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

Unless the contrary is explicitly stated, all numbers that are mentioned in these exercises are understood to be real. =Chapter 1=

1
If $$r$$ is rational ($$r\ne 0$$) and $$x$$ is irrational, prove that $$r+x$$ and $$rx$$ are irrational.

Solution. Let $$r+x=y$$. If $$y$$ was rational then $$x=y-r$$ would be too. Similarly $$rx$$ is irrational.

2
Prove that there is no rational number whose square is 12.

Solution. Let, if possible, $$p,q(\ne 0)\in \mathbb{Z}$$ such that $$(p,q)=1$$ and $$\frac{p^2}{q^2}=12$$. Now $$12q^2=p^2$$. By the fundamental theorem of arithmetic $$p^2$$ and therefore $$p$$ has both 2 and 3 in its factorization. So $$36k^2=12q^2$$ for some $$k$$. But now $$3k^2=q^2$$ and so $$3|q$$, a contradiction.

3
Prove Proposition 1.15.

Solution. The results follow from using the facts related to $$\mathbb{R}$$ being a field.

4
Let $$E$$ be a nonempty subset of an ordered set; suppose $$\alpha$$ is a lower bound of $$E$$ and $$\beta$$ is an upper bound of $$E$$. Prove that $$\alpha\le \beta$$.

Solution. For $$x\in E$$ note that $$\alpha\le x\le \beta$$ and the result follows.

5
Let $$A$$ be a nonempty set of real numbers which is bounded below. Let $$-A$$ be the set of all numbers $$-x$$, where $$x\in A$$. Prove that inf $$A$$=-sup$$(-A)$$.

Solution. Let $$x=$$inf$$A$$ and $$y=$$sup$$(-A)$$. We need to show that $$-x=y$$. We first show that $$-x$$ is the upper bound of $$-A$$. Let $$a\in -A$$. Then $$-a\in A$$ and so $$x\le -a$$ or $$-x\ge a$$ follow. We now show that $$-x$$ is the least upper bound of $$-A$$. Let $$z$$ be an upper bound of $$-A$$. Then $$\forall a\in -A$$, $$a\le z$$ or $$-a\ge -z$$. So $$-z$$ is a lower bound of $$A$$. Since $$x=$$inf$$A$$ so $$-z\le x$$ or $$-x\le z$$.

6
Fix $$b>1$$.

(a) If $$m,n,p,q$$ are integers, $$n>0$$, $$q>0$$, and $$r=m/n=p/q$$, prove that $$(b^m)^{1/n}=(b^p)^1/q$$. Hence it makes sense to define $$b^r=(b^p)^{1/q}$$.

(b) Prove that $$b^{r+s}=b^rb^s$$ if $$r$$ and $$s$$ are rational.

(c) If $$x$$ is real, define $$B(x)$$ to be the set of all numbers $$b^t$$, where $$t$$ is rational and $$t\le x$$. Prove that $$b^r=$$sup$$B(r)$$ when r is rational. Hence it makes sense to define $$b^x=$$sup$$B(x)$$ for every real $$x$$.

(d) Prove that $$b^{x+y}=b^xb^y$$ for all real $$x$$ and $$y$$.

Solution. (a) Suppose $$(m,n)=1$$. Then $$mq=pn$$ and the fundamental theorem of arithmetic imply that $$p=km$$ and $$q=kn$$ where $$k\in \mathbb{N}$$. So $$((b^m)^{1/n})^q=((b^m)^k=b^p$$ and so we are done. If $$(m,n)\ne 1$$ then reduce $$m/n$$ to lowest factors, say $$s/t$$. Clearly now $$(b^m)^{1/n}=(b^s)^1/t=(b^p)^1/q$$ by the already worked out case when the ratios are coprime.

(b) We will let $$r=m/n$$ and $$s=p/q$$ and equivalently show that $$b^{mq+pn}=(b^rb^s)^{nq}$$. Clearly $$(b^rb^s)^{nq}=(b^r)^{nq}(b^s)^{nq}=(b^{m/n})^{nq}(b^{p/q})^{nq}=b^{mq}+b^{pn}=b^{mq+pn}$$. The last equality holds as the exponents are integers.

(c) Clearly $$b^r\in B(r)$$. 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 p1 so bpbq-p=bq>bp or in other words for every bt in B(r) we have t≤r and so bt≤br, i.e. br is the upper bound.

(d) Suppose r is a rational number with r0. Choose a rational p such that x-δ<p<x and put q=r-p. Then q1, 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 $$n>\frac{b-1}{t-1}$$ then b1/ny, then bw-(1/n)>y for sufficiently large n.

(f) Let A be the set of all w such that bw 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), bn ≥ n(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 bn ≥ n(b - 1) + 1 > z. Hence the set {bn : n ∈ N} is unbounded. Now consider the function f : R → R 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 < w ≤ x. 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.

8
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.

9
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.

10
Suppose z = a + bi, w = u + iv and $$a=\Big(\frac{|w|+u}{2}\Big)^{1/2}$$, $$b=\Big(\frac{|w|-u}{2}\Big)^{1/2}$$. Prove that z2 = w if v ≥ 0 and that $$\bar z^2$$ = w if v ≤ 0. Conclude that every complex number (with one exception!) has two complex roots.