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

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


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.


Prove Proposition 1.15.

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


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.


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  .


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 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   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  ,  . Prove that z2 = w if v ≥ 0 and that   = w if v ≤ 0. Conclude that every complex number (with one exception!) has two complex roots.