ДискретнаяМатематика / Student Solutions Manual / chapter 1

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

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

By the Principle of Mathematical Induction, T = N.

19.Prove by induction for all n 0:

(a)3 divides n3 + 2n

(b)5 divides n5 n

(c)6 divides n3 n

(d)6 divides n3 + 5n

19. (a) Let n0 = 0: Let T = fn 2 N : 3 j (n3 + 2n)g: (Base step) Since 3j0, we have 0 2 T .

(Inductive step) Assume n 2 T : Now prove n + 1 2 T : That is, assume that n n0 and that 3 j (n3 + 2n) is true and prove that 3 j ((n + 1)3 + 2(n + 1)) and n + 1 n0 are true.

(n + 1)3 + 2(n + 1) = n3 + 3n2 + 3n + 1 + 2n + 2

=n3 + 3n2 + 5n + 3

=(n3 + 2n) + (3n2 + 3n + 3)

By the inductive hypothesis 3j(n3 + 2n): Clearly, 3j(3(n2 + n + 1)). Since 3 divides both terms of the sum, 3 divides the sum. Of course, n + 1 n0. Therefore, n + 1 2 T :

By the Principle of Mathematical Induction, T = N. 19. (b) Let n0 = 0: Let T = fn 2 N : 5 j (n5 n)g: (Base step) Since 5j0, we have 0 2 T .

(Inductive step) Assume n 2 T : Now prove n + 1 2 T : That is, assume that n n0 and that 5 j (n5 n) are true and prove that 5 j ((n + 1)5 (n + 1)) and n + 1 n0 are true.

(n + 1)5 (n + 1) = n5 + 5n4 + 10n3 + 10n2 + 5n + 1 n 1

=n5 + 5n4 + 10n3 + 10n2 + 4n

=(n5 n) + (5n4 + 10n3 + 10n2 + 5n)

By the inductive hypothesis 5j(n5 n): Clearly, 5j(5n4 +10n3 +10n2 +5n), so 5 divides the sum. Of course, n + 1 n0. Therefore, n + 1 2 T :

By the Principle of Mathematical Induction, T = N.

19. (c) Let n0 = 0: Let T = fn 2 N : 6 j (n3 n)g: (Base step) Since 6j0, we have 0 2 T .

(Inductive step) Assume n 2 T : Now prove n + 1 2 T : That is, assume that n n0 and that 6 j (n3 n) are true and prove that 6 j ((n + 1)3 (n + 1)) and n + 1 n0 are true.

(n + 1)3 (n + 1) = n3 + 3n2 + 3n n 1 = (n3 n) + (3n2 + 3n)

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

By the Principle of Mathematical Induction, T = N. 19. (d) Let n0 = 0: Let T = fn 2 N : 6 j (n3 + 5n)g: (Base step) Since 6j0, we have 0 2 T .

(Inductive step) Assume n 2 T : Now prove n + 1 2 T : That is, assume that n n0 and that 6 j (n3 + 5n) are true and prove that 6 j ((n + 1)3 + 5(n + 1)) and n + 1 n0 are true.

By the inductive hypothesis 6 j (n3 + 5n): Since 2 j n(n + 1) we have that 6 j 3n(n + 1), and clearly 6 j 6: Of course, n + 1 n0. Therefore, n + 1 2 T :

By the Principle of Mathematical Induction, T = N.

21.Prove by induction. The sum of the cubes of any three consecutive natural numbers is divisible by 9.

21. Let n0 = 0: Let T = fn 2 N : 9 j (n3 + (n + 1)3 + (n + 2)3)g: (Base step) For n = 0, (03 + 13 + 23) = 9: Clearly, 9j9 so 0 2 T :

(Inductive step) Assume n 2 T : Now prove n + 1 2 T : That is, assume that n n0 and that 9 j (n3 + (n + 1)3 + (n + 2)3) are true and prove that 9 j ((n + 1)3 + (n + 2)3 + (n + 3)3) and n + 1 n0 are true.

(n + 1)3 + (n + 2)3 + (n + 3)3

=n3 + 3n2 + 3n + 1 + n3 + 6n2 + 12n + 8 + n3 + 9n2 + 27n + 9

=3n3 + 18n2 + 42n + 18

=(3n3 + 9n2 + 15n + 9) + (9n2 + 27n + 9)

=(n3 + (n + 1)3 + (n + 2)3) + (9n2 + 27n + 9)

By the inductive hypothesis 9j(n3 + (n + 1)3 + (n + 2)3). Clearly, 9j(9(n2 + 3n+1)): Therefore, 9j((n+1)3 +(n+2)3 +(n+3)3). Of course, n+1 n0, Therefore, n + 1 2 T :

By the Principle of Mathematical Induction, T = N.

23.Prove by induction that the following identities are true for the Fibonacci numbers.

(Base step) For n = 0 we have F1 = 1 and F2 1 = 1: Therefore, 0 2 T .

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

By the Principle of Mathematical Induction, T = N.

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

L1 + L2 + + Ln = Ln+2 3

are true and prove that

L1 + L2 + + Ln + Ln+1 = Ln+3 3

and n + 1 n0 are true.

L1 + L2 + + Ln + Ln+1 = (L1 + L2 + + Ln) + Ln+1

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

By the Principle of Mathematical Induction, T = fn 2 N : n 1g. 25. (b) Let n0 = 2: Let

T = fn 2 N : n 2 and L21 + L22 + L23 + + L2n = Ln Ln+1 2g

(Base step) For n = 2 both the left-hand side and the right-hand side

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

By the Principle of Mathematical Induction, T = fn 2 N : n 2g. 25. (c) Let n0 = 1: Let

T = fn 2 N : n 1 and L2 + L4 + + L2 n = L2 n+1 1g:

(Base step) For n = 1 both the left-hand side and the right-hand side equal 3.

(Inductive step) Choose n such that n n0 and assume n 2 T : Now

prove that n +1 2 T : That is, assume that L2 +L4 + +L2n = L2n+1 1 and prove that L2 + L4 + + L2n + L2(n+1) = L2n+3 1:

L2 + L4 + + L2n + L2(n+1) = (L2 + L4 + + L2n) + L2(n+1)

= L2n+1 1 + L2(n+1)

= L2n+1 1 + L2n+2

= L2n+3 1

Therefore, n + 1 2 T :

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

27.Find a rational number representing each of the following repeating decimals.

(a)0.537537537537537537537537537.. .

(b)31.25469696969696969696969.. .

27. (a)

537

999

27. (b)

3094215

99000

29.(a) Prove by induction that 2n > n for n 0.

(b)Prove that 2n > n directly from Theorem 2 in Section 1.7.4 without explicit use of induction. (That is, Theorem 2 in Section 1.7.4 itself was proved using induction, but you should not have to do any additional induction.)

(c)Prove by induction that 2n > n3 for n 10:

29. (a) Let n0 = 0: Let T = fn 2 N : 2n > ng: (Base step) 20 = 1 > 0 so 0 2 T :

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

2n+1 = 2n + 2n

2n + 1 for n 0

> n + 1 (by ind. hyp.)

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

By the Principle of Mathematical Induction, T = N.

29. (b) Theorem 2 says that for any n 2 N there are 2n subsets in the power set of a set with n elements. Since each element of the set can be considered as a singleton subset and consequently one of the subsets, the number of singleton subsets is less than the number of subsets. (The empty set is a subset, but does not arise as a singleton subset.)Therefore, n < 2n.

29. (c) Let n0 = 10: Let

T = fn 2 N : n 10 and 2n > n3g:

(Base step) Let n0 = 10: Clearly 1024 > 1000: Hence, 10 2 T .

(Inductive step) Assume n 2 T : We assume that n n0 and 2n > n3 are true and we must prove 2n+1 > (n + 1)3 or n + 1 2 T and n + 1 n0 are true.

2n+1 = 2 2n

>(1 + 1=10)3 2n(1 + 1=n)3 2n

>((n + 1)=n)3 n3 = (n + 1)3

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

By the Principle of Mathematical Induction, T = fn 2 N : n 10g:

31.Let T = fn 2 N : sin (n ) = 0g: Prove that T = N: (Hint: sin(a + b) = sin(a) cos(b) + cos(a) sin(b):)

31. Let n0 = 0: Let T = fn 2 N : sin( n ) = 0g: (Base step) sin( 0 ) = 0: Therefore, 0 2 T :

(Inductive step) Assume that n 2 T : That is, assume that n n0 and sin(n ) = 0 are true and prove that sin((n + 1) ) = 0 and n + 1 n0 are true.

sin ((n + 1) ) = sin(n + ) = sin( n )cos( ) + cos( n )sin( ) = 0

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

By the Principle of Mathematical Induction, T = N.

33.(a) Suppose you take out a mortgage for A dollars at a monthly interest rate I and a monthly payment P: (To calculate I : if the annual interest rate is 12%, divide by 12 to get a monthly rate of 1%, then replace the percentage with the decimal fraction 0.01.) Let An denote the amount you have left to pay o after n months. So, A0 = A by de nition. At the end of each month, you are rst charged interest on all the money you owed during the month and then your payment is subtracted. So,

An+1 = An(1 + I) P

(b) Use this to calculate the monthly payment on a 30-year loan of $100,000 at 12% interest per year. (Note that the formula is inexact, since money is always rounded o to a whole number of cents. The derivation here does not do that. We use 12% to make the arithmetic easier. You should consult a local bank to nd a current value.)

33. (a) Let n0 = 0: Let

T = fn 2 N : n n0 and An = (A PI )(1 + I)n + PI g

(Base step) For n = 0 the left-hand side and the right-hand side both equal A0: Therefore, 0 2 T .

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

An = (A PI )(1 + I)n + PI

are true and prove that

An+1 = (A PI )(1 + I)n+1 + PI

and n + 1 n0 are true.

An+1 = An(1 + I) P

=((A PI )(1 + I)n + PI )(1 + I) P

=(A PI )(1 + I)n+1 + PI + P P

=(A PI )(1 + I)n+1 + PI

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

By the Principle of Mathematical Induction, T = N.

33. (b) The key for part (b) is that after the last payment the amount due is 0. In this case A360 = 0: The answer will be found by solving for P in the following expression:

0= (100000 P=0:01)(0:01)360 + P=0:01

35.Prove Theorem 4 of Section 1.5.4, in full generality. You may use Theorem 3 of Section 1.5.3, because it has already been proven. (Hint: Use induction on the number of sets.)

35.The proof will be by induction on n the number of sets. Let n0 = 2, and let

T = fn 2 N : n 2 and the Principle of Inclusion-Exclusion holds for n setsg

(Base step) The result is true for n = 1; 2; 3 by Theorem 3 of Section 1.5.3, so 2 2 T .

(Inductive step) Assume n 2 T and prove that n + 1 2 T : That is, assume that n 2 and that the Principle of Inclusion-Exclusion holds for n sets are true and prove that the Principle of Inclusion-Exclusion holds for n + 1 sets and n + 1 n0 are true.

A1 [ A2 [ [ An [ An+1 = (A1 [ [ An) [ An+1

Using the result for n = 2 gives

j A1 [ A2 [ [ An [ An+1 j

= j( A1 [ A2 [ [ An ) + An+1

= j A1 [ A2 [ [ An j + j An+1 j

j (A1 [ A2 [ [ An) \ An+1 j

but

j (A1 [ A2 [ [ An) \ An+1 j =

j (A1 \ An+1) [ (A2 \ An+1) [ [ (An \ An+1) j

Therefore,

j A1 [ A2 [ [ An [ An+1 j = j A1 [ A2 [ [ An j + j An+1 jj (A1 \ An+1) [ (A2 \ An+1) [ [ (An \ An+1) j

Two of the three expressions involve the union of n sets so the inductive hypothesis can be applied to each of them. The subscripts shown and always distinct between or among themselves.

j A1 [ A2 [ [ An j =

n

XX

X

+j Ai \ Aj \ Ak j +

i;j;k n

+ ( 1)n 1j A1 \ A2 \ \ An j

and

j (A1 \ An+1) [ (A2 \ An+1) [ [ (An \ An+1) j =

X

+j (Ai \ An+1) \ (Aj \ An+1) \ (Ak \ An+1) j +

i;j;k n

+ ( 1)n 1j (A1 \ An+1) \ (A2 \ An+1) \ \ (An \ An+1) j

Two simpli cations. The rst is that only one copy of An+1 need appear in the intersections of pairs of sets as generated in the right-hand side of the second equation above. The second simpli cation is to recognize that when only one copy of An+1 appears, two terms combine to form a subsum that can be simpli ed as follows:

and so on. Carrying out the details of these simpli cations yields the result.