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.


Chapter 1Edit


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.


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.


Prove Proposition 1.15.

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


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.


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=infA 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=infA so -z\le x or -x\le z.


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=supB(r) when r is rational. Hence it makes sense to define b^x=supB(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 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.


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 n>\frac{b-1}{t-1} 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.


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.


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.


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.