Chapter Suppose you wish to use the Principle of Mathematical Induction to prove that 1 1! + 2 2! + 3 3! n n! = (n + 1)! 1 for all n 1.

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1 Chapter 4. Suppose you wish to prove that the followig is true for all positive itegers by usig the Priciple of Mathematical Iductio: ( ) =. (a) Write P() (b) Write P(7) (c) Write P(73) (d) Use P(7) to prove P(73) (e) Write P(k) (f) Write P(k + ) (g) Use the Priciple of Mathematical Iductio to prove that P() is true for all positive itegers As: (a) =. (b) = 7. (c) = 73. (d) = ( ) + 45 = = = (7 + ) = 73. (e) (k ) = k. (f) (k + ) = (k + ). (g) P() is true sice =. P(k) P(k + ): (k + ) = k + (k + ) = (k + ).. Suppose you wish to use the Priciple of Mathematical Iductio to prove that! +! + 3 3! ! = ( + )! for all. (a) Write P() (b) Write P(5) (c) Write P(k) (d) Write P(k + ) (e) Use the Priciple of Mathematical Iductio to prove that P() is true for all As: (a)! =!. (b)! +! ! = 6!. (c)! +! k k! = (k + )!. (d)! +! (k + )(k + )! = (k + )!. (e) P() is true sice! = ad! =. P(k) P(k + ):!+!+...+ ( k+ )( k+ )! = ( k+ )! + ( k+ )( k+ )! = ( k+ )[! + ( k+ )] = ( k+ )(! k+ ) = ( k + )!. Page 55

2 3. Use the Priciple of Mathematical Iductio to prove that + 3 ( ) ( ) = for all positive itegers. 3 ( ) + As: P(): =, which is true sice both sides are equal to. P(k) P(k 3 + ): k+ k k+ k k+ k+ k+ k+ ( ) + k+ k+ ( ) + + 3( ) ( ) = + ( ) = 3 3 k + k k ( ) ( 3( )) + k k ( ) ( ) + k ( ) + = = = Use the Priciple of Mathematical Iductio to prove that + 3 for all. As: P(): + 3, which is true sice both sides are equal to 3. P(k) P(k + ): + k + = ( + k ) + k 3 k + k 3 k + 3 k = 3 k < 3 3 k = 3 k Use the Priciple of Mathematical Iductio to prove that 3 > + 3 for all. As: P(): 3 > + 3 is true sice 8 > 7. P(k) P(k + ): (k + ) + 3 = k + k = (k + 3) + k + < k 3 + k + k 3 + 3k k 3 + 3k + 3k + = (k + ) Use the Priciple of Mathematical Iductio to prove that ( + ) for all 0. As: P(0): 0 + 0, which is true sice 0. P(k) P(k + ): (k + ) + (k + ) = (k + k) + (k + ), which is divisible by sice k + k ad (k + ) Use the Priciple of Mathematical Iductio to prove that = for all 0. 3 As: P(0): =, which is true sice =. P(k) P(k + ): k+ k+ k+ k+ 3 k k = + 3 = =. Page 56

3 8. Use the Priciple of Mathematical Iductio to prove that (3 ) (3 ) = for all. As: P(): =, which is true sice =. P(k) P(k + ): k(3k ) (3( k+ ) ) = + (3k+ ) k(3k ) + (3k+ ) 3k + 5k+ = = (3k+ )( k+ ) ( k+ )(3( k+ ) ) = = 9. Use the Priciple of Mathematical Iductio to prove that ( + 3) for all. As: P(): + 3, which is true sice 4. P(k) P(k + ): (k + ) + 3(k + ) = (k + 3k) + (k + ), which is divisible by sice k + 3k ad (k + ). 0. Use the Priciple of Mathematical Iductio to prove that + 3 for all 4. As: P(4): , which is true sice 6. P(k) P(k + ): (k + ) + 3 = (k + 3) + k + k + k = k +.. Use the Priciple of Mathematical Iductio to prove that 3 ( ) for all. As: P(): , which is true sice 3 6. P(k) P(k + ): (k + ) 3 + 3(k + ) + (k + ) = (k 3 + 3k + k) + 3(k + 3k + ), which is divisible by 3 sice each of the two terms is divisible by 3.. Use the Priciple of Mathematical Iductio to prove that ay iteger amout of postage from 8 cets o up ca be made from a ifiite supply of 4-cet ad 7-cet stamps. As: P(8): use oe 4-cet stamp ad two 7-cet stamps. P(k) P(k + ): if a pile of stamps for k cets postage has a 7-cet stamp, replace oe 7-cet stamp with two 4- cet stamps; if the pile cotais oly 4-cet stamps (there must be at least five of them), replace five 4-cet stamps with three 7-cet stamps. 3. Suppose that the oly currecy were 3-dollar bills ad 0-dollar bills. Show that ay dollar amout greater tha 7 dollars could be made from a combiatio of these bills. As: P(8): Eightee dollars ca be made usig six 3-dollar bills. P(k) P(k + ): Suppose that k dollars ca be formed, for some k 8. If at least two 0-dollar bills are used, replace them by seve 3-dollar bills to form k + dollars. Otherwise (that is, at most oe 0-dollar bill is used), at least three 3-dollar bills are beig used, ad three of them ca be replaced by oe 0-dollar bill to form k + dollars. Page 57

4 4. Use mathematical iductio to prove that every amout of postage of six cets or more ca be formed usig 3-cet ad 4-cet stamps. As: P(6): Six cets postage ca be made from two 3-cet stamps. P(k) P(k + ): either replace a 3-cet stamp by a 4-cet stamp or else (if there are oly 4-cet stamps i the pile of stamps makig k cets postage) replace two 4-cet stamps by three 3-cet stamps. 5. Prove that ( j+ ) = 3 for all positive itegers. j= As: The basis case holds sice k j= k ( k + ) j= k+ ( j+ ) = 3k k ( j + ) = 3 = 3. Now assume that j= for some k. It follows that ( j+ ) = ( j+ ) (k+ ) + (4k+ ) + (4k+ 3) = 3k + 6k+ 3 = 3( k+ ). j= k 6. Use mathematical iductio to show that lies i the plae passig through the same poit divide the plae ito regios. As: The basis step follows sice oe lie divides the plae ito regios. Now assume that k lies passig through the same poit divide the plae ito k regios. Addig the (k + )st lie splits exactly two of these regios ito two parts each. Hece k + cocurret lies split the plae ito k + = (k + ) regios. 7. Let a =, a = 9, ad a = a + 3a for 3. Show that a 3 for all positive itegers. As: Let P() be the propositio that a 3. The proof uses the Secod Priciple of Mathematical Iductio. The basis step follows sice a = 3 = 3 ad a = 9 9 = 3. Now assume that P(k) is true for all k such that k <. The a k 3 k for k <. Hece a = a + 3a = = 3 3 = Floor borders oe foot wide ad of varyig legths are to be covered with ooverlappig tiles that are available i two sizes: 3 ad 5 sizes. Assumig that the supply of each size is ifiite, prove that every border ( > 7) ca be covered with these tiles. As: P(8): use oe of each type. P(k) P(k + ): If a 5 tile is used as part of the coverig of a k strip, replace a 5 tile with two 3 tiles to cover a ( k + ) strip. Otherwise, the tiles for the k strip must iclude three 3 tiles; replace three of these with two 5 tiles to cover a ( k + ) strip. Page 58

5 9. A T-omio is the tile that is pictured below. Prove that every ( > ) chessboard ca be tiled with T-omioes. As: P(): The figure below shows a tilig of a 4 4 board. P(k) P(k + ): Divide the k + k + board ito four quarters, each of which is a k k board. P(k) guaratees that each of these four k k boards ca be tiled. Put these four tiled boards together to obtai a tilig for the k + k + board. 0. Use the Priciple of Mathematical Iductio to prove that 4 (9 5 ) for all 0. As: P(0): 4 is true sice 4 0. P(k) P(k + ): 9 k + 5 k + = 9(9 k 5 k ) + 5 k (9 5). Each term is divisible by 4: 4 9 k 5 k (by P(k)) ad Use the Priciple of Mathematical Iductio to prove that 5 (7 ) for all 0. As: P(): 5 7 is true sice 5 5. P(k) P(k + ): 7 k + k + = 7(7 k k ) + k (7 ). Each term is divisible by 5: 5 7 k k (by P(k)) ad Prove that the distributive law A (A... A ) = (A A )... (A A ) is true for all 3. As: The secod form of mathematical iductio is used. P(3) is true sice it is the ordiary distributive law for itersectio over uio. P(3)... P() P( + ): A (A... A + ) = A ((A... A ) A + ) = [A (A... A )] (A A + ) = [(A A )... (A A )] (A A + ) = (A A )... (A A + ) Prove that = for all. 4 8 ( ) As: P(): =, which is true sice the right side is equal to /. P(k) P(k + ): k+ k+ k+ k+ 3 k+ k k+ 4 k+ k+ 3 k ( k+ ) = + = = = k+ k k+ k+ k+ k Page 59

6 4. Fid the error i the followig proof of this theorem : Theorem: Every positive iteger equals the ext largest positive iteger. Proof: Let P() be the propositio ' = + '. To show that P(k) P(k + ), assume that P(k) is true for some k, so that k = k +. Add to both sides of this equatio to obtai k + = k +, which is P(k + ). Therefore P(k) P(k + ) is true. Hece P() is true for all positive itegers. As: No basis case has bee show. Use the followig to aswer questios 5-33: I the questios below give a recursive defiitio with iitial coditio(s). 5. The fuctio f () =, =,, 3,... As: f () = f ( - ), f () =. 6. The fuctio f () =!, = 0,,,... As: f () = f ( - ), f (0) =. 7. The fuctio f () = 5 +, =,, 3,... As: f () = f ( - ) + 5, f () = The sequece a = 6, a = 3, a 3 = 0, a 4 = 7,. As: a = a 3, a = The Fiboacci umbers,,, 3, 5, 8, 3,. As: a = a + a, a =, a =. 30. The set {0,3,6,9, }. As: 0 S; x S x + 3 S. 3. The set {,5,9,3,7, }. As: S; x S x + 4 S. 3. The set {,/3,/9,/7, }. As: S; x S x/3 S. 33. The set {, 4,,0,,4,6, }. As: 0 S; x S x ± S. Use the followig to aswer questios 34-39: I the questios below give a recursive defiitio (with iitial coditio(s)) of {a } ( =,,3, ). Page 60

7 34. a =. As: a = a, a =. 35. a = 3 5. As: a = a + 3, a =. 36. a = ( + )/3. As: a = a + /3, a = / a =. As: a = a, a =. 38. a = /. As: a =, a =. a 39. a = +. As: a = a +, a =. Use the followig to aswer questios 40-44: I the questios below give a recursive defiitio with iitial coditio(s) of the set S. 40. {3,7,,5,9,3, }. As: 3 S; x S x + 4 S. 4. All positive iteger multiples of 5. As: 5 S; x S x + 5 S. 4. {, 5, 3,,,3,5, }. As: S; x S x ± S. 43. {0.,0.0,0.00,0.000}. As: 0. S; x S x/0 S. 44. The set of strigs,,,,. As: S; x S x S (or x S 00x + S). 45. Fid f () ad f (3) if f () = f ( ) + 6, f (0) = 3. As: f () = 30, f (3) = Fid f () ad f (3) if f () = f ( ) f ( ) +, f (0) =, f () = 4. As: f () = 5, f (3) =. Page 6

8 47. Fid f () ad f (3) if f () = f ( ) / f ( ), f (0) =, f () = 5. As: f () = 5/, f (3) = /. 48. Suppose {a } is defied recursively by As: a 3 = 63 ad a 4 = 3,968. a = ad that a 0 =. Fid a 3 ad a 4. a 49. Give a recursive algorithm for computig a, where is a positive iteger ad a is a real umber. As: The followig procedure computes a: procedure mult(a: real umber, : positive iteger) if = the mult(a,) : = a else mult(a,) : = a + mult(a, ). 50. Describe a recursive algorithm for computig 3 where is a oegative iteger. As: The followig procedure computes 3 : procedure power(: oegative iteger) if = 0 the power() : = 3 else power() : = power( ) power( ). 5. Verify that the program segmet a : = b : = a + c is correct with respect to the iitial assertio c = 3 ad the fial assertio b = 5. As: Suppose c = 3. The program segmet assigs to a ad the assigs a + c = + 3 = 5 to b. 5. Cosider the followig program segmet: i : = total : = while i < begi i : = i + total : = total + i ed. ii ( + ) Let p be the propositio total = ad i. Use mathematical iductio to prove that p is a loop ivariat. As: Before the loop is etered p is true sice total = ad i. Suppose p is true ad i < after a executio of the loop. Suppose that the while loop is executed agai. The variable i is icremeted by, ad hece i. The variable total was ( i ) i, which ow becomes ( i ) i i( i+ ) i + =. Hece p is a loop ivariat. Page 6

9 53. Verify that the followig program segmet: if x y the max : = y else max : = x. is correct with respect to the iitial assertio T ad the fial assertio (x y max = y) (x > y max = x). As: If x < y iitially, max is set equal to y, so (x y max = y) is true. If x = y iitially, max is set equal to y, so (x y max = y) is agai true. If x > y, max is set equal to x, so (x > y max = y) is true. Page 63