# 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

## 1Edit

If is rational () and is irrational, prove that and are irrational.

Solution. Let . If was rational then would be too. Similarly is irrational.

## 2Edit

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

Solution. Let, if possible, such that and . Now . By the fundamental theorem of arithmetic and therefore has both 2 and 3 in its factorization. So for some . But now and so , a contradiction.

## 3Edit

Prove Proposition 1.15.

Solution. The results follow from using the facts related to being a field.

## 4Edit

Let be a nonempty subset of an ordered set; suppose is a lower bound of and is an upper bound of . Prove that .

Solution. For note that and the result follows.

## 5Edit

Let be a nonempty set of real numbers which is bounded below. Let be the set of all numbers , where . Prove that inf =-sup.

Solution. Let inf and sup. We need to show that . We first show that is the upper bound of . Let . Then and so or follow. We now show that is the least upper bound of . Let be an upper bound of . Then , or . So is a lower bound of . Since inf so or .

## 6Edit

Fix .

(a) If are integers, , , and , prove that . Hence it makes sense to define .

(b) Prove that if and are rational.

(c) If is real, define to be the set of all numbers , where is rational and . Prove that sup when r is rational. Hence it makes sense to define sup for every real .

(d) Prove that for all real and .

Solution. (a) Suppose . Then and the fundamental theorem of arithmetic imply that and where . So and so we are done. If then reduce to lowest factors, say . Clearly now by the already worked out case when the ratios are coprime.

(b) We will let and and equivalently show that . Clearly . The last equality holds as the exponents are integers.

(c) Clearly . 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*. Now let

^{r}=(b^{m})^{1/n}>1*p*,

*q*be any rational numbers with

*p*<

*q*. As

*b*so

^{q-p}>1*b*or in other words for every

^{p}b^{q-p}=b^{q}>b^{p}*b*in

^{t}*B(r)*we have

*t*≤

*r*and so

*b*, i.e.

^{t}≤b^{r}*b*is the upper bound.

^{r}(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}.

## 7Edit

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 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}*- 1 ≥*

^{n}*n*(

*b*- 1).

^{1/n}(c) *b ^{1/n}* = (

*b*- 1) + 1 ≤ (

^{1/n}*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*or

^{-w}y*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*. If

^{x}*x*<

*y*then as

*B(x)*⊆

*B(y)*so

*b*<

^{x}*b*; i.e.

^{y}*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.

## 8Edit

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.

## 9Edit

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.

## 10Edit

Suppose *z* = *a* + *bi*, *w* = *u* + *iv* and , . Prove that z^{2} = w if v ≥ 0 and that = w if v ≤ 0. Conclude that every complex number (with one exception!) has two complex roots.