ДискретнаяМатематика / Student Solutions Manual / chapter 1

37.A common use of induction is to prove various facts that seem to be fairly obvious but are otherwise awkward or impossible to prove. These frequently involve expressions with ellipses. Use induction to show that

(a)X [(X1 \X2 \X3 \ \Xn) = (X [X1) \(X [X2) \ \(X [Xn):

(b)X \(X1 [X2 [X3 [ [Xn) = (X \X1) [(X \X2) [ [(X \Xn):

(c)(X1 \ X2 \ \ Xn) = X1 [ X2 [ [ Xn:

(d)(X1 [ X2 [ [ Xn) = X1 \ X2 \ \ Xn:

37. (a) Let n0 = 1: Let T = fn 2 N : n 1 and X [ (X1 \ X2 \ X3 \ Xn) = (X [ X1) \ (X [ X2) \ \ (X [ Xn)g

(Base step) For n = 1 the result is obvious, so 1 2 T .

(Inductive step) Assume n 2 T : Now prove that n + 1 2 T : That is, assume that n n0 and X [ (X1 \ X2 \ X3 \ Xk ) = (X [ X1) \ (X [ X2) \ \ (X [ Xn) are true and prove that

X [ (X1 \ X2 \ X3 \ Xn \ Xn+1) =

(X [ X1) \ (X [ X2) \ \ (X [ Xn) \ (X [ Xn+1) and n + 1 n0 are true.

X [ (X1 \ X2 \ \ Xn \ Xn+1) = X [ ((X1 \ X2 \ \ Xn) \ Xn+1) = (X [ (X1 \ X2 \ \ Xn)) \ (X [ Xn+1) =

((X [ X1) \ (X [ X2) \ \ (X [ Xn)) \ (X [ Xn+1)

Of course, n + 1 n0. Therefore, n + 1 2 T :

By the Principle of Mathematical Induction, T = fn 2 N : n 1g.

37. (b) Let n0 = 1: Let T = fn 2 N : n 1 and X \(X1 [X2 [X3 [ [

Xn) = (X \ X1) [ (X \ X2) [ [ (X \ Xn)g:

(Base step) The case n = 1 is obvious. Therefore, 1 2 T .

(Inductive step) Assume n 2 T : Now prove that n + 1 2 T : That is,

assume that n n0 and X \(X1[X2 [X3[ [Xn) = (X \X1)[(X \X2)[

[(X \Xn) are true and prove that X \(X1 [X2 [X3 [ [Xn [Xn+1) = (X \ X1) [ (X \ X2) [ [ (X \ Xn) [ (X \ Xn+1) and n + 1 > n0 are true.

X \ (X1 [ X2 [ [ Xn [ Xn+1) = X \ ((X1 [ X2 [ [ Xn) [ Xn+1) = (X \ (X1 [ X2 [ [ Xn)) [ (X \ Xn+1) =

((X \ X1) [ (X \ X2) [ [ (X \ Xn)) [ (X \ Xn+1) Of course, n + 1 n0. Therefore, n + 1 2 T :

By the Principle of Mathematical Induction, T = fn 2 N : n 1g.

37. (c) Let n0 = 1: Let

T = fn 2 N : (X1 \ X2 \ \ Xn) = X1 [ X2 [ [ Xng

(Base step) Let n = 1. The proof is obvious. Therefore, 1 2 T .

(Inductive step) Assume n 2 T : Now prove that n + 1 2 T : That is, assume that n n0 and

(X1 \ X2 \ \ Xn) = X1 [ X2 [ [ Xn

are true and prove that

(X1 \ X2 \ \ Xn \ Xn+1) = X1 [ X2 [ [ Xn [ Xn+1

Of course, n + 1 n0. Therefore, n + 1 2 T :

By the Principle of Mathematical Induction, T = fn 2 N : n 1g. 37. (d) Let n0 = 1: Let

T = fn 2 N : (X1 [ X2 [ [ Xn) = X1 \ X2 \ \ Xng

(Base step) The case n = 1 is obvious. Therefore, 1 2 T .

(Inductive step) Assume n 2 T : Now prove that n + 1 2 T : That is, assume that n n0 and

(X1 [ X2 [ [ Xn) = X1 \ X2 \ \ Xn are true and prove that

(X1 [ X2 [ [ Xn [ Xn+1) = X1 \ X2 \ \ Xn \ Xn+1 and n + 1 n0 are true.

X1 \ X2 \ \ Xn \ Xn+1

Of course, n + 1 n0. Therefore, n + 1 2 T :

By the Principle of Mathematical Induction, T = fn : n 1g.

39. Refer to the Square Root II algorithm.

p

= (Rn 17)2 2Rn

2

= n 2Rn

39. (c) The problem is asking for an upper bound on j 8 j. By part (b),

j 8 j < ( 12 )6 28 3 = ( 12 )1533

Thus log10(j 8 j) is approximately 461:5. So, the approximation is correct to at least 461 decimal places past the decimal point. (Of course, if you calculate this on the computer, you probably won't get 461 decimal places of accuracy. The problem is round-o error. Normally, a computer rounds o each real number, producing round-o errors. So, there is a programming-language induced limit on the accuracy of your answer. Also, as calculations are made with rounded-o numbers, the round-o errors may increase.) Note that this does not mean that if you print out the number, the rst 461 decimal places are accurate. For example, 0.99999, as an approximation to 1.00000, is accurate to 4 decimal places, but the actual digits in all those decimal places are di erent from those of 1.0000. . . .

As before, let R denote the value of Root after the loop has been executed

n p

n times. Let n = 17 Rn; the error in the calculation.

41.Using the Principle of Mathematical Induction, prove the following forms of the Principle of Mathematical Induction:

(a)Induction with a possibly negative starting point: Suppose that S Z;

that some integer n0 2 S; and that, for every n 2 Z; if n 2 S and n n0; then n + 1 2 S: Then, every integer n n0 is in S.

(b)Induction downward: Suppose that S Z; that some integer n0 2 S; and that, for every n 2 Z; if n 2 S and n n0; then n 1 2 S: Then, for every integer n n0, we have n 2 S:

(c)Finite induction upward: Let n0; n1 2 Z, n0 n1. Suppose that

S Z; that n0 2 S; and that for every n 2 Z; if n 2 S and n0 n n1, then n + 1 2 S: Then, every integer n where n0 n n1 is in S.

(Base step) Since n0 2 S, 0 2 S0. Suppose k 2 S0. So n0 + k 2 S.

(Inductive step) By the assumption of the problem, n0 + k + 1 2 S. Hence, k +1 2 S0. So, by the principle of mathematical induction, S0 = N.

Now prove that every integer n n0 is in S. Pick any such n. Since n n0, n n0 2 N. Hence, by the paragraph above, n n0 2 S0. So, by de nition of S0, n 2 S.

41. (b) Let S0 = fn 2 N : n0 n 2 Sg: The proof is now analogous to that in part (a).

41. (c) Let S0 = fn 2 N : n0 + n 2 S or n0 + n > n1g. The proof is now analogous to that of part (a), with separate cases depending upon

whether n0 + n n1 or n0 + n > n1.

41. (d) Let n 2 N. Then some n0 n 2 S, for otherwise S f0; 1; : : : ; n 1g, and S would be nite. Let k = n0 n.

Now let

S0 = fm 2 N : m > n0 or n0 m 2 Sg

Prove, analogously to part (a), that S0 = N. Then k 2 S0, so n 2 S.

1.11 Exercises

Assume all variables not given an explicit domain are elements of N:

1. The terms of a sequence are given recursively as a0 = 2; a1 = 6; and an = 2 an 1 + 3 an 2 for n 2: Find the rst eight terms of this sequence.

1.

3.The terms of a sequence are given recursively as a0 = 0; a1 = 4; and an = 8 an 1 16 an 2 for n 2: Find the rst eight terms of this sequence.

3.

nan

5.The terms of a sequence are given recursively as p0 = 1, p1 = 2, and

pn = 2 pn 1 pn 2 for n 2: Write out the information that the inductive step assumes and what the step must prove in proving bn = 2 3n is a closed form for the sequence. Suppose n0 = 0 is given and the base cases are 0 and 1.

5. There are two base cases. Let n > n0 + 1 so that n is not a base case and assume that for all k such that n0 k < n that k 2 T : That is, for each such k, bk = 2 3k is a closed form for pk . Now prove that n 2 T : That is, bn = 2 3n is a closed form for pn = 2pn 1 pn 2:

7.The terms of a sequence are given recursively as a0 = 0; a1 = 4; and

an = 8 an 1 16 an 2 for n 2: Write out the information that the inductive step assumes and what the step must prove in proving bn = n 4n is a closed form for the sequence. Suppose n0 = 0 and the base cases are 0 and 1.

7. There are two base cases. Let n > n0 +1 so that n is not a base case and assume that for all k such that n0 k < n that k 2 T : That is, for each

such k, bk = k 4k is a closed form for ak = 8ak 1 16ak 2. Now prove that n 2 T : That is, bn = n 4n is a closed form for an = 8an 1 16an 2:

9.Given that bn 1 = 2n+1 1 and bn 2 = 2n 1, prove that if bn = 3bn 1 2bn 2, then bn = 2n+2 1 provided n 2:

9.

bn = 3bn 1 2bn 2

=3(2n+1 1) 2(2n 1) (by ind. hyp.)

=3 2n+1 3 2n+1 + 2

=2 2n+1 1

=2n+2 1

11.The terms of a sequence are given recursively as a0 = 2, a1 = 6, and an = 2an 1 + 3an 2 for n 2: Prove by induction that bn = 2 3n is a closed form for the sequence.

11.Let n0 = 0. Let T = fn 2 N : an = 2 3n gives the nth term of the sequenceg.

(Base step) b0 = 2 and b1 = 6: Since a0 = b0 and a1 = b1, we have 0; 1 2 T :

(Inductive step) Now let n > n0 + 1 so that n is not a base case and assume for all k such that n0 k < n we have k 2 T : Prove n 2 T :

an = 2 an 1 + 3 an 2

=2 (2 3n 1) + 3 (2 3n 2) (by ind. hyp.)

=4 3n 1 + 2 3n 1

=6 3n 1 = 2 3n

Therefore, n 2 T :

By the Strong Form of Mathematical Induction, T = N.

13.The terms of a sequence are given recursively as a0 = 0, a1 = 4, and an = 8an 1 16an 2 for n 2: Prove by induction that bn = n 4n is a closed form for the sequence.

13.Let n0 = 0. There are two base cases. Let T = fn 2 N : an = n 4n gives the nth term of the sequenceg.

(Base step) b0 = 0 and b1 = 4: Since a0 = b0 and a1 = b1, we have 0; 1 2 T :

(Inductive step) Now, let n > n0 + 1 so that n is not a base case and assume for all k such that 1 k < n that k 2 T : Prove n 2 T :

an = 8 an 1 16 an 2

=8 ((n 1) 4n 1) 16 ((n 2) 4n 2) (by ind. hyp.)

=8 n 4n 1 16 n 4n 2 8 4n 1 + 32 4n 2

=2 n 4n n 4n 2 4n + 2 4n

=n 4n

Therefore, n 2 T :

By the Strong Form of Mathematical Induction, T = N.

15.(a) Prove that with just 3-cent and 5-cent stamps, you can make any amount of postage (any natural number of cents) except 1 cent, 2 cents, 4 cents, and 7 cents. (Hint: That you can make 0-cent postage is obvious. You need to prove two things: (i) that you can assemble any amount of postage except 1 cent, 2 cents, 4 cents, and 7 cents, and (ii) that you cannot assemble these four amounts. Be careful about whether you use the Principle of Mathematical Induction or the Strong Form of Mathematical Induction.)

(b)What amounts of postage can be assembled with 4-cent and 7-cent stamps only?

(c)What amounts of postage can be assembled with 8-cent and 10-cent stamps only?

(d)What amounts of postage can be assembled with 7-cent, 8-cent, and 10-cent stamps only?

(e)What amounts of postage can be assembled with 2-cent and 5-cent stamps only?

15. (a) It is obvious that no sum of 3's and 5's can equal 1, 2, 4, or 7. It is obvious that 8, 9, 10, 11, and 12 are possible. Let n0 = 8. There are three base cases. Let T = fn 2 N : you can make any amount of postage greater than or equal 8 cents with only 3 and 5 cent stamps.g

(Base step) It is clear that 8; 9; 10 2 T :

(Inductive step) Choose n such that n > n0 + 2 = 10 so that it is not a base case and assume for all k such that 7 k < n that k 2 T : We must now prove that n 2 T : Use the inductive hypothesis for n 3 and add one more 3 cent stamp. Therefore, n 2 T :

15. (b) It is obvious that no sum of 4's and 7's can equal 1, 2, 3, 5, 6, 9, 10, 13, 17. It is obvious that 18, 19, 20, and 21 are possible. Let n0 = 18. Let T = fn : you can make any amount of postage greater than or equal 18 cents with only 4 and 7 cent stamps.g

(Base step) It is clear that 18; 19; 20; and 21 2 T :

(Inductive step) Choose n such that 21 = n0 + 3 < n so that n is not a base case and assume for all k such that n0 k < n that k 2 T : We must now prove that n 2 T : Use the inductive hypothesis for n 4 cents and add one more 4 cent stamp. Therefore, n 2 T :

By the Strong Form of Mathematical Induction, T fn 2 N : n 18g. Adding the other initial cases gives

T= fn 2 N : n = 4; 7; 8; 11; 12; 14; 15; 16; or n 18g

15.(c) It is clear that no odd amount of postage can be made with 8 and

10.It is easy to show that the even numbers 2, 4, 6, 12, 14, and 22 cannot be formed as a sum of 8's and 10's.

(c) Fn2 + Fn2+1 is a Fibonacci number

17. Fix n. Let n0 = 1. There are two base cases. Let T = fm 2 N :

Fn+m = Fn Fm + Fn 1 Fm 1 and m 1g:

(Base step) m = 1: The lhs is Fn+1. The rhs is Fn F1 +Fn 1 F0 = Fn+1: For m = 2 the lhs is Fn+2: For the rhs

Fn F2 + Fn 1 F1 = 2Fn + Fn 1 = Fn + Fn+1 = Fn+2

Therefore, 1; 2 2 T .

(Inductive step) Choose m such that 2 = n0 + 1 < m so that n is not a base case and assume for all k such that n0 k < m that k 2 T : Now prove that m 2 T :

Fn+m = Fn+m 1 + Fn+m 2

=FnFm 1 + Fn 1Fm 2 + FnFm 2 + Fn 1Fm 3 (ind. hyp.)

=Fn(Fm 1 + Fm 2) + Fn 1(Fm 2 + Fm 3) collecting terms)

=FnFm + Fn 1Fm 1 (using Fibonacci relation)

Therefore, m 2 T :

By the Strong Form of Mathematical Induction, T = N f0g:

17. (a) F2n 1 = Fn 1(Fn + Fn 2)

17. (b) F3n 1 = F2n 1Fn +F2n 2Fn 1: By part (a) the conclusion follows.

17. (c) Fn+n = F 2 + F 2

n n 1

19. The Lucas numbers are de ned as L0 = 2, L1 = 1, and Ln = Ln 1 + Ln 2 for n 2: Prove that Ln+1 = Fn 1 + Fn+1 for n 2:

19. Let n = 2. There are two base cases. Let T = fn : n 2 and Ln =

Fn 1 + Fn+1g:

(Base step) For n = 2 we have the lhs and the rhs equaling three. For n = 3 we have

L3 = L2 + L1 = F1 + F3 + 1 = 4

Therefore, 2; 3 2 T .

(Inductive step) Choose n such that 3 = n0 + 1 < n so that n is not a base case and assume for all k such that n0 k < n that k 2 T : Now prove that n 2 T : That is, prove that Ln = Fn 1 + Fn+1 is true.

Ln+1 = Ln + Ln 1

=Fn 1 + Fn+1 + Fn 2 + Fn (ind. hyp.)

=Fn + Fn+2

Therefore, n 2 T :

By the Strong Form of Mathematical Induction, T = fn 2 N : n 2g.

21.What exactly is wrong with the following \proof" that for every real number x 0; x = 2x:

Suppose the result is true for all real numbers y where 0 y < x:

Case 1: x = 0: Then, 2x = 2 0 = 0 = x:

Case 2: x > 0: Then, 0 < x=2 < x: So, by hypothesis, x=2 = 2(x=2) = x: Doubling both sides, deduce that x = 2x: So, the result holds for every real number x 0 by the Strong Form of Mathematical Induction.

21. Both forms of the Principle of Mathematical Induction refer to sets of integers, not sets of reals. This exercise demonstrates that the analogous principle for real numbers is false.

23.The Binary Search of Phone Directory algorithm in Section 1.10.4 looks for any page (if any) containing a Name in a telephone book City. The portion of the algorithm used in searching for the page is called BinarySearch. Prove that the algorithm works correctly.

23 See the discussion in Chapter 9.

1.12 End of Chapter Materials

Starting to Review

1.Which of the following set descriptions gives the set f2, 8, 14, 20, 26, 32g?

(a)fn 2 N : n = 2x + 6 for some integer x such that 1 x 6g

(b)fn 2 N : n = 6x + 2 for some integer x such that 1 x 6g

(c)fn 2 N : n = 6x + 2 for some integer x such that 0 x < 6g

(d)None of the above

1.c

3.What is the contrapositive of the statement \If the sun is shining, then it is time to go outside."

(a) If the sun is shining, then it is not time to go outside.

(b) If it is time to go outside, then the sun is shining.

(c) If it is not time to go outside, then the sun is not shining.

(d) None of the above

3.c