Iductio proofs - practice! SOLUTIONS 1. Prove that f ) = 6 + + 15 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 ) = 4 + + 13 for some Z +. That meas that 4 + + 13 = 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 = 4 +8+4)++1)+13 = 4 ++13)+8+14. But 4 + + 13 is odd, ad therefore ca be writte as k + 1 for some k Z +. So f + 1) = k + 1 + 8 + 14 = k + 4 + 7) + 1, re-arragig a bit. Sice,k Z +, k + 4 + 7 Z + ad therefore, k + 4 + 7) + 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 ) = 6 + + 15 is odd for all Z +.. Prove that! > for a iteger greater tha 4. Base case: For = 4,! = 4! = 4 ad = 4 = 16. 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
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 + 5 4 + 10 3 + 10 + 5 + 1 1 = 5 ) + 5 4 + 3 + + 1). Now, 5 is divisible by 5, so we ca write it as 5 = 5k for some k Z +. The + 1) 5 + 1) = 5k + 5 4 + 3 + + 1). = 5k + 4 + 3 + + 1). 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 5. 4. For each Z +, it follows that +1 1 1. 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, +1 1 1 holds. Iductive hypothesis: Assume that +1 1 1 for some Z +. Iductive step: Show true for + 1, that is, that +1 + 1. +1 =. but +1 1 1. Therefore, +1 +1 1 1) + ad the sice < 1, + < + 1. Thus, +1 + 1
We have show that, if +1 1 1 for some Z +, it follows that +1 + 1. Sice the base case, that +1 + 1 for = 1 is true, ad sice we have show that if +1 1 1 for some Z +, it follows that +1 + 1, we have prove that +1 1 1 for each Z +. 5. Prove by iductio that 1 + + 3 + 4 +... + = + 1)/ for Z +. Base case: For = 1: LHS = 1++3+...+=1. RHS = 11 + 1)/ = 1. Sice LHS=RHS, 1 + + 3 + 4 +... + = + 1)/ for = 1. Iductive hypothesis: Assume true for some 1, that is, that 1++3+4+...+ = +1)/. Iductive step: Show true for some + 1, that is, that 1 + + 3 + 4 +... + + + 1 = + 1) + )/. LHS=1 + + 3 + 4 +... + + + 1 = 1 + + 3 + 4 +... + ) + + 1. But 1 + + 3 + 4 +... + = + 1)/. + 1) LHS=1 + + 3 + 4 +... + + + 1 = + + 1 ) = + 1) + 1 ) 1 = + 1) + ) + 1) + ) = = RHS. + 1, we have show that Z +, 1 + + 3 + 4 +... + = + 1)/. 6. Prove that 1 + + 3 + + ) = 1 3 + 3 + 3 3 + + 3 for every Z +. Base case: For = 1: LHS = 1) = 1. RHS = 1 3. Sice LHS=RHS, 1 + + 3 + + ) = 1 3 + 3 + 3 3 + + 3 for = 1. Iductive hypothesis: Assume true for some 1, that is, that 1 + + 3 + + ) = 1 3 + 3 + 3 3 + + 3. Iductive step: Show true for some + 1, that is, that 1 + + 3 + + + + 1)) = 1 3 + 3 + 3 3 + + 3 + + 1) 3 3
LHS = 1 + + 3 + + + + 1)) = 1 + + 3 + + ) + + 1)1 + + 3 + + ) + + 1) From the iductive hypothesis, 1++3+ +) = 1 3 + 3 +3 3 + + 3. Ad remember that 1 + + 3 +... + = + 1)/ SEE #5). The LHS = 1 3 + 3 + 3 3 + + 3 ) + + 1) + 1)/) + + 1) = 1 3 + 3 + 3 3 + + 3 ) + + 1) + + 1) = 1 3 + 3 + 3 3 + + 3 ) + + 1) + 1) = 1 3 + 3 + 3 3 + + 3 + + 1) 3 = RHS + 1, we have show that Z +, 1 + + 3 + + ) = 1 3 + 3 + 3 3 + + 3. 7. Prove by iductio that 1 + 3 + 5 + 7 +... + 1) = for Z +. Base case: For = 1: LHS = 1. RHS = 1) = 1. Sice LHS=RHS, 1 + 3 + 5 +... + 1) = for = 1. Iductive hypothesis: Assume true for some 1, that is, that 1 + 3 + 5 + 7 +... + 1) =. Iductive step: Show true for some + 1, that is, that 1 + 3 + 5 + 7 +... + 1) + + 1) = + 1). LHS = 1 + 3 + 5 + 7 +... + 1) + + 1) = + + 1), by the iductive hypothesis. But + + 1 = + 1). Therefore, LHS = + 1) = RHS + 1) LHS=1 + + 3 + 4 +... + + + 1 = + + 1 ) = + 1) + 1 ) 1 = + 1) + ) + 1) + ) =. 4
+ 1, we have show that Z +, 1 + 3 + 5 + 7 +... + 1) =. 8. Show by iductio that 1 + + 3 +... + = +1 for Z +. Base case: For = 1: LHS = 1 =. RHS = ) 1+1 =. Sice LHS=RHS, 1 + + 3 +... + = +1 for = 1. Iductive hypothesis: Assume true for some 1, that is, that 1 + + 3 +... + = +1. Iductive step: Show true for some + 1, that is, that 1 + + 3 +... + + +1 = +. LHS = 1 + + 3 +... + ) + +1 = +1 ) + +1, sice by the iductive hypothesis, 1 + + 3 +... + = +1. The LHS = +1 = + = RHS. + 1, we have show that Z +, 1 + + 3 +... + = +1. 9. 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 > 1 + 1 ) for some 3. Iductive step: Show true for some + 1, that is, that + 1 > 1 + 1 +1) +1. 5
This time start o the right had side. RHS = 1 + 1 ) +1 + 1 < 1 + 1 +1, ) = 1 + 1 ) 1 + 1 ) < 1 + 1 ) sice 1 > 1 +1 for 1 ad here, 3). The RHS < 1 + 1 ) 1 + 1 ) < 1 + 1 ), sice 1 + 1 ) < iductive hypothesis. Fially, distributig the, RHS < 1 + 1 ) = + 1 = LHS LHS > RHS that is, + 1 > 1 + 1 ) +1 + 1 Thus we have show that, assumig > 1 + 1 ) is true, it follows that 1 + 1 +1) +1. Ad sice the > 1 + 1 ) holds for = 3, we have show by iductio that > 1 + 1 ) 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 +1 +1 chessboard with ay oe square removed ca be tiled by the 3-square L-tiles. Divide the +1 +1 board as follows, where A, B, C, D are each a board: 6
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 3 + 1. 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