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 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^pb^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^pb^q≤b^xb^y. So b^xb^y is an upper bound for {b^r:r≤x+y} or b^x+y≤b^xb^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^pb^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^xb^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, bⁿ-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 suffficiently 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 then 1 and summing them and applying the forumla 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^-wy = t (say). Choose n > (b - 1)/(t - 1) then by (c), b^1/n < b^-wy or b^{w + (1/n)} < y for sufficiently large n.

(e) Choose t = b^w/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) = 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 {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 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 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.