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1 When k, +!! # + + # + F Therefore, () holds when k Thuswemayassumethatk In this case, F k+ F k + F k +! k! k # + +! k! k # +! k + +! k #! k +! k # +! k +!#! k!# +! k +! #! k! # +! k+! k+ # By the Strong Principle of Mathematical Induction, it follows that [( F n + ) n ( ) n ] for every positive integer n Solution The primary error is not being careful when attempting to give aproof Thesecond line of the proof should read: Assume, for an arbitrary nonnegative integer k, thati 0for every integer i with 0 i k Suppose that k 0 Thenwemustshowthat(k +)0 where k 0 Thenextsentencestates: Leti and j be integers such that 0 i k and 0 j k and i + j k + When k 0(whichisapossiblevalueofk), the only nonnegative integers i and j for which i + j k + are i 0andj (ori andj 0) However, there is no integer j satisfying 0 j k when j andk 0 Sonosuchintegersi and j exist Hence we cannot show that (k +)0 for an arbitrarynonnegative integer k (because it s not true) (a) s,s,s ands 8 (b) Conjecture: s n F n+ for every positive integer n Proof We proceed by the Strong Principle of Mathematical Induction First, s F Assume,foranarbitrarypositiveintegerk, thats i F i+ every integer i with i k We show that s k+ F k+ First, s F,sowemayassumethatk when covering the (k +) squares of the (k +) board by k + dominos, either the top two squares are covered by a horizontal domino or the top four squares are covered by two vertical dominos Thus s k+ s k + s k By the induction hypothesis, s k+ F k+ +F k F k+ BytheStrongPrincipleofMathematicalInduction,s n F n+ for every positive integer n Supplementary Exercises for Chapter Proof We proceed by induction Since ( + )( + ),

2 the formula holds for n Assumethat k(k +) for an arbitrary positive integer k Weshowthat (k +)(k +) (k +)(k +) k(k +)(k +) [ (k +)(k +)(k +) ] k(k +) (k +)(k +) + k(k +)(k +) (k +)(k +) + k(k +)(k +)+(k +)(k +) (k +)(k +)(k +) for every positive integer n Let Then and so n(n +) S (n ) n(n +)(n +) (n ) + (n ) + +n(n ) S [+n )] + [ + (n )] + +[(n ) + ] n(n ) for every positive integer n (n ) n(n ) Proof We use induction Since ( ), the formula holds for n Assumethat (k ) where k is an arbitrary positive integer We show that (k +) k(k ), (k +)((k +) ) (k +)(k +) (k +) [++7+ +(k )] + (k +) k(k ) k(k ) + (k +) +(k +) k +k + (k +)(k +) for every positive integer n (n ) n(n )

3 Proof We verify this formula by induction Since the formula holds for n Assumethat + + +(k ) ( )( +), where k is an arbitrary positive integer We show that + + +(k +) k(k )(k +), (k +)(k +)(k +) + + +(k +) [ + + +(k ) ]+(k +) for every positive integer n + + +(n ) Proof We verify this formula by induction Since the formula holds for n Assumethat k(k )(k +) +(k +) k(k )(k +)+(k +) (k +)[k(k ) + (k +)] (k +)(k +k +) (k +)(k +)(k +) n(n )(n +) ( + )( + )( + ) k (k +) (k +) for an arbitrary positive integer k We show that + + +(k +)(k +)(k +) + + +(k +)(k +)(k +), k(k +)(k +)(k +) (k +)(k +)(k +)(k +) [ k (k +) (k +)]+(k +)(k +)(k +) k(k +)(k +)(k +) +(k +)(k +)(k +) k(k +)(k +)(k +)+(k +)(k +)(k +) (k +)(k +)(k +)(k +)

4 + + + n (n +) (n +) for every positive integer n Proof We use induction Since the formula holds for n Assumethat (!) ( + )!!, (!) + (!) + + k(k!) (k +)! for an arbitrary positive integer k We show that n(n +)(n +)(n +) (!) + (!) + +(k +)[(k +)!](k +)! (!) + (!) + +(k +)[k +)!] [(!)+(!)+ + k(k!)] + (k +)[k +)!] [(k +)! ] + (k +)[(k +)!](k +)[(k +)!] (k +)! for every positive integer n (!) + (!) + + n(n!) (n +)! Proof We use induction Since > ++, the inequalityholds forn Assume that k >k + k +foranarbitraryintegerk We show that k+ > (k +) +(k +)+k +k ++k +k +k + k+ k > (k + k +)k +k +k + k +k + k + k k +k + k +k +k + k +k +k +>k +k + n >n + n +foreveryintegern 7 Proof # for n, as desired n+ < + n +(n +) n, Proof # We use induction Since < +(+),theinequalityholdsfor n Assumethat k+ < +(k +) k for an arbitrary positive integer k Weshowthat k+ < +(k +) k+

5 k+ k+ < [ +(k +) k] +(k +) k+ +(k +) k+ k+ +(k +) k+ + ( k+) < +(k +) k+ n+ < +(n +) n for every positive integer n 8 Proof We employ induction Since >,theinequalityholdsforn Assumethat k >k,wherek is an arbitrary positive integer We show that k+ > (k +) Since >, it follows that k+ > (k +) when k Henceweassumethatk k+ k > k k +k k +k k k +k k +k +k k +k +>k +k +(k +) n >n for every positive integer n 9 Proof Since 0 0 > 0 000,theinequalityholdswhenn 0 Assumethat k >k where k 0 is an arbitrary integer We show that k+ > (k +) Observethat k+ k > k k + k k +0k k +k +7k > k +k +7k k +k +k +k > k +k +k +(k +) n >n for every integer n 0 0 Proof We proceed by induction Since >,theinequalityholdswhenn Assume that k >k for an arbitrary positive integer k Weshowthat k+ > (k +) For k,wehave > 8 and so k+ > (k +) when k Fork, observe that k+ k > k k +k k +(k)k k +k k +k +k k +k +(k)k k +k +k k +k +k +k > k +k +k +(k +) n >n for every positive integer n Proof We proceed by induction Since >!, the statement is true for n Assume that k k >k!foranarbitraryintegerk We show that (k +) k+ > (k +)! (k +) k+ (k +)(k +) k > (k +)k k > (k +)k! (k +)! n n >n!foreveryintegern Proof We proceed by induction Since,theinequalityholdsforn Assume that k k for an arbitrary positive integer k We show that (k +) k +

6 (k +) ( k ) + (k +) + k + (k +) + (k +) + k k(k +) k + k + k(k +) < k + k k(k +) k(k +) k(k +) (k +) n integer n Proof We use induction Since n for every positive > > 7 > /, the inequality holds for n Assumethat k > (k +) / for an arbitrary integer k We show that k+ > (k +) / k + ( k ) + > (k +) / + k + k + k + k + k +(k +) / k + > k +(k +) / k + for every integer n (k +) / > (n +) / n

7 7 Proof We proceed by induction If n 0,then0 +00 is even Assume that k + k is even for an arbitrary nonnegative integer k Thus k + k x for some integer x We show that (k +) +(k + ) is even Therefore, (k +) +(k +) k +k ++k +(k + k)+k + x +k +(x + k +) Since x + k + is aninteger, (k +) +(k +)is even By the Principleof Mathematical Induction, n + n is even for every nonnegative integer n Proof We proceed by induction Since ( + x) +x, theinequalityholdswhenn Assume that ( + x) k +kx, wherek is an arbitrary positive integer We show that ( + x) k+ +(k +)x since + x>0 Thus ( + x) k+ (+x)( + x) k ( + x)( + kx) ( + x) k+ ( + x)( + kx) +(k +)x + kx +(k +)x since kx 0 (+x) n +nx for every positive integer n Proof We proceed by induction Since the formula holds for n 0 Assumethat a +0a (0 + )(a +0), k (a + i) i0 (k +)(a + k) for an arbitrary nonnegative integer k We show that k+ (a + i) i0 k+ (a + i) i0 (k +)(a + k +) [ k ] (a + i) +(a + k +) i0 (k +)(a + k)+(a + k +) for every nonnegative integer n n (a + i) (n +)(a + n) i0 7 Proof We proceed by induction Since k +ak +a +k + (k +)(a + k) a +0 b a (0 + )(a +0 b), +(a + k +) k +ak +a +k +

8 8 the formula holds for n 0 Assumethat k (a + ib) i0 (k +)(a + kb) for some nonnegative integer k We show that k+ (a + ib) i0 k+ (a + ib) i0 (k +)[a +(k +)b] [ k ] (a + i) +[a +(k +)b] i0 (k +)(a + kb)+(a + kb + b) for every nonnegative integer n bk +ak +bk +a +b (k +)(a + kb) n (a + ib) (n +)(a + nb) i0 +[a +(k +)b] bk +ak +bk +a +b 8 Proof We proceed by the Strong Principle of Mathematical Induction Since a +, the statement is true for n Assumeforapositiveintegerk that a i i + for every integer i with i k Weshowthat a k+ (k +)+k + Since a 7 +,itfollowsthata k+ k +fork Thuswemayassumethat k a k+ a k a k (k +) (k +) k +9 k k + By the Strong Principle of Mathematical Induction, a n n +for everypositive integer n 9 Proof We proceed by induction Since a 0 00,theformulaholdsforn 0 Assume that a k k k for a nonnegative integer k Weshowthata k+ (k +) k Notethat a k+ a k + k (k k )+ k k k + k (k +) k a n n n for every nonnegative integer n 0 Proof We proceed by the Strong Principle of Mathematical Induction Since a, the formula holds for n Assumeforapositiveintegerk that a i i for every integer i with i k Weshowthata k+ k+ Since a,itfollowsthata k+ k+ for k Thuswemayassumethatk a k+ a k +a k k + k k + k k k+ By the Strong Principle of Mathematical Induction, a n n for every positive integer n

9 9 Proof We proceed by the Strong Principle of Mathematical Induction Since a, the formula holds for n Assumeforapositiveintegerk that a i i(i +) for every integer i with i k We show that a k+ (k +)(k +) Since a, it follows that a k+ (k +)(k +)fork Thuswemayassumethatk a k+ a k a k +k(k +) (k )k + k[(k +) (k )] + k(k +)+ k +k +(k +)(k +) By the Strong Principle of Mathematical Induction, a n n(n +)foreverypositiveinteger n (a) a a a,a a a 8,a a a 8,a a a 8 (b) a, a, a, a 8, a, a 8 A reasonable conjecture would be that a n Fn,whereF n is the nth Fibonacci number for n (c) Proof We proceed by the Strong Principle of Mathematical Induction Since a F,thestatementistrueforn Assumeforapositiveintegerk that a i Fi for every integer i with i k Weshowthata k+ F k+ Fork,a k+ a F Thus a k+ F k+ for k Hencewemayassumethatk a k+ a k a k F k F k F k +F k F k+ By the Strong Principle of Mathematical Induction, a n Fn for every positive integer n Proof We proceed by the Strong Principle of Mathematical Induction Since F > ( ) ( ) i + /, the inequality holds for n Assumeforanintegerk, that Fi > + ( + for every integer i with i k WeshowthatF k+ > ) k ( + SinceF > ), ( + it follows that F k+ > ) k when k Hencewemayassumethatk Observe that F k+ F k + F k > +! k + +! k +! k + +! +! k +! k +! +! k +! ( + By the Strong Principle of Mathematical Induction, F n > ) n for every positive integer n (a) The integer x +y can equal 7 for nonnegative integers x and y only when x and y However, in all such cases, x +y 7 (b) Proof We use induction Since 8 +, the statement is true when n 8 Assume for an integer k 8thatk a +b for some nonnegative integers a and b We show that k +x +y for some nonnegative integers x and y Ifb, then k + a +b +a +b + +( ) (a +)+(b )

10 70 If b 0,thenk a for some integer a Thus k +a + ( ) + (a ) + for every integer n 8, there exist nonnegative integers x and y such that n x +y (a) S is a subset of itself and so S does not satisfy the definition of a well-ordered set (b) (0, ] [0, ] and (0, ] does not have a least element Proof First we assume that the Well-Ordering Principle is true Thus every nonempty subset of N has a least element We show that the Principle of Mathematical Induction is true Let P (n) beastatementforeachn N Assume that () P () is a true statement and that () if P (k), then P (k +)istrueforeveryk N Assume,tothecontrary,thatP (n) isnottrue for every positive integer n Let S be the set of positive integers n such that P (n) isfalse Thus S BytheWell-OrderingPrinciple,S contains a least element m SinceP () is true, m Hence m N and m / S Therefore,P (m ) is true By (), P (m) isalso true This, however, is a contradiction Next, we verify the converse Assume that the Principle of Mathematical Induction is true We show that the Well-Ordering Principle is true Assume, to the contrary, that the Well- Ordering Principle is not true Then there is a nonempty subset S of N having no least element For each n N, letp (n) beastatement P (n): S contains no element less than or equal to n Since S contains no least element, / S and so P () is true If P (k) istrueforsomepositive integer k, thenp (k + ) is certainly true; for otherwise, k +istheleastelementofs Thus the hypothesis of the Principle of Mathematical Induction is true,which implies that P (n) is true for every positive integer n ThusS, whichproducesacontradiction 7 Proof For n 0, let P (n) beastatement P (n): There exists a subset of S, thesumofwhoseelementsisn We prove that P (n) istrueforeveryintegern with n 0 by applying the Principle of Finite Induction Since the sum of the element(s) of the subset {} is, the basis step is established Assume, for an integer k with k<0, that there exists a subset T of S, the sum of whose elements is k WeshowthatthereisasubsetofS, thesumofwhoseelementsis k + If / T,thenT {} has the desired property Hence we may assume that T Since k<0, it follows that T S Thusthereexistsasmallestintegerm suchthatm/ T However then, the sum of the elements of (T {m }) {m} is k + Bythe Principleof Finite Induction, there exists a subset of S, the sum of whose elements is n for every integer n with n 0 8 Proof We use the Strong Principle of Mathematical Induction Since F 7 + F + F,thestatementistrueforn Assumeforanarbitraryintegerk that F i+ F i+ + F i for every integer i with i k We show that F k+ F k+ + F k Since this statement is true for k,wemayassumethatk Now F k+ F k+ + F k+ (F k+ + F k )+(F k+ + F k ) (F k+ + F k+ )+(F k + F k )F k+ + F k

11 7 By the Strong Principle of Mathematical Induction, for every positive integer n F n+ F n+ + F n 9 Proof We proceed by induction Since cos π, the statement holds for n Assume that π cos k+, where the number occurs k times in the expression on the left We show that π cos k+, where the number occurs k +timesintheexpressionontheleft Since π cos k+, where the number occurs k times in the expression on the left, it follows that π +cos () k+ Since cos x cos x foreveryrealnumberx, itfollowsthatcos x +cosx and so cos π π +cos k+ Takingthesquarerootofbothsidesof(),weobtain k π cos k+, where the number occurs k + times in the expression on the left By the Principle of Mathematical Induction, the statement holds for every positive integer n