# 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 ${\displaystyle r}$ is rational (${\displaystyle r\neq 0}$) and ${\displaystyle x}$ is irrational, prove that ${\displaystyle r+x}$ and ${\displaystyle rx}$ are irrational.

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

## 2

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

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

## 3

Prove Proposition 1.15.

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

## 4

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

Solution. For ${\displaystyle x\in E}$ note that ${\displaystyle \alpha \leq x\leq \beta }$ and the result follows.

## 5

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

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

## 6

Fix ${\displaystyle b>1}$.

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

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

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

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

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

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

(c) Clearly ${\displaystyle b^{r}\in B(r)}$. 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.

## 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 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 ${\displaystyle n>{\frac {b-1}{t-1}}}$ 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.

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