Homework 9. (n + 1)! = 1 1

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1 . Chapter : Questio 8 If N, the Homewor 9 Proof. We will prove this by usig iductio o. 2! + 2 3! + 3 4! + + +! +!. Base step: Whe the left had side is. Whe the right had side is 2! 2 +! 2 which proves the equality is true whe. Iductive step: Let. We will assume that Our goal is to 2! 3! 4! +! +! show that Usig our iductive assumptio we have 2! 3! 4! ++! ++! 2! + 2 3! + 3 4! ! + +! + + +! + 2! +! + + 2! + +! + 2 +! !. By iductio we have thus show that ! 3! 4!. 2. Chapter : Questio For ay iteger, it follows that for all atural umbers +! +! Proof. We will assume to the cotrary that there is some iteger where 3 does ot divide 5 2. This meas that there is some smallest iteger such that Sice is smallest for ay iteger i with i < we ow 3 5 2i. Note that sice if we have 5 2 ad 3. Sice we ca assume that >. Sice > we ow that ad is less tha so by our previous assumptios By defiitio of divides we have for some iteger m that 5 2 3m m m m m + 8 which cotradicts our assumptio that Hece for all itegers. 3. Chapter : Questio 8 Suppose A, A 2,..., A are sets i some uiversal set U, ad 2. Prove that A A 2 A A A 2 A.

2 Proof. We will prove this by usig iductio o. Base step: 2. We ow wheever we have two sets A ad A 2 i the uiversal set U by DeMorga s law A A 2 A A 2. Iductive step: Let 2. We will assume that give ay subsets B, B 2,..., B of U that B B 2 B B B 2 B. Our goal is to show that give ay + subsets A, A 2,..., A + of U that A A 2 A + A A 2 A +. So let A, A 2,..., A + be ay + sets which are subsets of the uiversal set U. Note that the A, A 2,..., A is a collectio of subsets of U so we ca apply our iductive assumptio to this collectio of sets. Usig DeMorga s law ad our iductive assumptio we have that which proves our statemet by iductio. 4. Chapter : Questio 22 If N, the 2 4 A A 2 A A + A A 2 A A + 8 Proof. We will prove this by usig iductio o. A A 2 A A + A A 2 A A Base step: Whe the left had side is. Whe the right had side is ad sice the iequality is true for. 2 2 Iductive step: Let. We will assume that Our goal is to show that Usig our iductive assumptio we have [ 2 ] [ ] Because we ow so Thus we have show for all N Prove that if m N is a multiple of 4 the F m is a multiple of Hece by iductio The Fiboacci umbers are defied to be F, F 2, ad F F + F 2 for > 2.

3 Proof. We will show that whe m 4 for ay iteger we have that F m F 4 is a multiple of 3. We will prove this by iductig o. Base step: Whe we have m 4 4 ad F m F 4 F 3 +F 2 F 2 +F +F which is a multiple of 3. Iductive step: Let we will assume that F 4 is a multiple of 3 ad show that F 4+ is also a multiple of 3. Sice F 4 is a multiple of 3 we ow F 4 3x for some iteger x. Usig the Fiboacci recurrece we have F 4+ F 4+4 F F 4+2 F F 4+ + F 4+ + F 4 F 4+ + F 4 + F 4+ + F 4+ + F 4 3F F 4 3F x 3F x. Hece F 4+ is also a multiple of 3. Therefore, by iductio we ca coclude that F 4 is a multiple of 3 for all N. 6. Show usig iductio for ay iteger ad x, y R that x + y x y. We defie for itegers ad the biomial coefficiet to be {!!! ad otherwise where factorial is! with!. Proof. Let x ad y be real umbers. Base step: Whe the left had side is x + y. The right had side is x y x y, so we ca see the equality is true whe. Iductive step: Let. We will assume that x + y x y. Our goal is to show that x + y x y +. First ote that by the give defiitio of biomial coefficiets that + +. Also by the defiitio we have that! +! +! +!!!!!! + +! +!! +!! +! + +.

4 Usig our iductive assumptio ad the facts we established above we have x + y + x + y x + y x y x + y x + y + x y + + x y + + x y + x y + + x + y + x y + + x y + x y + + x y + + x y + + x + y x y + + x y + + x y + + x + y y x y + + x x y + + x y + + x + y x y +. Hece, by iductio x + y x y for all itegers. 7. The Triboacci umbers are defied as T, T 2, T 3, ad T T + T 2 + T 3 for > 3. Show that for all N that T 2. Proof. We will prove the statemet by iductig o. Base steps: Whe we have T which is less tha 2 2. Whe 2 we have T 2 ad 2 2. Whe 3 we have T 3 ad 2 3. Hece, the statemet is true for equal to, 2, ad 3. Iductive step: Let 3. We will assume for all positive itegers that T 2. Our goal is to show that T Usig our iductive hypotheses ad the defiitio of Triboacci umbers we have T + T + T + T Hece, T This proves by iductio that T 2 for all positive itegers.

5 8. For N let B be the set of all -digit biary umbers. For example B 3 {,,,,,,, }. a Show that B 2. Proof. We will prove this by iductio o. First a small otatioal ote. If we have two biary words x ad y we will let xy deote their cocateatio. Meaig if x ad y the xy. Further this will mae x ad x. Base steps: Whe the set B {, } has cardiality 2 2. Thus, the statemet is true for. Iductive step: Let. We will assume that B 2 ad will show that B Let S {x B + : x starts with a } {x : x B } ad S {x B + : x starts with a } {x : x B }. Ay word i the set B + will start with a or start with a so S S B +. Sice a word caot both start with a ad a we ow S S. We ca see that S {x : x B } has the same umber of elemets as B so S B 2. Similarly, S {x : x B } has the same umber of elemets as B so S B 2. Fact: If a fiite set C ca be writte as a uio C A B with A B the C A + B. Sice S S B + ad S S we ow B + S + S Hece, by iductio B 2 for all atural umbers. b Chapter : Questio 32 Prove that the umber of -digit biary umbers that have o cosecutive s is the Fiboacci umber F +2. Proof. We will use the set B ad the otatio defied for cocateatio of biary words itroduced i part a. We will let C {x B : x has o cosecutive s}. Our goal is to prove that C F +2 which we will prove by usig iductio o. Base steps: Whe the set C {, } has cardiality 2 F 3 F +2. Whe 2 the set C 2 {,, } has cardiality 3 F 4 F 2+2. Thus, the statemet is true for ad 2. Iductive step: Let 2. We will assume for all positive itegers that C F +2 ad will show that C + F ++2 F +3. Let R {x C + : x starts with a }. We will show that R {x : x C }. Let y R the certaily y x for some biary word x of legth. Sice y C + we ow y does ot have ay cosecutive s so x does t have ay cosecutive s so x C. Thus, y {x : x C } ad R {x : x C }. Let y {x : x C } the y x for some x C. Sice x has o cosecutive s ad puttig a at the frot of x does t create ay cosecutive s we ow y does t have ay cosecutive s, starts with a, ad has legth +. Hece, y R ad R {x : x C }. We ca see ow that R C F +2 by our iductive assumptio. Let R {x C + : x starts with a }.

6 We will show that R {x : x C }. Let y R the certaily y x for some biary word x of legth. Sice y C + we ow y does ot have ay cosecutive s meaig that the secod umber i the word y must be a else we have cosecutive s. Now we have y z for some z C which implies that y {x : x C } ad R {x : x C }. Say istead that y {x : x C }. The y x for some x C. Sice x has o cosecutive s ad placeig a before x does t create ay cosecutive s we ow y x has o cosecutive s, y starts with a, ad has legth +. Hece y R ad R {x : x C }. We ca see ow that R C F +2 F + by our iductive assumptio. Ay word i the set C + will start with a or start with a so R R C +. Sice a word caot both start with a ad a we ow R R. Fact: If a fiite set C ca be writte as a uio C A B with A B the C A + B. Sice R R C + ad R R we ow C + R + R F +2 + F + F +3. Hece, by iductio C F +2 for all atural umbers. c Use parts a ad b to show F +2 2 where F meas the th Fiboacci umber. Proof. We will use the same otatio for sets itroduced i parts a ad b. I part a we had B the collectio of all biary words legth ad B 2. I part b we had C the collectio of all biary words legth with o cosecutive s ad C F +2. Sice ay biary word of legth with o cosecutive s is a biary word legth we have C B for all atural umbers. Sice these are fiite sets their cardialities satisfy the iequality C B which implies that F +2 2 for all atural umbers.