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 4cet ad 7cet stamps. As: P(8): use oe 4cet stamp ad two 7cet stamps. P(k) P(k + ): if a pile of stamps for k cets postage has a 7cet stamp, replace oe 7cet stamp with two 4 cet stamps; if the pile cotais oly 4cet stamps (there must be at least five of them), replace five 4cet stamps with three 7cet stamps. 3. Suppose that the oly currecy were 3dollar bills ad 0dollar 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 3dollar bills. P(k) P(k + ): Suppose that k dollars ca be formed, for some k 8. If at least two 0dollar bills are used, replace them by seve 3dollar bills to form k + dollars. Otherwise (that is, at most oe 0dollar bill is used), at least three 3dollar bills are beig used, ad three of them ca be replaced by oe 0dollar 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 3cet ad 4cet stamps. As: P(6): Six cets postage ca be made from two 3cet stamps. P(k) P(k + ): either replace a 3cet stamp by a 4cet stamp or else (if there are oly 4cet stamps i the pile of stamps makig k cets postage) replace two 4cet stamps by three 3cet 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 Tomio is the tile that is pictured below. Prove that every ( > ) chessboard ca be tiled with Tomioes. 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 533: 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 3439: 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 4044: 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
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