Copyright © Peter Cappello Mathematical Induction Goals Explain & illustrate construction of proofs of a variety of theorems using mathematical induction.

Copyright © Peter Cappello Motivation Mathematics uses 2 kinds of arguments: deductive inductive Proposition: P( n ): … + n = n( n + 1 )/2. Observe that P(1), P(2), P(3), & P(4). Conjecture:  n  N P( n ). Mathematical induction is a finite proof pattern for proving propositions of the form  n  N P( n ).

Copyright © Peter Cappello The Principle of Mathematical Induction Let P( n ) be a predicate function:  n  N P( n ) is a proposition. To prove  n  N P( n ), it suffices to prove: 1.P( 1 ) is true. 2.  n  N ( P( n )  P( n + 1 ) ). This is not magic. It is a recipe for constructing a finite proof for arbitrary n  N.

Copyright © Peter Cappello Proving P( 3 ) Given P( 1 )   n  1 ( P( n )  P( n + 1) ). Proof: 1. P( 1 ).[premise 1] 2. P( 1 )  P( 2 ). [U.S. of premise 2 for n = 1] 3. P( 2 ).[step 1, 2, & modus ponens] 4. P( 2 )  P( 3 ). [U.S. of premise 2 for n = 2] 5. P( 3 ).[step 2, 3, & modus ponens] Construct a finite proof for P( 1,999,765 ).

Copyright © Peter Cappello Mathematical Induction as the Domino Principle If the 1 st domino falls over and the n th domino falls over implies that the ( n + 1 ) st domino falls over then domino n falls over for all n  N.

Copyright © Peter Cappello Mathematical Induction as the Domino Principle 123nn +1

Copyright © Peter Cappello The 3-Step Method The implication in step 2 typically is proved directly. The proof pattern thus has 3 steps: 1.Prove P( 1 ). [called the basis] 2.Assume P( n ) [called the induction hypothesis] 3.Prove P( n + 1 ) [called the inductive step] The last 2 steps are for arbitrary n  N. Using P( n ) to prove P( n + 1 ) implies a recursive formulation of P( n ).

Copyright © Peter Cappello Induction as a Creative Process Mathematical induction is similar to, but not identical to, scientific induction. In both cases, a “theory” is created. Look at specific cases; perceive a pattern. Hypothesizing a pattern, a theory, is a creative process (only people who are bad at mathematics say otherwise). With mathematical induction, a “theory” can be proved.

Copyright © Peter Cappello Scientific theories cannot be proved. They can be disproved. A scientific theory can be based on a mathematical model. Propositions can be proved within the model. Like axioms, the relationship between: the mathematical model physical reality cannot be proven correct.

Copyright © Peter Cappello Example 1 = 1 3 = = = What is a general formula, if any, for … + n? Let F( n ): n. A recursive formulation: F( n ) = F( n - 1 ) + n.

Copyright © Peter Cappello 1: 2: 3: Put these blocks, which represent numbers, together to form sums: = = A Geometric Interpretation

Copyright © Peter Cappello n n Area is n 2 /2 + n/2 = n(n + 1)/2

Copyright © Peter Cappello … + n = n(n + 1)/2 A Mathematical Induction Proof F( 1 ) = 1( )/2 = 1. Assume F( n ) = n( n + 1 )/2 Show F( n + 1 ) = ( n + 1 )( n + 2 )/2. F( n + 1 ) = n + ( n + 1 ) [Definition] = F( n ) + n + 1 [Recursive formulation] = n( n + 1 ) / 2 + n + 1 [Induction hyp.] = n( n + 1 ) / 2 + ( n + 1 ) (2/2) = ( n + 1 ) ( n + 2 ) / 2.

Copyright © Peter Cappello In finding a recursive formulation, we focused on the: similarities differences for successive values of n. Sometimes, it is useful to: Note the difference between F( n ) & F( n – 1 ). Find a pattern in this sequence of differences.

Copyright © Peter Cappello Example: n 3 = ? Let F( n ) = n 3. What is a formula for F(n)? 1 = = = = Do you see a pattern?

Copyright © Peter Cappello Prove that  n F( n ) = [ n( n + 1 )/2 ] 2 1. F( 1 ) = 1 3 = 1 = [ 1( 2 ) / 2 ] Assume F( n ) = [ n( n + 1 ) / 2 ] 2.  I.H. 3. Prove F( n + 1 ) = [ ( n + 1 )( n + 2 ) / 2 ] 2. F ( n + 1 ) = n 3 + ( n +1 ) 3  Defn. of F( n + 1 ) = F( n ) + ( n + 1 ) 3  Recursive formulation = [ n( n + 1 )/ 2 ] 2 + ( n + 1 ) 3  Use I. H. = ( n + 1 ) 2 [ ( n / 2 ) 2 + ( n + 1 ) ] = ( n + 1 ) 2 [ n 2 / 4 + ( 4 / 4 )( n + 1 ) ] = ( n +1 ) 2 [ ( n 2 + 4n + 4 ) / 4 ] = [ ( n + 1) (n + 2) / 2 ] 2.

Copyright © Peter Cappello Translating the starting point If we: know P( n ) is false for 1  n  9 think P( n ) is true for n > 9. Then define Q( n ) = P( n + 9 ). Use mathematical induction to show that  n  N Q( n ). We thus can start the induction at any natural number.

Copyright © Peter Cappello Example: Stamps Suppose the US Post Office prints only 5 & 9 cent stamps. Prove  n > 34, you can make postage for n cents, using only 5 & 9 cent stamps. Let S( n ) denote the statement: You can make postage for n cents using only 5-cent & 9-cent stamps. 1. Basis: For n = 35: Use 7 5-cent stamps. 2. I.H.: Assume S( n ).

Copyright © Peter Cappello 3. Prove S( n + 1 ). Case: For S( n ), # of 9-cent stamps used = 0: Only 5-cent stamps are used for S( n ). # of 5-cent stamps ≥ 7. Replace 7 5-cent stamps with 4 9-cent stamps. Case: For S( n ), # of 9-cent stamp used > 0: Replace 1 9-cent stamp with 2 5-cent stamps.

Copyright © Peter Cappello Generalizing the Basis To prove  n  N P( n ), if suffices to show: P( 1 )  P( 2 ).  n  N( [ P( n )  P( n + 1 ) ]  P( n + 2 ) ) If: We can push over the first 2 dominos; Pushing over any 2 adjacent dominos implies pushing over the next domino. then we can push over all the dominos.

Copyright © Peter Cappello The Fibonacci Formula Define the n th Fibonacci number, F( n ), as: F( 0 ) = 0, F( 1 ) = 1, F( n ) = F( n – 1 ) + F( n – 2 ). Prove F( n ) = 5 -1/2 ( [ ( /2 ) / 2] n - [ ( /2 ) / 2 ] n ). Basis: F( 0 ) = 5 -1/2 ( [ ( /2 ) /2 ] 0 - [ ( /2 ) / 2 ] 0 ) = 0. F( 1 ) = 5 -1/2 ( [ ( /2 ) / 2 ] 1 - [ ( /2 ) / 2 ] 1 ) = ( 5 -1/2 / 2 ) ( / /2 ) = 1.

Copyright © Peter Cappello The Fibonacci Formula F( n ) = 5 -1/2 ( [ ( /2 ) / 2] n - [ ( /2 ) / 2 ] n ) Let a = ( /2 ) / 2 b = ( /2 ) / 2. Note: a + 1 = a 2 & b + 1 = b 2. Induction hypotheses: F( n ) = 5 -1/2 ( a n – b n ) F( n +1 ) = 5 -1/2 ( a n + 1 – b n + 1 ). Induction step: Show F( n + 2 ) = 5 -1/2 ( a n b n + 2 ).

Copyright © Peter Cappello F( n ) = 5 -1/2 ( [ ( /2 ) / 2] n - [ ( /2 ) / 2 ] n ) Proof of Induction Step F( n + 2 ) = F ( n + 1 ) + F ( n ) [Definition] = 5 -1/2 ( a n + 1 – b n + 1 ) /2 ( a n – b n ) [I.H.] = 5 -1/2 ( a n a n – b n + 1 – b n ) = 5 -1/2 ( a n ( a + 1 ) – b n ( b + 1 ) ). = 5 -1/2 ( a n + 2 – b n + 2 )

Copyright © Peter Cappello Generalizing this... If P( 1 )  P( 2 ) ... P( k )  n { [ P( n + 1 )  P( n + 2 ) ... P( n + k ) ]  P( n + k + 1 ) } then  n  N P( n ).

Copyright © Peter Cappello End

Copyright © Peter Cappello Strong Mathematical Induction If P( 1 )  P( 2 ) ...  P( k ) and for n  k, [ P( 1 )  P( 2 ) ...  P( n ) ]  P( n + 1 ) then,  n  N P(n).

Copyright © Peter Cappello Example: Fundamental Theorem of Arithmetic Prove that all natural numbers  2 can be represented as a product of primes. Basis: 2: 2 is a prime. Assume that 1, 2,..., n can be represented as a product of primes.

Copyright © Peter Cappello Show that n + 1can be represented as a product of primes. Case n + 1 is a prime: It can be represented as a product of 1 prime, itself. Case n + 1 is composite: n = ab, for some a,b < n. Therefore, a = p 1 p 2... p k & b = q 1 q 2... q l, where the p i s & q i s are primes. Represent n = p 1 p 2... p k q 1 q 2... q l.