (It is easy to see that bw > 1 if w > 0, since there is a positive rational number r = 7: with 0 < r < w, and br = (bm )^1 /n. Then bm > 1 since b > 1, and (bm)lfn > 1 since 1n = 1 < bm.)

10 CHAPTER 1. THE REAL AND COMPLEX NUMBER SYSTEMS

Hence

2ab = 2( (~)

2

r

12

Now (x^2 )^112 = x if x 2: 0 and (x^2 )^112 = -x if x ~ 0. We conclude that 2ab = v

if v 2: 0 and 2ab = -v if v ~ 0. Hence z^2 = w if v 2: 0. Replacing b by -b, we

find that (z) 2 = w if v ~ 0. Hence every non-zero complex number has (at least) two complex square roots.

Exercise 1.11 If z is a complex number, prove that there exists an r > 0 and a

complex number w with lwl = 1 such that z = rw. Are wand r always uniquely

determined by z?

Solution. If z =' 0, we take r = 0, w = 1. (In this case w is not unique.)

Otherwise we taker= !zl and w = z/lzl, and these choices are unique, since if

z = rw, we must haver= r!wl = !rw! = !z!, zjr.

Exercise 1.12 If z1 , ... , Zn are complex, prove that

Solution. The case n = 2 is Part (e) of Theorem 1.33. We can then apply this result and induction on n to get

!(zl + Z2 + · · · + Zn-l) + Znl

< !zl + Z2 + · · · + Zn-ll + lzn I

< lz1l + lz2l + · · · + lzn-11 + lznl·

Exercise 1.13 If x, yare complex, prove that

j!xl-iYII ~ !x- Yl·

Solution. Since x = x- y + y, the triangle inequality gives

lxl ~ !x- Yl + IYI,

so that lxl- !YI ~ !x- Yl· Similarly IY!-!xl ~ !x- Yl· Since !xl- IYI is a real

number we have either j!x!-IYII = !xi-IYI or j!xi-IY!I = IYI-!xl. In either

case, we have shown that jlxl - IYII ~ lx- Yl·

Exercise 1.14 If z is a complex number such that \z\ = 1, that is, such that

zz = 1, compute

Solution. \1 + z\ 2 = (1 + z)(1 + z) = 1 + z + z + zz = 2 + z + z. Similarly

\1- z\ 2 = (1- z)(1- z) = 1- z- z + zz = 2- z- z. Hence

\1 + z\ 2 + \1- z\ 2 = 4.

Exercise 1.15 Under what conditions does equality hold in the Schwarz in- equality? Solution. The proof of Theorem 1.35 shows that equality can hold if B = 0 or

if Baj- Cb.i = 0 for all j, i.e., the numbers aj are proportional to the numbers

bj. (In terms of linear algebra this means the vectors a = (a1, a2, ... , an) and

b = (bll b2, ... , bn) in complex n-dimensional space are linearly dependent. Con- versely, if these vectors are linearly independent, then strict inequality holds.)

Exercise 1.16 Suppose k 2: 3, x, y E Rk, \x- y\ = d > 0, and r > 0. Prove: (a) If 2r > d, there are infinitely many z E Rk such that

\z - x\ = \z - y\ = r.

(b) If 2r = d, there is exactly one such z.

(c) If 2r < d, there is no such z. How must these statements be modified if k is 2 or 1? Solution. (a) Let w be any vector satisfying the following two equations:

w·(x-y)

\w\

From linear algebra it is known that all but one of the components of a solution w of the first equation can be arbitrary. The remaining component is then uniquely determined. Also, if w i~ any non-zero solution of the first equation,

there is a unique positive number t such that tw satisfies both equations. (For

example, if x 1 =/:. y1 , the first equation is satisfied whenever

Z2(X2- Y2) + ... + Zk(Xk - Yk)

Zl =. Yl- Xl

If (z 1 , z2, ... , zk) satisfies this equation, so does (tz1 , tz2, ... , tzk) for any real

number t.) Since at least two of these components can vary independently, we

can find a solution with these components having any prescribed ratio. This

If x and y are the sides of a parallelogram, then x + y and x - y are its diagonals. Hence this result says that the sum of the squares on the diagonals of a parallelogram equals the sum of the squares on the sides.

Exercise 1.18 If k 2:: 2 and x E Rk, prove that there exists y E Rk such that

y =I= 0 but x · y = 0. Is this also true if k = 1? Solution. If x has any components equal to 0, then y can be taken to have the corresponding components. equal to 1 and all others equal to 0. If all the components of x are nonzero, y can be taken as ( -x2 , x (^) 1 , 0, ... , 0). This is, of course, not true when k = 1, since the product of two nonzero real numbers is nonzero.

Exercise 1.19 Suppose a E Rk, bE Rk. Find c E Rk and r > 0 such that

lx - ai = 2jx- hi

if and only if lx- cl = r. (Solution: 3c = 4b- a, 3r = 2lb- a!.)

Solution. Since the solution is given to us, all we have to do is verify it, i.e., we need to show that the equation

lx-al = 2lx- bl

is equivalent to lx- cl = r, which says

I x- 4 -b+^ -a=^1 I^ -lb-al^2 3 3 3.

If we square both sides of both equations, we an equivalent pair of equations, the first of which reduces to

3lxl 2 + 2a · x- 8b · x- lal 2 + 4lbl 2 = 0,

and the second of which reduces to this equation divided by 3. Hence these equations are indeed equivalent.

Exercise 1.20 With reference to the Appendix, suppose that property (III) were omitted from the definition of a cut. Keep the same definitions of order and addition. Show that the resulting ordered set has the least-upper-bound property, that addition satisfies axioms (A1) to (A4) (with a slightly different zero element!) but that (A5) fails. Solution. We are now defining a cut to be a proper subset of the rational numbers that contains, along with each of its elements, all smaller rational

14 CHAPTER 1. THE REAL AND COMPLEX NUMBER SYSTEMS

numbers. Order is defined by containment. Now given a set A of cuts having an upper bound {3, let a be the union of all the cuts in A. Obviously a is properly contained in {3, and so is a proper subset of the rationals. It also obviously satisfies the property that if p E a and q < p, then q E a; hence a is a cut. It is further obvious that a contains each elements of A, and so is an upper bound

for A. It remains to prove that there is no smaller upper bound.

To that end, suppose, "/ < a, then a contains an element x not in "Y· By definition of a, x must belong to some cut b in A. But then "Y < b, and so "Y is

not an upper bound for A. Thus a is the least upper bound.

The proof given in the text goes over without any change to show that (Al), (A2), and (A3) hold. As for (A4) let 0 = {r : r ~ 0}. We claim 0 +a = a. The proof is easy. First, we obviously have 0 +a~ a. For r + s ~ s if r ~ 0.

Hence r + s E a if s E a. Conversely a ~ 0 +a, since each s in a can be written

as 0 + s.

Unfortunately, if 0' = { r : r < 0}, there is no element a such that a + 0' =

1. For a + 0' has no largest element. If x = r + s E a + 0', where r E a and s E 0', there is an element t E 0' with t > s, and so r+t E a+O' and r+t > s.

Since 0 has a largest element (namely 0), these two sets cannot be equal.