Induction proofs - practice! SOLUTIONS

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1 Iductio proofs - practice! SOLUTIONS 1. Prove that f ) = is odd for all Z +. Base case: For = 1, f 1) = 41) + 1) + 13 = 19. Sice 19 is odd, f 1) is odd - base case prove. Iductive hypothesis: Assume f ) = for some Z +. That meas that = k + 1 for some k Z +. Iductive step: Show true for + 1, that is, that f + 1) is odd. f +1) = 4+1) ++1)+13 = )++1)+13 = ) But is odd, ad therefore ca be writte as k + 1 for some k Z +. So f + 1) = k = k ) + 1, re-arragig a bit. Sice,k Z +, k Z + ad therefore, k ) + 1 is odd. Thus we have show that, if f ) is odd, f + 1) must be odd, too. Sice the base case, that f 1) is odd, is true, ad sice we have show that if the statemet is true for, it is true for + 1, we have prove that f ) = is odd for all Z +.. Prove that! > for a iteger greater tha 4. Base case: For = 4,! = 4! = 4 ad = 4 = > 16, so 4! > 4. Base case prove. Iductive hypothesis: Assume that! > for some 4. Iductive step: Show true for + 1, that is, that + 1)! > +1. Startig with the left-had side, + 1)! = + 1)! > + 1), sice! > iductive hypothesis). Now sice 4, + 1 5; we ca therefore safely say that + 1 > ad so, + 1)! > = +1. Thus we have show that, if! >, it follows that + 1)! > +1. Sice the base case, that 4! > 4 is true, ad sice we have show that if! >, it follows that + 1)! > +1, we have prove that! > for a iteger greater tha 4. 1

2 3. If is a o-egative iteger, show that 5 is divisible by 5. Base case: For = 1, 5 1 = 1) 5 1 = 0, which is divisible by 5. Base case prove. Iductive hypothesis: Assume that 5 is divisible by 5 for some a o-egative iteger. This meas that 5 = 5k for some k Z +. Iductive step: Show true for + 1, that is, that + 1) 5 + 1) is divisible by 5. Expad the expressio: + 1) 5 + 1) = = 5 ) ). Now, 5 is divisible by 5, so we ca write it as 5 = 5k for some k Z +. The + 1) 5 + 1) = 5k ). = 5k ). This shows +1) 5 +1) is a iteger multiple of 5, so +1) 5 +1) is divisible by 5. We have ow show that, if 5 is divisible by 5, it follows that + 1) 5 + 1) is divisible by 5 too. The base case, that 5 is divisible by 5 for = 1 is true, ad we have show that if 5 is divisible by 5, it follows that + 1) 5 + 1) is divisible by 5 too. Thus we have prove that if is a o-egative iteger, show that 5 is divisible by For each Z +, it follows that Show usig iductio. Base case: For = 1, LHS= 1 = ad RHS= 1+1) 1 1) 1 = 0 1 =. LHS=RHS for ad therefore for = 1, holds. Iductive hypothesis: Assume that for some Z +. Iductive step: Show true for + 1, that is, that =. but Therefore, ) + ad the sice < 1, + < + 1. Thus,

3 We have show that, if for some Z +, it follows that Sice the base case, that for = 1 is true, ad sice we have show that if for some Z +, it follows that , we have prove that for each Z Prove by iductio that = + 1)/ for Z +. Base case: For = 1: LHS = =1. RHS = )/ = 1. Sice LHS=RHS, = + 1)/ for = 1. Iductive hypothesis: Assume true for some 1, that is, that = +1)/. Iductive step: Show true for some + 1, that is, that = + 1) + )/. LHS= = ) But = + 1)/. + 1) LHS= = ) = + 1) + 1 ) 1 = + 1) + ) + 1) + ) = = RHS. + 1, we have show that Z +, = + 1)/. 6. Prove that ) = for every Z +. Base case: For = 1: LHS = 1) = 1. RHS = 1 3. Sice LHS=RHS, ) = for = 1. Iductive hypothesis: Assume true for some 1, that is, that ) = Iductive step: Show true for some + 1, that is, that )) = ) 3 3

4 LHS = )) = ) + + 1) ) + + 1) From the iductive hypothesis, ) = Ad remember that = + 1)/ SEE #5). The LHS = ) + + 1) + 1)/) + + 1) = ) + + 1) + + 1) = ) + + 1) + 1) = ) 3 = RHS + 1, we have show that Z +, ) = Prove by iductio that ) = for Z +. Base case: For = 1: LHS = 1. RHS = 1) = 1. Sice LHS=RHS, ) = for = 1. Iductive hypothesis: Assume true for some 1, that is, that ) =. Iductive step: Show true for some + 1, that is, that ) + + 1) = + 1). LHS = ) + + 1) = + + 1), by the iductive hypothesis. But = + 1). Therefore, LHS = + 1) = RHS + 1) LHS= = ) = + 1) + 1 ) 1 = + 1) + ) + 1) + ) =. 4

5 + 1, we have show that Z +, ) =. 8. Show by iductio that = +1 for Z +. Base case: For = 1: LHS = 1 =. RHS = ) 1+1 =. Sice LHS=RHS, = +1 for = 1. Iductive hypothesis: Assume true for some 1, that is, that = +1. Iductive step: Show true for some + 1, that is, that = +. LHS = ) + +1 = +1 ) + +1, sice by the iductive hypothesis, = +1. The LHS = +1 = + = RHS. + 1, we have show that Z +, = For a iteger greater or equal to 3, show that +1 > + 1). First ote that the statemet +1 > + 1) is equivalet to > 1 + 1, 3. ) We ll prove this latter form. Base case: For = 3. LHS=3 ad RHS=1 + 1/3) 3 = 64/7 < 3 = LHS. Base case doe. Iductive hypothesis: Assume true for some 3, that is, that > ) for some 3. Iductive step: Show true for some + 1, that is, that + 1 > ) +1. 5

6 This time start o the right had side. RHS = ) < , ) = ) ) < ) sice 1 > 1 +1 for 1 ad here, 3). The RHS < ) ) < ), sice ) < iductive hypothesis. Fially, distributig the, RHS < ) = + 1 = LHS LHS > RHS that is, + 1 > ) Thus we have show that, assumig > ) is true, it follows that ) +1. Ad sice the > ) holds for = 3, we have show by iductio that > ) holds for > 3. Therefore, the equivalet formulatio +1 > + 1) holds for 3, Z. 10. Prove that for Z +, a chessboard with ay oe square removed ca be tiled by these 3-square L-tiles: Base case: For = 1, we have a board with 1 square removed, which ca be tiled by 1 L-tile. Iductive hypothesis: Assume true for some 1, that is, that a chessboard with ay oe square removed ca be tiled by the 3-square L-tiles. Iductive step: Show true for some +1, that is, that a chessboard with ay oe square removed ca be tiled by the 3-square L-tiles. Divide the board as follows, where A, B, C, D are each a board: 6

7 Without loss of geerality, suppose that the oe square has bee removed from B. The by the iductive hypothesis, B ca be tiled. Remove the ceter corers of A, C, ad D, such that each of their remaiders ca be tiled by the iductive hypothesis. Tile the remaiig 3 squares i the ceter with a sigle L-tile, ad we have completed the tilig. + 1, we have show that Z +, a chessboard with ay oe square removed ca be tiled by the 3-square L-tiles. 11. Collatz cojecture: Take ay atural umber. If is eve, divide it by to get /. If is odd, multiply it by 3 ad add 1 to obtai Repeat the process idefiitely. The Collatz cojecture states that, o matter what umber you start with, you will always evetually reach 1. Here s a hit. Proof: There is t a kow proof of the Collatz cojecture! that s why it s a cojecture). 7