An introduction to mathematical thinking

Chapter Solutions An Introduction to Mathematical Thinking: Algebra and Number Systems William J Gilbert and Scott A Vanstone, Prentice Hall, 2005 Solutions prepared by William J Gilbert and Alejandro Morales Exercise 4-1: Calculate the following Solution: = = 5·4·3 3·2·1 60 = 10 Exercise 4-2: Calculate the following 10 Solution: 10 = 10 = 10 · · · = 210 4·3·2·1 Exercise 4-3: Calculate the following 8! (4!)2 Solution: 8! (4!)2 = = = 8·7·6·5 4·3·2·1 70 4.1 Exercise 4-4: Calculate the following 100! − 99! Solution: 100! − 99! = 99!(100 − 1) = (98!)(99)2 Exercise 4-5: Show that n n r = n−1 r r−1 Solution: For n, r ≥ 1, we have n n r n! n · r!(n − r)! (n − 1)! , since n! = n(n − 1)!, n ≥ r!(n − r)! n−1 (n − 1)! = r(r − 1)!(n − r)! r r−1 = = = establishing the result Exercise 4-6: Show that n r r s n s = n−s r−s Solution: n r r s = = = = r! n! · r!(n − r)! s!(r − s)! n! (n − s)! · s!(n − r)!(r − s)! (n − s)! n! (n − s)! · s!(n − s)! (r − s)!((n − s) − (r − s))! n n−s s r−s Exercise 4-7: Find n if n+2 n = 36 Solution: 4.2 By definition, n+2 n = (n + 2)! 2!n! Hence, we must solve n2 + 3n + = 72 or n2 + 3n + 70 = By the quadratic formula, we find n = or n = −10 Exercise 4-8: Write the following in sigma notation 99 + + + ··· + 100 Solution: 99 99 r + + + ··· + = 100 r=1 r + Exercise 4-9: Write the following in sigma notation + 15 + 24 + 35 + · · · + (n2 − 1) Solution: n + 15 + 24 + 35 + · · · + (n2 − 1) = (r2 − 1) r=3 Exercise 4-10: Write the following in sigma notation ak + a2k + a4k + a8k + a16k + · · · + a256k Solution: r ak + a2k + a4k + a8k + a16k + · · · + a256k = a2 k r=0 Exercise 4-11: Prove, by induction, the following results for all n ∈ P 12 + 22 + 32 + · · · + n2 = n(n+1)(2n+1) Solution: (i) When n = the assertion is true since 4.3 1(1+1)(2·1+1) = 6 = (ii) Suppose that 12 + 22 + 32 + · · · + k = k(k + 1)(2k + 1) Then 12 + 22 + 32 + · · · + k + (k + 1)2 k(k + 1)(2k + 1) 6(k + 1)(k + 1) + = 6 (k + 1)(k(2k + 1) + 6(k + 1) = (k + 1)(k(2k + 1) + 2(2k + 1) + 2(k + 2)) = (k + 1)(k + 2)(2k + + 2) (k + 1)(k + 2)(2(k + 1) + 1) = = 6 Hence the assertion is true for n = k + Therefore, by the principle of mathematical induction, the assertion is true for all n ∈ P Exercise 4-12: Prove, by induction, the following results for all n ∈ P n(n + 1) 13 + 23 + 33 + · · · + n3 = Solution: (i) If n = 1, then 13 = 1(1+1) 2 Hence the proposition is valid for n = (ii) Assume that 13 + 23 + · · · + k = k(k+1) 2 for some k ∈ P Then 13 + 23 + · · · + k + (k + 1)3 = = = k2 k(k + 1) + (k + 1)3 = (k + 1)2 + (k + 1) k + 4k + (k + 2)2 (k + 1)2 = (k + 1)2 22 (k + 1)(k + 2) 2 Thus the proposition is valid for n = k + Therefore, by the principle of mathematical induction, the proposition is true for all n ∈ P Exercise 4-13: Prove by induction the following results for all n ∈ P 14 + 24 + 34 + · · · + n4 = n(n + 1)(6n3 + 9n2 + n − 1) 30 4.4 Solution: 30 (i) For the base case n = 1, we have 14 = and 1(2)(6+9+1−1) = 30 = 30 Hence the formula is true when n = (ii) As induction hypothesis, suppose that the formula holds when n = k, that is suppose 14 + 24 + 34 + · · · + k = k(k + 1)(6k + 9k + k − 1) 30 Then 14 + 24 + · · · + k + (k + 1)4 k(k + 1)(6k + 9k + k − 1) = + (k + 1)4 30 (k + 1)(6k + 9k + k − 1) + 30(k + 1)(k + 1)3 = 30 If we add and subtract 2(6k + 9k + k − 1) to the numerator it will have the terms (6k + 9k + k − 1)(k + 2)(k + 1) and (k + 1)30(k + 1)2 − 2(k + 1)(6k + 9k + k − 1) Working with these last two terms we have, (k + 1)(30k + 90k + 90k + 30 − 12k − 18k − 2k + 2) = (k + 1)(18k + 72k + 88k + 32) = (k + 1)(k + 2)(18k + 36k + 16) Putting the terms together we get 14 + 24 + 34 + · · · + k + (k + 1)4 = (k + 1)(k + 2)(6k + 9k + k − + 18k + 36k + 16) 30 (k + 1)(k + 2) (6k + 18k + 18k + 6) + 9k + 18k + 10 + k − = 30 (k + 1)(k + 2) 6(k + 1)3 + (9k + 18k + 9) + (k + 1) − = 30 (k + 1)(k + 2)(6(k + 1)3 + 9(k + 1)2 + (k + 1) − 1) = 30 That is, the result is true for n = k + whenever it is true for n = k Therefore, by the Principle of Mathematical Induction, the result is true for all n ∈ P 4.5 Exercise 4-14: Prove by induction the following results for all n ∈ P 12 + 32 + 52 + · · · + (2n − 1)2 = n(2n − 1)(2n + 1) Solution: = (i) For the base case n = 1, we have 2n − = 1, 12 = and 1(1)(3) Hence the formula is true when n = (ii) As induction hypothesis, suppose that the formula holds when n = k, that is suppose 12 + 32 + 52 + · · · + (2k − 1)2 = k(2k − 1)(2k + 1) Then 12 + · · · + (2k − 1)2 + (2(k + 1) − 1)2 = k(2k − 1)(2k + 1) + (2k + 1)2 , (2k + 1)(k(2k − 1) + 3(2k + 1)) = Adding and subtracting 4k and inside the parenthesis gives = = = = (2k + 1)(k(2k + 3) − 4k − + 3(2k + 3)) (2k + 1)(k(2k + 3) − 2(2k + 3) + 3(2k + 3)) (2k + 1)(2k + 3)(k + 1) (k + 1)(2(k + 1) − 1)(2(k + 1) + 1) That is, the result is true for n = k + whenever it is true for n = k Therefore, by the Principle of Mathematical Induction, the result is true for all n ∈ P Exercise 4-15: Prove by induction the following results for all n ∈ P · + · + · + · · · + n(n + 1) = n(n + 1)(n + 2) Solution: (i) For the base case n = 1, we have · = and formula is true when n = 4.6 1(2)(3) = Hence the (ii) As induction hypothesis, suppose that the formula holds when n = k, that is suppose · + · + · + · · · + k(k + 1) = k(k + 1)(k + 2) Then · + · · · + k(k + 1) + (k + 1)(k + 2) = = k(k + 1)(k + 2) + (k + 1)(k + 2) (k + 1)(k + 2)(k + 3) That is, the result is true for n = k + whenever it is true for n = k Therefore, by the Principle of Mathematical Induction, the result is true for all n ∈ P Exercise 4-16: Prove, by induction, the following results for all n ∈ P n n+2 + + + ··· + n = − 2 2 2n Solution: (i) If n = then 21 = − 32 (ii) Suppose the result is true for n Then, n n+1 + + + + n + n+1 22 2 n+1 n+2 n+1 n+2 + n+1 = − + = 2− 2n 2n 2n n+1 (n + 1) + = 2− n n+2− =2− 2 2n+1 as required Therefore, by the principle of mathematical induction, the proposition is true for all n ∈ P Exercise 4-17: A set with n elements contains 2n subsets (including the set itself and ∅) Solution: Let S be the set containing n elements (i) For the base case n = 0, S has no elements so S = ∅ Therefore, its only subset is itself There are 20 subsets (ii) As induction hypothesis, suppose that a set S with n = k elements has 2k subsets 4.7 Then for a set S with n = k + elements, S = {a1 , a2 , , ak , ak+1 } A subset of S either contains ak+1 or it does not Notice that S is of the first type and that ∅ is of the second So we can partition the subsets of S into those that contain ak+1 and those that not contain it If they not contain ak+1 then they are subsets of {a1 , a2 , , ak } By our induction hypothesis there are 2k of these If G is a subset of S containing ak+1 then G − {ak+1 } is also a subset of {a1 , a2 , , ak } Similarly if H is a subset of {a1 , a2 , , ak } then H with ak+1 is a subset of S So the number of subsets of S containing ak+1 is equal to the number of subsets of S not containing ak+1 We already know that there are 2k of these Then a set S with k + elements has in total # of subsets = # of subsets with ak+1 + # of subsets without ak+1 = 2k + 2k = 2k+1 That is, the result is true for n = k + whenever it is true for n = k Therefore, by the Principle of Mathematical Induction, the result is true for all n ∈ P + {0} Exercise 4-18: Prove, by induction, the following results for all n ∈ P 6|(2n3 + 3n2 + n) Solution: (i) If n = 1, then 2n3 + 3n2 + n = and 6|6 Hence the proposition is valid for n = (ii) Assume 6|(2k + 3k + k) for k ∈ P Now 2(k + 1)3 + 3(k + 1)2 + (k + 1) = 2k + 6k + 6k + + 3k + 6k + + k + = (2k + 3k + k) + (6k + 12k + 6) 6|2k + 3k by hypothesis and 6|6k + 12k + Therefore 6|2(k + 1)3 + 3(k + 1)2 + (k + 1) Thus the proposition is valid for n = k + whenever it is valid for n = k Therefore, by the principle of mathematical induction, the proposition is true for all n ∈ P i.e 6|(2n3 + 3n2 + n) for all n ∈ P Exercise 4-19: Find an expression for n r=1 r(r!) and prove that it is correct 4.8 Solution: The first few sums give 1 + 2(2!) + 2(2!) + 3(3!) + 2(2!) + 3(3!) + 4(4!) = 1=2−1 = 5=6−1 = 23 = 24 − = 119 = 120 − By observing the above pattern, we can formulate the following general result n r(r!) = (n + 1)! − r=1 We shall prove that this result is true for all n ∈ P by induction on n (i) For the base case n = 1, = 2! − 1, so the result holds (ii) As induction hypothesis, suppose that the formula holds for n = k, that is k r(r!) = (k + 1)! − r=1 Then k+1 r(r!) = (k + 1)! − + (k + 1)(k + 1)! r=1 = (k + 1)!(k + 2) − = (k + 2)! − So the formula holds for n = k + every time it holds for n = k Therefore, by the Principle of Mathematical Induction, n r(r!) = (n + 1)! − for all n ∈ P r=1 Exercise 4-20: Are the following true for all positive integer values of n? If so, prove the result; if not, give a counterexample n! ≥ 2n Solution: It is true for all n ∈ P with n ≥ For n = 1, we have 1! < and < 4, which are two counterexamples Exercise 4-21: Are the following true for all positive integer values of n? If so, prove the result; if not, give a counterexample 3|(22n − 1) 4.9 Solution: It is true To prove it we use induction (i) For the base case n = 1, 22 − = and 3|3 The result is true for n = (ii) As induction hypothesis, suppose that the result holds for n = k, that is 3|(22k − 1) Then 22(k+1) − = (22k · 4) − By induction hypothesis 22k ≡ (mod 3) multiplying by ≡ (mod 3) gives 22(k+1) ≡ (mod 3) This shows that 3|(22(k+1) − 1) That is, the result is true for n = k + whenever it is true for n = k Therefore, by the Principle of Mathematical Induction, the result is true for all n ∈ P Exercise 4-22: Are the following true for all positive integer values of n? If so, prove the result; if not, give a counterexample 7|(5n + n + 1) Solution: It is false For n = 53 + + = 129 and 129 = 7(18) + so | 129 Exercise 4-23: Are the following true for all positive integer values of n? If so, prove the result; if not, give a counterexample (a + b)|(a2n − b2n ) Solution: It is true, we prove it by induction on n (i) For the base case n = 1, (a + b)|(a2 − b2 ) The result is true for n = (ii) As induction hypothesis, suppose that the result holds for n = k, that is (a + b)|(a2k − b2k ) Then a2(k+1) − b2(k+1) = a2k+2 − b2k+2 adding and subtracting a2 b2k gives = a2 a2k − a2 b2k + a2 b2k − b2 b2k = a2 (a2k − b2k ) + b2k (a2 − b2 ) 4.10 Solution: We will prove this by induction on n (i) For n = r=1 fr2 = f12 = 12 = and f1 f2 = · = So the result holds for n = (ii) As induction hypothesis suppose that the result holds for n = k, that is k fr2 = fk fk+1 r=1 Then for n = k + 1, k + ≥ so we can use the inductive formula Thus, k fr2 + fk+1 = fk fk+1 + fk+1 r=1 = fk+1 (fk + fk+1 ) = fk+1 fk+2 Therefore, the result holds for n = k + By the Principle of Mathematical Induction the result is true for all n ∈ P Problem 4-66: Prove that n r=1 f2r−1 = f2n for all n ∈ P Solution: We will prove this by induction on n (i) For n = 1, r=1 f2r−1 = f2(1)−1 = f1 = and f2·1 = So the result holds for n = (ii) As induction hypothesis suppose that the result holds for n = k, that is k f2r−1 = f2k r=1 Then for n = k + 1, 2k + ≥ so we can use the inductive formula Thus, k f2r−1 + f2(k+1)−1 = f2k + f2k+1 r=1 = f2k+2 Therefore, the result holds for n = k + By the Principle of Mathematical Induction the result is true for all n ∈ P Problem 4-67: Prove that fn+5 ≡ 3fn (mod 5) for all n ∈ P Solution: 4.37 We will prove this by induction on n (i) The first seven numbers of the Fibonacci Sequence are 1, 1, 2, 3, 5, 8, 13 For n = 1, f5+1 = ≡ · ≡ · f1 (mod 5) We cannot use the inductive formula until n ≥ 3, so for n = 2, f2+5 = 13 ≡ · ≡ · f2 (mod 5) So the result holds for n = 1, (ii) As induction hypothesis suppose that the result holds whenever ≤ r ≤ k, that is fk+5 ≡ 3fk (mod 5) Then for n = k + 1, k + ≥ so we can use the inductive formula Thus, f(k+1)+5 = fk+6 = fk+5 + fk+4 , and for ≤ k − 1, k ≤ k fk+5 ≡ · fk fk+4 ≡ · fk−1 (mod 5) (mod 5) Adding these two congruences gives fk+5 + fk+4 ≡ · fk−1 + · fk (mod 5) ≡ 3(fk−1 + fk ) (mod 5) ≡ 3fk+1 (mod 5) Therefore the result is true for n = k + By the Principle of Strong Induction the result is true for all n ∈ P Problem 4-68: The downtown portion of a city consists of a rectangular area m blocks long and n blocks wide If all the streets in each direction are through streets, find the number of different shortest routes from one corner of the downtown area to the opposite corner Solution: O q m Bq q C qA .n Let the required number be a(m, n) To get to corner C from corner O we must first get to either A and B We can get to A in a(m, n − 1) different ways and we can get to B in a(m − 1, n) different ways Thus a(m, n) = a(m − 1, n) + a(m, n − 1) Calculation of a few values suggests the conjecture a(m, n) = 4.38 m+n n This can be proved by two-way induction (ugh!) or by the following direct argument Since we are counting shortest routes, all “steps” must be either north or east We must make exactly m + n steps, since each step increases either our x coordinate or our y coordinate, and, at C, x + y = m + n Hence we can specify a route uniquely by specifying which steps are made to the east, i.e by choosing n of the m + n steps This can be done in m+n ways n Problem 4-69: At a party of logicians, the host attached either a gold star or a blue star to the back of each of the guests After all the guests had arrived and mingled with each other, the host announced that there was at least one gold star, and anybody who could prove they had one on their back, without peeking, should claim the prize No one came forward to claim the prize Every five minutes after this, the host again announced anybody who could prove they had a gold star should claim the prize, and no one came forward Finally after the twentieth time the host asked, everyone rushed forward to claim the prize How many guests were at the party? Solution: We will prove that if among a party of m logicians n guests had a gold star on their back, then all those logicians with a gold star would know after the host asked for the nth time We will prove this by induction on n (i) If only logician of the party has a gold star, after mingling with all the other guests he would not see any gold star With the first announcement he would know that there is at least one gold star He would conclude that there is one gold star and that he has it (ii) As inductive hypothesis assume that the result is true for n = r, whenever ≤ r ≤ k That is if r logicians have a gold star, only after the rth announcement from the host would those r logicians know that they have one Then if there are n = k + gold stars in the party By the inductive hypothesis,in the first k announcements no logician would be sure about having a gold star k + logicians see exactly k gold stars among the party, while the other m − k − see k + gold stars The first k + logicians know the inductive hypothesis (because they are logicians), so if there had been only k golds stars in the party, those k logicians with gold stars they see would have rushed to get the prize after the kth announcement Because this did not happen, they would know that there are more than k gold stars but not more than those k gold stars and the one they possibly have Therefore they are all aware that they have a gold star Then the result is true for n = k + By the principle of Strong Induction the result is true for all n ∈ P After the twentieth announcements, all logicians rushed, therefore there were 20 logicians in the party 4.39 Problem 4-70: (Trinomial Theorem) Prove that (a + b + c)n = n! p q r a b c p!q!r! p+q+r=n for all n ∈ P, where the sum is taken over all nonnegative values of p, q and r for which p + q + r = n Solution: We apply the Binomial Theorem twice n (a + b + c)n = r=0 n = r=0 n r n−r q=0 n (a + b)n−r cr r n−r q (Note that when r = n, this is correct, with n n−r r=0 q=0 = an−r−q bq 0 = 0! 0!0! cr =1) n! (n − r)! an−r−q bq cr r!(n − r)! q!(n − r − q)! Now let p = n − r − q, so p + q + r = n, and p, q, r ≥ (for p, since q ≤ n − r) This gives n! p q r a b c (a + b + c)n = p!q!r! p+q+r=n Problem 4-71: What is the maximum number of regions that a plane can be divided into by n straight lines? Solution: We shall prove that the maximum number of regions that a plane can be divided into by n straight lines is n(n + 1) n2 + n + +1= 2 Lemma If a straight line is added to a collection of straight lines in the plane, and intersects the collection in r points, then the number of regions into which the collection of lines divides the plane is increased by r + Proof Let a line intersect the collection in the r distinct points a1 , a2 , , ar in sequence along the line These points could already be intersection points of the collection 4.40 ❈ qa1 ❈ ❈ ❈ q✄a2 ♣ ♣ ♣ ♣ ❈✄ ✄❈ ❈ qar−1 ❈ ❈ ✄ar ✄q ✄ Each of the r − finite segments a1 a2 , a2 a3 , , ar−1 ar , plus the two infinite segments of ending with a1 and ar , will split one region into two Hence, adding the line to the collection of lines will increase the number of regions by r + If we add another line to a collection of k lines in the plane, then the maximum number of intersection points on will be k, since there can be at most one intersection point on each of the original lines Moreover, for any collection, there is a line that meets the collection in exactly k points The collection of k lines will have a finite number of slopes (at most k) Choose the slope of to be different, so that will not be parallel to any line, and will intersect every line There are only a finite number of intersection points in the original collection (at most k2 ), so the line can be chosen to avoid these Now we can prove the problem by induction on the number of lines (i) For n = 1, one straight line divides the plane into two regions, and = 1·2 +1 So the result holds for n = (ii) As induction hypothesis suppose that the result is true for n = k That is, the maximum number of regions that a plane can be divided into by k straight lines is k(k+1) + 1, and this maximum can be achieved If we have n = k + lines, choose one line and call it By the induction hypothesis, the other k lines divide the plane into at most k(k+1) + regions and, by the Lemma, the line adds at most k + points Hence the maximum number of regions is k(k + 1) + + (k + 1) = (k + 1)(k + 2) + Therefore the result for the maximum holds for n = k + This maximum can be achieved by choosing a configuration of k lines that has the maximum number of regions, and then adding a (k + 1)st line that is not parallel to any of the lines, and does not go thorough any of the intersection points By the Principle of Mathematical Induction, the result is true for all n ∈ P Problem 4-72: What is the maximum number of regions that three dimensional space can be divided into by n planes? Solution: We shall prove that the maximum number of regions that a three dimensional space can be divided into by n planes is (n + 1) (n(n − 1) + 6) 4.41 Lemma If a plane is added to a collection of planes in the space, and intersects the collection in r lines, then the number of regions into which the collection of planes divides the space is increased by s(r), where s(r) is the number of regions that the r lines of intersection divide the added plane Proof Let a plane P intersect the collection in the r lines , , , r in the plane These lines are not necessarily distinct and could be intersection lines of the collection P ✡✡ ✡✡ ✁ ✡ ❏✁ ❆❆✡✡ ✡ ✡ ✁ ❏ r−1 ✡ ❆ r ✡ ✡ ✡ These r lines divide the plane P into s(r) regions Each of these s(r) regions of the plane will split one region of the space into two Hence, adding the plane P to the collection of planes will increase the number of regions of the space by s(r) From Lemma we see that by adding an additional plane to the collection of k planes, the maximum increase to the number of regions is achieved when r and s(r) are maximum and the r lines are distinct The number of lines, r is at most k, since there can be at most one intersection line with each of the original k planes And given the r distinct lines in P, by Problem 4-71 the maximum value of s(r) is r(r + 1)/2 + We will now show that such a configuration is possible Lemma There is a collection of k planes in space, such that any plane intersects all the others in k − different lines that divide that plane into the maximum number of regions possible Proof This means that if we take one plane P in the collection, then it has k − lines of intersection that, by Problem 4.71, meet the following conditions (i) no three of the k − lines intersect in one point, (ii) no two lines are parallel This means that (iii) Each plane intersects the other k − planes, (iv) No three planes of the collection intersect in the same line, (v) No three planes of the collection intersect in parallel lines ★ ★ ★ ★ ❏★ ★ ✡ ✡ ❏❏ ✡✡✡ ✡ ❏ ✡★❏ ★ ❏ ❏✡ ❏ ✡❏ ❏ ✡★✡ ❏ ★ ★ ✡ ★ ★ ❏★ ◗ ✑ ◗✑ ✑◗ ✑ ◗ ✑ ◗ ✑ ◗ ✑ ◗ ❨ ✯ ✟ ★❍ ★ ✑◗ ❍✑ ◗✟ ★ ✑ ★ ◗ ★ ❄ ★ ◗ ✑ ◗✑ ✑◗ ✑ ◗ ✑ ◗ ✑ ◗ ✑ ◗ 4.42 Three planes intersect in the same line, or in parallel lines, if and only if their normal vectors are coplanar (lie in the same plane) If k = then the desired configuration could be, ★ ★ ✡ ★❏ ❏ ★ ✡✡ ❏ ❏❏ ❏ ✡✡✡ ✡ ❏✡ ✡❏ ❏ ❏✡★ ✡ ❏ ★❏ ❏❏ ❏✡ ✡✡ ❏ ❏ ❏❏ ✡ ❏ ✡★ ✡ ❏ ❏❏★ ✡ ❏★ ★ If a collection, C, of k − planes satisfy conditions (iii), (iv) and (v) then different lines and no three of their k − normal the planes intersect in k−1 vectors are coplanar It is possible to choose a kth vector v that is not coplanar with any two of these k − normal vectors If P is a plane with such a normal vector v, then P intersects the other k − planes P can be translated so that it does not contain any of the k−1 lines of intersection of the other k − planes So C and such a translated plane P is a collection of k planes also satisfying conditions (iii), (iv) and (v) Hence, by induction, such a configuration is possible Now we can prove the problem by induction on the number of planes (i) For n = 1, one plane divides the space into two regions, and = 2(0 + 6)/6 So the result holds for n = (ii) As induction hypothesis suppose that the result is true for n = k That is, the maximum number of regions that the space can be divided into by k planes is (k+1) (k(k − 1) + 6), and this maximum can be achieved If we have n = k + planes, choose one plane and call it P By the induction hypothesis, the other k planes divide the space into at most (k+1) (k(k − 1) + 6) regions and, by the discussion after the Lemma 1, the plane P adds at most k(k + 1)/2 + regions Hence the maximum number of regions of the space is (k + 1) k(k + 1) (k(k − 1) + 6) + +1 k(k + 1) (k + 1)k(k − 1) + (k + 2) + = k−1 = (k + 1)k + + (k + 2) k+2 (k + 2) = (k + 1)k + (k + 2) = ((k + 1)k + 6) 6 Therefore the result for the maximum holds for n = k + By the Principle of Mathematical Induction, the result is true for all n ∈ P 4.43 Problem 4-73: Let a1 , a2 , , an be real numbers such that ≤ ≤ for ≤ i ≤ n Prove that (1 − a1 )(1 − a2 ) · · · (1 − an ) ≥ − (a1 + a2 + · · · + an ) Solution: We will prove this by induction on n (i) For n = we have (1 − a1 ) = − a1 so the result holds (ii) As induction hypothesis assume that the result is true for n = k, that is for a1 , a2 , · · · , an real numbers such that ≤ ≤ for ≤ i ≤ k we have that (1 − a1 )(1 − a2 ) · · · (1 − ak ) ≥ − (a1 + a2 + · · · + ak ) Then for n = k + 1, with ≤ ak+1 ≤ 1, because (1 − ak+1 ) ≥ (1 − a1 )(1 − a2 ) · · · (1 − ak )(1 − ak+1 ) ≥ − (a1 + a2 + · · · + ak ) (1 − ak+1 ) ≥ − (a1 + a2 + · · · + ak ) − ak+1 + ak+1 (a1 + a2 + · · · + ak ) Because ≤ ≤ for ≤ i ≤ k + then ak+1 (a1 + a2 + · · · + ak ) ≥ So 1−(a1 +a2 +· · ·+ak +ak+1 )+ak+1 (a1 +a2 +· · ·+ak ) ≥ 1−(a1 +a2 +· · ·+ak +ak+1 ) it follows that (1 − a1 )(1 − a2 ) · · · (1 − ak )(1 − ak+1 ) ≥ − (a1 + a2 + · · · + ak + ak+1 ) The result holds for n = k + whenever it is true for n = k By the Principle of Mathematical Induction the result is true for all n ∈ P Problem 4-74: A sequence x1 , x2 , x3 , of real numbers is defined by x1 = and xn+1 = Prove that xn = n xn + 1, n+1 n+1 for all positive integers n Solution: (i) When n = the assertion is true since (ii) Suppose that xk = k+1 Then xk+1 = if n ≥ 1+1 = = x1 k k k+1 k (k + 1) + xk + = · +1= +1= k+1 k+1 2 Hence the result is true for n = k + 4.44 Therefore, by the principle of mathematical induction, xn = n ≥ n+1 for all Problem 4-75: A sequence y1 , y2 , y3 , of integers is defined by y1 = 4, and yn+1 = yn2 − 2, if n ≥ Prove that √ n−1 √ n−1 yn = (2 + 3)2 + (2 − 3)2 for all positive integers n Solution: We will prove this by induction on n √ 1−1 √ √ √ 1−1 = 2+2+ 3− = (i) For n = we have y1 = and (2+ 3)2 +(2− 3)2 So the result holds (ii) As induction hypothesis assume that the result is true for n = k, that is yk = (2 + √ 3)2 k−1 + (2 − √ 3)2 k−1 Then for n = k + 1, k + ≥ so yk+1 = yk2 − √ k−1 √ k−1 = (2 + 3)2 + (2 − 3)2 −2 √ √ 2k √ 2k √ k−1 = (2 + 3) + (2 − 3) + (2 + 3)(2 − 3)]2 −2 √ 2k √ 2k k−1 −2 = (2 + 3) + (2 − 3) + 2(4 − 3)2 √ 2k √ 2k = (2 + 3) + (2 − 3) + The result holds for n = k + whenever it is true for n = k By the Principle of Mathematical Induction the result is true for all n ∈ P Problem 4-76: Prove that 1− 1− 1− 16 ··· − n2 = n+1 2n for all integers n ≥ Solution: We will prove this by induction on n (i) For n = we have − 14 = 43 = 2+1 So the result holds (ii) As induction hypothesis assume that the result is true for n = k for k ≥ 2, that is 1 1 k+1 1− 1− 1− ··· − = 16 k 2k 4.45 Then for n = k + 1, 1− 1− 1 · 1− k2 (k + 1)2 k+1 · 1− 2k (k + 1)2 k + k(k + 2) · 2k (k + 1)2 k+2 2(k + 1) ··· − = = = The result holds for n = k + whenever it is true for n = k ≥ By the Principle of Mathematical Induction the result is true for all n ∈ P Problem 4-77: Show that 1+ 1 1 + + ··· + ≤ − 22 n n for all integers n ≥ Solution: We will prove this by induction on n (i) For n = we have = − 11 so the result is holds (ii) As induction hypothesis assume that the result is true for n = k for k ≥ 1, that is 1 1 + + + ··· + ≤ − k k Then for n = k + 1, 1+ 1 1 1 + + ··· + + ≤ , 2− + 22 k (k + 1)2 k (k + 1)2 And for k ≥ 1, k2 + k k +k k(k + 1)2 k+1 ≤ k2 + k + (k + 1)2 − k ≤ k(k + 1)2 1 ≤ − k (k + 1)2 Adding two and swapping the terms of the inequality we get 2− 1 + ≤2− , k (k + 1)2 k+1 so, 1+ 1 1 1 + + ··· + + ≤, − + ≤2− 22 k (k + 1)2 k (k + 1)2 k+1 4.46 The result holds for n = k + whenever it is true for n = k ≥ By the Principle of Mathematical Induction the result is true for all n ∈ P Problem 4-78: Find the value of each recursive mystery function myst (x1 , x2 , , xn ) on any n-tuple (x1 , x2 , , xn ) and prove that your value is correct myst (x1 , x2 , , xn ) = x1 xn − myst (x1 , x2 , , xn−1 ) if n = if n > Solution: Claim: For all n ≥ we have n (−1)n+i xi myst(x1 , , xn ) = i=1 We will prove it by induction on n (i) When n = the assertion is true since i=1 (−1)1+i xi = x1 = myst(x1 ) k (ii) Suppose that myst(x1 , , xk ) = i=1 (−1)k+i xi Then for n = k + 1, k+1>1 myst(x1 , , xk+1 ) = xk+1 − myst(x1 , , xk ) k (−1)k+i xi = xk+1 − i=1 k (−1)k+1+i xi = xk+1 + i=1 k+1 (−1)k+1+i xi , = i=1 2k+2 k+1+k+1 since = (−1) , so xk+1 = (−1) xk+1 Hence the assertion is true for n = k + Therefore the claim follows by the principle of mathematical induction Problem 4-79: Find the value of each recursive mystery function myst (x1 , x2 , , xn ) on any n-tuple (x1 , x2 , , xn ) and prove that your value is correct myst (x1 , x2 , , xn ) = x1 xn · myst (x1 , x2 , , xn−1 ) Solution: Claim: For all n ≥ we have n myst(x1 , , xn ) = xn xn−1 · · · x2 x1 = xi i=1 4.47 if n = if n > We will prove it by induction on n (i) When n = the assertion is true since myst(x1 ) = x1 k (ii) Suppose that myst(x1 , , xk ) = i=1 xi Then for n = k + 1, k + > myst(x1 , , xk+1 ) = xk+1 · myst(x1 , , xk ) k = xk+1 · k+1 xi = i=1 xi i=1 Hence the assertion is true for n = k + Therefore the claim follows by the principle of mathematical induction Problem 4-80: Find the value of each recursive mystery function myst (x1 , x2 , , xn ) on any n-tuple (x1 , x2 , , xn ) and prove that your value is correct  if n =  x1 xn if xn > myst (x1 , x2 , , xn−1 ) myst (x1 , x2 , , xn ) =  myst (x1 , x2 , , xn−1 ) otherwise Solution: Claim: for all n ∈ P, myst (x1 , x2 , , xn ) = max (x1 , x2 , , xn ) We will prove this by induction on the number of elements (i) For n = 1, myst (x1 ) = x1 = max (x1 ) So the claim is true for n = (ii) As induction hypothesis, assume that the claim is true for n = k, k ≥ 1, that is myst (x1 , x2 , , xk ) = max (x1 , x2 , , xk ) Then for n = k + 1, k + > by definition of the mystery function we have two cases: Case 1: If xk+1 > myst (x1 , x2 , , xk ) then myst (x1 , x2 , , xk , xk+1 ) = xk+1 Also max (x1 , x2 , , xk+1 ) = max (max (x1 , x2 , , xk ), xk+1 ) = max (myst (x1 , x2 , , xk ), xk+1 ) = xk+1 Case 2: If xk+1 ≤ myst (x1 , x2 , , xk ) then myst (x1 , x2 , , xk , xk+1 ) = myst (x1 , x2 , , xk ) Also max (x1 , x2 , , xk+1 ) = max (max (x1 , x2 , , xk ), xk+1 ) = max (myst (x1 , x2 , , xk ), xk+1 ) = myst (x1 , x2 , , xk ) In both cases myst (x1 , x2 , , xk+1 ) = max (x1 , x2 , , xk+1 ) So the claim holds for n = k + Therefore, by the Principle of Mathematical Induction, the claim is true for all n ∈ P 4.48 Problem 4-81: Find the value of each recursive mystery function myst (x1 , x2 , , xn ) on any n-tuple (x1 , x2 , , xn ) and prove that your value is correct myst (x1 , x2 , , xn ) = x1 x1 − myst (x2 , x3 , , xn ) if n = if n > Solution: Claim: for all n ∈ P, myst (x1 , x2 , , xn ) = x1 + (−2)x2 + · · · + (−2)n−1 xn We will prove this by induction on the number of elements (i) For n = 1, myst (x1 ) = x1 So the claim is true for n = (ii) As induction hypothesis, assume that the claim is true for n = k, k ≥ 1, that is myst (x1 , x2 , , xk ) = x1 + (−2)x2 + · · · + (−2)k−1 xk Then for n = k + 1, k + > by definition of the mystery function we have myst (x1 , x2 , , xk+1 ) = x1 − 2myst (x2 , x3 , , xk ) = x1 − 2[x2 + (−2)x3 + · · · + (−2)k−1 xk+1 ] = x2 + (−2)x2 + (−2)2 x3 + · · · + (−2)k xk+1 So the claim holds for n = k + Therefore, by the Principle of Mathematical Induction, the claim is true for all n ∈ P Problem 4-82: Find the value of each recursive mystery function myst (x1 , x2 , , xn ) on any n-tuple (x1 , x2 , , xn ) and prove that your value is correct  if n =  x1 myst (x1 , x2 , , xn−1 ) myst (x1 , x2 , , xn ) =  +myst (x2 , x3 , , xn ) if n > Solution: n−1 Claim: for all n ∈ P, myst (x1 , x2 , , xn ) = n−1 x1 + x2 + · · · + n−1 x n n−1 We will prove this by induction on the number of elements (i) For n = 1, myst (x1 ) = x1 = 00 x1 So the claim is true for n = (ii) As induction hypothesis, assume that the claim is true for n = k, k ≥ 1, that is k−1 k−1 k−1 myst (x1 , x2 , , xk ) = x1 + x2 + · · · + xk k−1 Then for n = k + 1, k + > by definition of the mystery function we have myst (x1 , x2 , , xk+1 ) = myst (x1 , x3 , , xk ) + myst (x2 , x3 , , xk+1 ) k−1 k−1 k−1 k−1 = x1 + · · · xk + x2 + · · · xk+1 k−1 k−1 4.49 Adding equal terms we have = = k−1 + k−1 k−1 x1 + k x1 + x2 + · · · + k−1 r k−1 r−1 x2 + · · · + k r x2 + · · · + xr+1 + · · · + k−1 k−1 k−1 k−2 xk + xr+1 + · · · + k k−1 xk + xr+1 + · · · + xk k−1 k−1 xk+1 xk+1 using Proposition 4.32 So the claim holds for n = k + Therefore, by the Principle of Mathematical Induction, the claim is true for all n ∈ P Problem 4-83: Find a recursive definition for the function e(n) = + 1 + + ··· + 1! 2! n! that gives an approximation to e, the base of natural logarithms Solution: Lets define the recursive function f (n), f (n) = 1 n! if n = if n > Solution: We will prove that f (n) = e(n) using induction on n ∈ P + {0} (i) For n = 0, e(0) = = f (0) So the result is true for n = (ii) As induction hypothesis, assume that the claim is true for n = k, k ≥ 0, that is f (k) = e(k) Then for n = k + 1, k + > so f (k + 1) + f (k) (k + 1)! = + e(k) (k + 1)! 1 1 = + + + + ··· + (k + 1)! 1! 2! k! = e(k + 1) = So the result holds for n = k + whenever it is true for n = k Therefore, by the Principle of Mathematical Induction, f (n) = e(n) for all n ∈ P + {0} Problem 4-84: Discuss the following recursive definition of the GCD for a ≥ and b ≥ Is it correct? Is it efficient?  if b =  a gcd(b, a) if a < b gcd(a, b) =  gcd(b, a − b) if a ≥ b > 4.50 Solution: To prove that the recursive definition is correct we need the following result: gcd(a, b) = gcd(b, a − b) for a, b ∈ Z Let d = gcd(a, b) then d|a and d|b It follows that d|(a − b) so d is a common divisor of b and a − b Let c be a common divisor of b and a − b, it also follows that c|(a − b) + b so c|a Hence c is also a common divisor of a and b Since d is the greatest common divisor of these two numbers, c ≤ d This shows that d is the gcd(b, a − b) Let f (a, b) be the given recursive function for a ≥ and b ≥ We will use induction on a + b = k (i) For k = 0, both a, b = and f (a, b) = f (0, 0) = = gcd(0, 0) So the function is correct for k = (ii) As induction hypothesis, assume that the function is correct for a + b = k, k ≥ that is f (a, b) = gcd(a, b) Then for a + b = k + 1, k + > 0, if a = then b > so f (0, b) = f (b, 0) = b = gcd(0, b), if b = then a > so f (a, 0) = a = gcd(a, 0) If a and b are not zero, because the function commutes, then we can assume that a ≥ b in which case f (a, b) = f (b, a − b) Because a − b ≥ and a − b + b = a ≤ k, by induction hypothesis f (b, a − b) = gcd(b, a − b) But by the result at the beginning of the problem gcd(b, a − b) = gcd(a, b) so f (a, b) = gcd(a, b) So the function is the gcd(a, b) for a + b = k + Therefore, by the Principle of Mathematical Induction, f (a, b) = gcd(a, b) for all a ≥ and b ≥ with a + b = k for all k ∈ P + {0} Although correct, the algorithm has more recursive calls than the one covered in the chapter using the integer function MOD To see this assume that a ≥ b > 0, then MOD gcd(a, b) f (a, b) gcd(a, b) = gcd(b, a − b) gcd(a, b) = gcd(b, a MOD b) = gcd(b, a − qb) If q > then a−qb < a−b so the second algorithm will most probably terminate faster For example f (a, b) = gcd(a, b) MOD gcd(a, b) gcd(66, 39) gcd(39, 66) = gcd(66, 39) = gcd(39, 27) gcd(39, 27) = gcd(27, 12) = gcd(12, 15) gcd(15, 12) = gcd(12, 3) = gcd(3, 9) = gcd(9, 3) = gcd(3, 6) = gcd(6, 3) = gcd(3, 3) = gcd(3, 0) = gcd(66, 39) = gcd(27, 12) = gcd(12, 3) = gcd(3, 0) = However the algorithm only uses subtraction (and recursive calls), but not the MOD function, which most probably involves division Hence a comparison of efficiencies will depend on the efficiency of the MOD function 4.51 ... 1, to transfer one disc to peg B in 21 − = move then the first and only move is transferring it to peg B For n = 2, to transfer two discs to peg B in 22 − = moves, the first disc should be transferred... would be done (i) For n = one disc has to be transferred to another peg in at least 21 − = move (ii) As induction hypothesis suppose that to transfer k discs to another peg, at least 2k − moves are... move in order to transfer the discs to peg B in 2n − moves is to move the first disc to peg C if n is even, or peg B if n is odd Notice that if instead the discs should be transferred to peg C, the