Mathematical Induction Krzysztof R. Apt (so not Krzystof and - PDF document

Mathematical Induction Krzysztof R. Apt (so not Krzystof and definitely not Krystof ) CWI, Amsterdam UvA 1

A puzzle Can one be 1 / 3 Indian? 2

Mathematical Induction and Functions f (1) = 2 , f ( n + 1) = 2 · f ( n ) . So f (2) = 2 · f (1) = 4 , f (3) = 2 · f (2) = 8 , f (4) = 2 · f (3) = 16 , and so on. It looks that f ( n ) = 2 n . It is true for n = 1 , 2 , 3 , 4 . But how to prove it for all n ≥ 1 ? 3

Principle of mathematical induction: • First we prove the property for n = 1 . So-called induction base. • Subsequently we assume that it is true for n and prove the property for n + 1 . So-called induction step. If we prove these two facts, then we proved f ( n ) = 2 n for all n ≥ 1 . 4

Proof f (1) = 2 , f ( n + 1) = 2 · f ( n ) . f ( n ) = 2 n ? 1. It holds for n = 1 . 2. Suppose it holds for some n . Then f ( n + 1) = 2 · f ( n ) = 2 · 2 n = 2 n +1 . 3. Conclusion: f ( n ) = 2 n holds for all n ≥ 1 . 5

Hugs Hugs> :load f.hs Main> f(10) 1024 where f.hs is simply the definition of f : f(1) = 2 f(n+1) = 2*f(n). 6

Mathematical Induction and Relations Note: 1 = 1 1 + 3 = 4 1 + 3 + 5 = 9 1 + 3 + 5 + 7 = 16 1 + 3 + 5 + 7 + 9 = 25 7

Theorem The sum of the first n odd numbers = n 2 . Proof Denote this sum by S ( n ) . Induction step. Note: S ( n + 1) = S ( n ) + 2 n + 1 . But S ( n ) = n 2 , so S ( n + 1) = n 2 + 2 n + 1 = ( n + 1) 2 . 8

Wimbledon S. Williams S. Williams V. Zvonareva S. Williams P. Kvitova V. Zvonareva T. Pironkova 4 players, 3 matches. Theorem In a binary tree with n ≥ 1 levels: The number of leaves (players): 2 n − 1 . The number of internal nodes (matches): 2 n − 1 − 1 . Example n = 3 . 2 n − 1 = 4 , 2 n − 1 − 1 = 3 . Example n = 8 . 2 n − 1 = 128 , 2 n − 1 − 1 = 127 . 9

Question Suppose there are 107 players. How many matches are needed to select a winner? 10

Sometimes we start with n = 5 Prove n 2 < 2 n . n left right side side 1 1 2 2 4 4 3 9 8 4 16 16 5 25 32 Theorem n 2 < 2 n holds for n ≥ 5 . 11

Homework 1 Prove that for n ≥ 1 the last digit of 6 n is 6 . 12

Homework 2 Prove that for n ≥ 1 the last digit of 7 n is not 5 . 13

Hanoi Towers 1 2 3 4 5 14

1 2 3 2 3 1 3 1 2 1 3 2 15

1 3 2 3 1 2 2 1 3 1 2 3 16

Theorem For all n ≥ 1 there is a solution to the towers of Hanoi puzzle. Proof • We first prove the property for n = 1 . Induction base: n = 1 . Move a disk from A to B. • Induction step: from n to n + 1 . – move the upper n disks from A to C using B. – move the largest disk from A to B, – move the upper n disks from C to B using A. 17

1 2 3 4 5 1 2 3 5 4 1 2 3 5 4 1 2 3 4 5 18

Prolog Program in Prolog: move(1,A,B,C) :- write(’move top disk from ’), write(A), write(’ to ’), write(B), nl. % newline move(M,A,B,C) :- M>1, N is M-1, move(N,A,C,B), move(1,A,B,C), move(N,C,B,A). 19

Prolog move(3,A,B,C). move top disk from A to B move top disk from A to C move top disk from B to C move top disk from A to B move top disk from C to A move top disk from C to B move top disk from A to B 20

Triomina Question: Is it possible to cover an arbitrary chessboard with one field removed? 21

Yes Theorem (Golomb) For each n ≥ 1 it is possible to cover an ar- bitrary chessboard of size 2 n by 2 n with one field removed. 22

Can one be 1 / 3 Indian? Theorem 1. For each n ≥ 1 and k ≤ 2 n one can be k/ 2 n Indian. 2. If one is f Indian, then for some n ≥ 1 and k ≤ 2 n we have f = k 2 n . 23

Mixing Paint | | | | | | | | | | | | |_____| |_____| |_____| | | | | | | | B | | W | | R | |_____| |_____| |_____| 24

Theorem For each paint in each bucket there is k/ 2 n litre of this paint, for some k ≤ 2 n . If k/ 2 n = 1 / 3 , then 3 k = 2 n , which can’t be true. 25

Mixing Drinks Can one get gin and tonic with 1 / 3 gin? | | | | | | | | |_____| | T | | | | | | G | | | |_____| |_____| 26

Recursion in Dutch history 27