j* v sji[* n sj $ (r\a]'sa(:\{, EL u (s\) = \ 8, Let x f (A\B) U B\A). Then x f A\l3 or x E B\A. If

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Alfred Carpenter
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$Chapter 0 PRELIMINARIES \ This chapter contains the basic material on sets, functions, relations and induction, as$ You will need to make some choices about what material from this chapter you cover in detail.

You will need to make some choices about what material from this chapter you cover in detail.

of real numbers as a complete ordered field.

oes for Solutions to Exercises 3, The proof is very similar to that of (v) of Theorem 0.. 4.

$Hence, A fl B C A. If x E A, then x E A U B, hence A C A U B. 6. If x E C\B, then x E C and x f B.$ $Here C\A = C\B but A $ k and B A. 7. A\(A\B) = B if and only if B C A.$ $To show this, you may want to prove that A\(A\B) = A fl 0. \ 8, Let x f (A\B) U B\A). Then x f A\l3 or x E B\A.$ $If x E B\A, a similar argument shows that % x E A U 0 and x f A fl B.$

1 Chapter 0 PRELIMINARIES \ This chapter contains the basic material on sets, functions, relations and induction, as well as the syater of real numbers. Your students may have seen a good share of this material before. You will need to make some choices about what material from this chapter you cover in detail. My experience has been that most students at t h i ~ level have little appreci'ation of the structure of the system of real numbers as a complete ordered field. For that reason, I have found that a comprehensive coverage of Section 0.5 is very important. oes for Solutions to Exercises 3, The proof is very similar to that of (v) of Theorem The proof is very similar to that of (i) of Theorem , If x E A n B, then x f A and x E 3, in particular, x E A. Hence, A fl B C A. If x E A, then x E A U B, hence A C A U B. 6. If x E C\B, then x E C and x f B. Since A C B and x j! B, then x f! A. Therefore, x E C\A. Thus, C\B C C\A. The converse is false as evidenced by the example C = J, A = -1,,3) and B = (-3,,3). Here C\A = C\B but A $ k and B A. 7. A\(A\B) = B if and only if B C A. To show this, you may want to prove that A\(A\B) = A fl 0. \ 8, Let x f (A\B) U B\A). Then x f A\l3 or x E B\A. If x E A\B, then x l A and x f! B, hence, x E A U B and I x f! A P 3. If x E B\A, a similar argument shows that % x E A U 0 and x f A fl B. In either case, x E (A U B \ A U B, Now assume x f (A U B)\(A fl 0). Then x E h S B and x f A n B. If xea,then x$b since x f A n B. If xeb, then xfa since x f A fl B. But x E A or x f 3 hence, x E A\B or x E B\A. Thus, x E $A\B) U $B\A~. We have shown that A\B) U B\A) C A U j* v sji[* n sj $ (r\a]'sa(:\*{, EL u (s\*) = A U B \ A n B. 9. The point of Russell's paradox is to show the student that using a rule to define a set can lead to logical difficulties. However, reassure your students that no such

$One distinction is that if x E S\( fl A1), then x E S 1 A and x 6 fl AX, hence x A for some p E A. Therefore, AEA P x E S\Ap, etc., etc. 1. a) R\{o} *b) R\C1,] 13.$ im f = {m E J: a is odd). f is 1-1, but not onto J. f has an inverse defined by for each m E im f. 14. dolm f = {x: x # -). f is I - 1 and as in example 0.

im f = {m E J: a is odd). f is 1-1, but not onto J. f has an inverse defined by for each m E im f. 14. dolm f = {x: x # -). f is I - 1 and as in example 0.

iu f = {,3,4,6,6) =D. f-':~ + A is defined by f' 1 (x) = r - t for each x E D.

There is b E B such that (a,b) E f, i.c., f(a) = b. Therefore, (b,a) E f-l, hence f-'(b) = a. Then (ilof)(a) = f-l(f(a)) = fd1(b) = a. The.proof for (f oi I) (b) = b is similar.

2 problems will arise in this book. 11, This proof is very similar to that of (i) of Theorem 0.4. One distinction is that if x E S\( fl A1), then x E S 1 A and x 6 fl AX, hence x A for some p E A. Therefore, AEA P x E S\Ap, etc., etc. 1. a) R\{o} *b) R\C1,] 13. im f = {m E J: a is odd). f is 1-1, but not onto J. f has an inverse defined by for each m E im f. 14. dolm f = {x: x # -). f is I - 1 and as in example 0.8, the best way to find ir f is to find the domain of the inverse. We find f- 1 (x) = x L d dom f-.- 1 = {x: x # I) = im f. 16. f(x) = 7 for all x f A. 16. f(x) = x + 1 fox each x E A. iu f = {,3,4,6,6) =D. f-':~ + A is defined by f' 1 (x) = r - t for each x E D. 17, f:a + B defined by f(x = x + I for each x E A; g:b -r C defined by g() = a, g(4) = b, g(4) = C, g(5) = d, g(6) = e, g(7) - e. 18. Suppoae a E A. There is b E B such that (a,b) E f, i.c., f(a) = b. Therefore, (b,a) E f-l, hence f-'(b) = a. Then (ilof)(a) = f-l(f(a)) = fd1(b) = a. The.proof for (f oi I) (b) = b is similar. Proofs by induction consist of verifying a staterent for one or more initid values and then showing that the truth of the statemnt for k or for all m k implies the truth of the statement for k + 1. This last proceas is often called the hd~cti~n step. For the exercises that involve induction, we will offer only the induction step. I $9. Assuming *. +n =, we hove

zn+31 0. Assuming 1 + 3 +... t (n - 1) = n, we have I +*3 +... + (m - 1) + (n + 1) = n + n + 1 = (n + 1) 1.

, We will use the result of Example 0.11 here, that is, n + 1 < " for all n) 3. Notice that n < " is false for n =, 3 and 4.

There is a smallest no E J such that P(no) is false, by the well-ordering principle, and by the hypothesis, no > r.

3 zn Assuming t (n - 1) = n, we have I +* (m - 1) + (n + 1) = n + n + 1 = (n + 1) 1. Assuming n3 + 5n is divisible by 6, we have nj + 5n is divisible by 6, as is 6, and since either n or n + 1 is even, 3n(n + 1) is divisible by 6., We will use the result of Example 0.11 here, that is, n + 1 < " for all n) 3. Notice that n < " is false for n =, 3 and 4. Now assume that n > 5 and n < " Then (a+l)=n+n+l<~n+n+~<n+"= p+i 3. Assume the hypotheses of 0.9 and assuple that P(n) is, false for some n E J. There is a smallest no E J such that P(no) is false, by the well-ordering principle, and by the hypothesis, no > r. But then P(no - 1) is true and P(k) is true for 1 < k 5 no - 1, hence P(no) is true, a contradiction..4. Assume f(k) < ' for all k 3 n. Then f(n + I) = f(n) + f(n - ) + f(n - ) < zn + "' + n- I I \ p+ w 1 + p- 1 = " + n = p+l,. 5. Assume f(k) <.4 for some k >. Then f(k + I) = i " < - 4 = 4- <.4. k 6- Assume f (k) = -5.3 t 5 k-1+$+3 for l(k<n. Then f(n + 1) = B*f(n) - 15=f(n - I) + 6. n+l = 8[-5-3" + 5 n- 1 +

7. Assure the hypotheses of 0.10 and assme P(n) is false forsome n>no. Let S=(nEZ: n>no) and T = {n E S: P(n) is false).

Then f(ko) is false, k, > no, so kg - 1 no and P(ko - 1) is true. But then P(ko) is true by the hypothesis, a contradiction. 8.

Then T is a non- empty subset 'of S and hence has a smallest member, call it ko.

Assume S is a countable set and T C S. If S is finite, then T is also finite and hence countable. If S.is countably infinite, there is a 1-1 function f:s + J that is onto.

4 7. Assure the hypotheses of 0.10 and assme P(n) is false forsome n>no. Let S=(nEZ: n>no) and T = {n E S: P(n) is false). T is a non- enpty subset of S, hence by the modified well-orderin principle, T has a smallest member, call it kg. Then f(ko) is false, k, > no, so kg - 1 no and P(ko - 1) is true. But then P(ko) is true by the hypothesis, a contradiction. 8. A~sume the hypotheses of the theorem and assume P(n$ is false for some n > no. As in Exercise 7, define and - T. Then T is a non- empty subset 'of S and hence has a smallest member, call it ko. Then P(ko) is false, kg no and for no 5 k < kg - 1, P(k) is true, hence P(kO) is true by (b) of the theorem. This is a contradiction. k 8. Assume f(k) = 5. + (-3)k for 0 5 k < n. Then f(n + 1) = 6*f(n - 1) - f(n) = "-'[ (-3)"-'[ = 5*"+' + (-3) n+ l Just a reminder: Be sure to check f(o) and f(1). 30. Assume S is a countable set and T C S. If S is finite, then T is also finite and hence countable. If S.is countably infinite, there is a 1-1 function f:s + J that is onto. Define g:t + J by g(t) = f(t) for each t ' T. g is 1-1, hence T is equivalent to in g which is a subeet of J. Hence T ie countable as ir g is a subset of J, hence countable by 0.14 and T is equivalent to im g. 3. The proof is by induction. Po is the set of all constant polynomials with integer coefficients, and of course that7s just, hence it is countable. Some people don't view the zero polynomial as having degree zero, so include it in Po anyway and the countability of Po isn't affected one way

5 or the other. Assume now that Pn is countable. Let A = (: z E, z # 0) and define f:ax Pn+P,,l as follows: P E P,. f(a,p) =ax n-t-1 +p foreach aea and This is a 1-1 function from A x Pn onto P,+i (there are a few details to fill in), hence P,+l is countable since both A and P, are countable as is A x P, by Theorem The set of all polynomials with integer coefficients is the union of the family {Pn)nEJ. This is a countable set by Theorem 0, f If S is countable, then it is equivalent to a subset of J, let f:s + J be the function that gives that equivalence. Define an indexed family {BnInEim as follows : B, = A, where f (s) = n Then U B = U A, is countable by nfim f " s S 35. For each p E P,, B(p) is finite since a polynomial of degree n has at most n real zeros. P, is countable, hence by Exercise 34, U B(p) is countable. PEP, 36. Let Q, = U B(p) as in Exercise 35. Q, is countable and p Pn U On is the set of all algebraic nurbers. nej 37. If f:a -, P(A) define C = jx: x A and x f! f(x)). * Furth~r, suppose that there 1s c E X,-uch that f(c) = C, ' Is cec? If c E C, then cff(c) = C whichisa contradiction. So it must be t at c f C, in which case c E f(c) = C, again a contradiction. o it must be the case that there is no such c E A, that is, C p ia f. x- & 38. f(x) = - a is s 1-1 function from [a,b] onto [O,i. Theorem 0.11 then shows that [a,b] is equivalent to {=,dl.

a 40. 0 5 (6- fi1 = x - ZR + y since x 0 and y > 0. Thus, fi 59. 41. Since 0 < a < b, then a = a a < ab < b b = b. Since 0 < a < b, then 0 < md 0 < fi.

So xy + xb < xy + ay or x(y + b) < y(x + a) and hence -* The manipulations with the + y < mrn inequalities were justified by the fact that x, y, a, and b are positive real numbers.

We may as well assume r O since a11 we need to establish is that there is a f A such that r < a and if r < 0, then 0 serves the purpose.

6 a (6- fi1 = x - ZR + y since x 0 and y > 0. Thus, fi Since 0 < a < b, then a = a a < ab < b b = b. Since 0 < a < b, then 0 < md 0 < fi. Since a#b, then # and so either f i e fi or fi>@. However, if fi > fi, then by the first part of this exercise, a > b. Thus, 0 < 6 < fi. x a 4. Since c and x, y, a, and b are positive, then xb < ay. So xy + xb < xy + ay or x(y + b) < y(x + a) and hence -* The manipulations with the + y < mrn inequalities were justified by the fact that x, y, a, and b are positive real numbers. The remaining inequality is established in a similar way. 43. Assume r is a rational number and r <. We may as well assume r O since a11 we need to establish is that there is a f A such that r < a and if r < 0, then 0 serves the purpose. low if r < and r > 0, choose a CL rational number 6 such that 0 < 6 < 1 and 6 < - m r'. we will show that (r + 6) <. Aote:' Since 0 < 6 < 1, then 6 c Suppose A is nonvoid and' x = sup A. If e > 0, then 'x - r c x, and henc~ x - r is not an upper bound for A, hence there is a E A such that x - E < a, but x is an qpper bound for A, hence 46. This proof is very similar t o that of Exercise 44. If E > 0, then y < y + E, SO y + r is not a lower bound for A, hence there is a E A such that a < y + e. As y is a lower bound for A,

PRINCIPLE OF MATHEMATICAL INDUCTION

Chapter 4 PRINCIPLE OF MATHEMATICAL INDUCTION 4.1 Overview Mathematical induction is one of the techniques which can be used to prove variety of mathematical statements which are formulated in terms of