# 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 ($r\neq 0$) 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, $p,q(\neq 0)\in \mathbb {Z}$ 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 $\alpha \leq \beta$.
Solution. For $x\in E$ note that $\alpha \leq x\leq \beta$ 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 $x\leq -a$ or $-x\geq a$ follow. We now show that $-x$ is the least upper bound of $-A$. Let $z$ be an upper bound of $-A$. Then $\forall a\in -A$, $a\leq z$ or $-a\geq -z$. So $-z$ is a lower bound of $A$. Since $x=$inf$A$ so $-z\leq x$ or $-x\leq z$.

## 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 $b^{r+s}=b^{r}b^{s}$ 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 $t\leq x$. 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 $b^{x+y}=b^{x}b^{y}$ 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 $(m,n)\neq 1$ 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 $b^{mq+pn}=(b^{r}b^{s})^{nq}$. Clearly $(b^{r}b^{s})^{nq}=(b^{r})^{nq}(b^{s})^{nq}=(b^{m/n})^{nq}(b^{p/q})^{nq}=b^{mq}+b^{pn}=b^{mq+pn}$. The last equality holds as the exponents are integers.
(c) Clearly $b^{r}\in B(r)$. We need merely show that b$^r$ is an upper bound for B(r) since being in B(r) it then automatically becomes its supremum.
Clearly b$^{1/n}$>1. Now if r=m/n is any positive rational then b$^r$=(b$^m$)$^{1/n}$>1. Now let p,q be any rational numbers with p<q. As b$^{q-p}$>1 so b$^p$b$^{q-p}$=b$^q$>b$^p$ or in other words for every b$^t$ in B(r) we have t≤r and so b$^t$≤b$^r$, i.e. b$^r$ 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) b$^r$=b$^{p+q}$=b$^p$b$^q$≤b$^x$b$^y$. So b$^x$b$^y$ is an upper bound for {b$^r$:r≤x+y} or b$^{x+y}$≤b$^x$b$^y$.
Now suppose p, q are rationals with p≤x and q≤y. Then b$^{p+q}$ is in B(x+y) and so b$^p$b$^q$=b$^{p+q}$≤b$^{x+y}$ by (b) and so b$^p$≤b$^{x+y}$/b$^q$. Now b$^p$ is in B(x). So for all q b$^{x+y}$/b$^q$ is an upper bound for B(x) as p can be chosen arbitrarily. By definition b$^x$≤b$^{x+y}$/b$^q$ and so b$^q$≤b$^{x+y}$/b$^x$. Again q can be chosen arbitarily so that b$^{x+y}$/b$^x$ is an upper bound for B(y). As before this leads to b$^y$≤b$^{x+y}$/b$^x$ or b$^x$b$^y$≤b$^{x+y}$.

## 7

Fix b>1, y>0 and prove that there is a unique real x such that b$^x$=y by completing the following outline. (This x is called the logarithm of y to the base b.)
(a) For any positive integer n, b$^n$-1≥n(b-1).
(b) b-1≥n(b$^{1/n}$-1)
(c) If t>1 and $n>{\frac {b-1}{t-1}}$ then b$^{1/n}$<t.
(d) If w is such that b$^w$<y then b$^{w+(1/n)}$<y for sufficiently large n.
(e) If b$^w$>y, then b$^{w-(1/n)}$>y for sufficiently large n.
(f) Let A be the set of all w such that b$^w$<y and show that x=sup A satisfies b$^x$=y.
(g) Prove that this x is unique.
Solution. (a) Clearly each of b$^{n-1}$, b$^{n-2}$,...b is greater than 1 and summing them and applying the formula of the finite sum of a geometric series gives the result.
(b) As b$^{1/n}$ > 1 so by (a), (b$^{1/n}$)$^n$ - 1 ≥ n(b$^{1/n}$ - 1).
(c) b$^{1/n}$ = (b$^{1/n}$ - 1) + 1 ≤ (b - 1)/n + 1 < t.
(d) Note that 1 < b$^{-w}$y = t (say). Choose n > (b - 1)/(t - 1) then by (c), b$^{1/n}$ < b$^{-w}$y or b$^{w+(1/n)}$ < y for sufficiently large n.
(e) Choose t = b$^w$/y > 1. The rest is similar.
(f) From (a), b$^n$ ≥ 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 b$^n$ ≥ n(b - 1) + 1 > z. Hence the set {b$^n$ : n ∈ N} is unbounded. Now consider the function f : R → R defined by f(x) = b$^x$. If x < y then as B(x) ⊆ B(y) so b$^x$ < b$^y$; i.e. f is an increasing function.
Define A = {w : b$^w$ < y} as in the problem. The set {b$^n$ : n ∈ N} being unbounded gaurantees the existence of a n such that b$^n$ > y. Thus n is an upper bound for A. Let x = sup A.
Suppose b$^x$ < y. By (d), for sufficiently large n, b$^{x+(1/n)}$ < y, i.e. x + 1/n is in A. But this is impossible as x = sup A. So b$^x$ < y is not possible. Suppose b$^x$ > y. By (e), for sufficiently large n, b$^{x-(1/n)}$ > 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, b$^{x-1/n}$ < b$^w$ < y, a contradiction as b$^{x-(1/n)}$ > y. So b$^x$ > y is not possible.
Hence b$^x$ = 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 z$^2$ = 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.