Solutions To Mathematics Textbooks/Principles of Mathematical Analysis (3rd edition) (ISBN 0070856133)/Chapter 1

< Solutions To Mathematics Textbooks‎ | Principles of Mathematical Analysis (3rd edition) (ISBN 0070856133)

Unless the contrary is explicitly stated, all numbers that are mentioned in these exercises are understood to be real.

Contents

Chapter 1Edit

1Edit

If is rational () and is irrational, prove that and are irrational.

Solution. Let . If was rational then would be too. Similarly is irrational.

2Edit

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

Solution. Let, if possible, such that and . Now . By the fundamental theorem of arithmetic and therefore has both 2 and 3 in its factorization. So for some . But now and so , a contradiction.

3Edit

Prove Proposition 1.15.

Solution. The results follow from using the facts related to being a field.

4Edit

Let be a nonempty subset of an ordered set; suppose is a lower bound of and is an upper bound of . Prove that .

Solution. For note that and the result follows.

5Edit

Let be a nonempty set of real numbers which is bounded below. Let be the set of all numbers , where . Prove that inf =-sup.

Solution. Let inf and sup. We need to show that . We first show that is the upper bound of . Let . Then and so or follow. We now show that is the least upper bound of . Let be an upper bound of . Then , or . So is a lower bound of . Since inf so or .

6Edit

Fix .

(a) If are integers, , , and , prove that . Hence it makes sense to define .

(b) Prove that if and are rational.

(c) If is real, define to be the set of all numbers , where is rational and . Prove that sup when r is rational. Hence it makes sense to define sup for every real .

(d) Prove that for all real and .

Solution. (a) Suppose . Then and the fundamental theorem of arithmetic imply that and where . So and so we are done. If then reduce to lowest factors, say . Clearly now by the already worked out case when the ratios are coprime.

(b) We will let and and equivalently show that . Clearly . The last equality holds as the exponents are integers.

(c) Clearly . We need merely show that br is an upper bound for B(r) since being in B(r) it then automatically becomes its supremum.

Clearly b1/n>1. Now if r=m/n is any positive rational then br=(bm)1/n>1. Now let p,q be any rational numbers with p<q. As bq-p>1 so bpbq-p=bq>bp or in other words for every bt in B(r) we have tr and so bt≤br, i.e. br is the upper bound.

(d) Suppose r is a rational number with r<x+y. WLOG let x<y and set δ=x+y-r>0. Choose a rational p such that x-δ<p<x and put q=r-p. Then q<y. By parts (b) and (c) br=bp+q=bpbq≤bxby. So bxby is an upper bound for {br:r≤x+y} or bx+y≤bxby.

Now suppose p, q are rationals with px and qy. Then bp+q is in B(x+y) and so bpbq=bp+q≤bx+y by (b) and so bp≤bx+y/bq. Now bp is in B(x). So for all q bx+y/bq is an upper bound for B(x) as p can be chosen arbitrarily. By definition bx≤bx+y/bq and so bq≤bx+y/bx. Again q can be chosen arbitarily so that bx+y/bx is an upper bound for B(y). As before this leads to by≤bx+y/bx or bxby≤bx+y.

7Edit

Fix b>1, y>0 and prove that there is a unique real x such that bx=y by completing the following outline. (This x is called the logarithm of y to the base b.)

(a) For any positive integer n, bn-1≥n(b-1).

(b) b-1≥n(b1/n-1)

(c) If t>1 and then b1/n<t.

(d) If w is such that bw<y then bw+(1/n)<y for sufficiently large n.

(e) If bw>y, then bw-(1/n)>y for suffficiently large n.

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

(g) Prove that this x is unique.

Solution. (a) Clearly each of bn-1, bn-2,...b is greater then 1 and summing them and applying the forumla of the finite sum of a geometric series gives the result.

(b) As b1/n > 1 so by (a), (b1/n)n - 1 ≥ n(b1/n - 1).

(c) b1/n = (b1/n - 1) + 1 ≤ (b - 1)/n + 1 < t.

(d) Note that 1 < b-wy = t (say). Choose n > (b - 1)/(t - 1) then by (c), b1/n < b-wy or bw + (1/n) < y for sufficiently large n.

(e) Choose t = bw/y > 1. The rest is similar.

(f) From (a), bnn(b - 1) + 1 for all n. For which each z in R choose an n so that n(b - 1) > z - 1 or n(b - 1) + 1 > z. Hence for all z we have an n such that bnn(b - 1) + 1 > z. Hence the set {bn : n ∈ N} is unbounded. Now consider the function f : RR defined by f(x) = bx. If x < y then as B(x)B(y) so bx < by; i.e. f is an increasing function.

Define A = {w : bw < y} as in the problem. The set {bn : n ∈ N} being unbounded gaurantees the existence of a n such that bn > y. Thus n is an upper bound for A. Let x = sup A.

Suppose bx < y. By (d), for sufficiently large n, bx + (1/n) < y, i.e. x + 1/n is in A. But this is impossible as x = sup A. So bx < y is not possible. Suppose bx > y. By (e), for sufficiently large n, bx - (1/n) > y, i.e. x - 1/n is not in A. Since x - 1/n cannot possibly be the sup of A so there is a w in A such that x - 1/n < wx. But then as f was increasing, bx - 1/n < bw < y, a contradiction as bx - (1/n) > y. So bx > y is not possible.

Hence bx = y.

(g) The function f described in (f) is increasing and hence 1-1.

8Edit

Prove that no order can be defined in the complex field that turns it into an ordered field.

Solution. Suppose an order < had been defined. Now (i)2 = -1 > 0 by Proposition 1.18. This violates 1 > 0.

9Edit

Suppose z = a + bi, w = c + di. Define z < w if a < c, and also if a = c but b < d. Prove that this turns the set of all complex numbers into an ordered set. Does this ordered set have the least-upper-bound property?

Solution. Clearly if a < c then x < y. If a = c then either of the cases exist: b < d implies x < y, b > d implies x > y, b = d implies x = y. If a > c then x > y. Also if x = (a,b), y = (c,d) and z = (e,f) and x < y, y < z then we can establish x < y by considering the various cases. For example if a < c and c < e then clearly x < z. Similarly other cases may be handled. This set doesn't have the least upper bound property as the x-axis, a set bounded above by (1,0) doesn't have a least upper bound.

10Edit

Suppose z = a + bi, w = u + iv and , . Prove that z2 = w if v ≥ 0 and that = w if v ≤ 0. Conclude that every complex number (with one exception!) has two complex roots.