Lecture 15 21 127 Concepts of Math 10 03 2012 General Induction Example from homework Recall Problem 5 from Homework 2 Mo Brendans Mo Problems That was a General Induction problem Claim With currency notes of denominaton 3 and 7 Brendans we can definitely make payments of anything over and including 12 Brendans Proof Let P n be the statement We can pay n with denominations 3 and 7 Base Cases We can show P 12 P 13 P 14 easily 12 4 3 13 2 3 7 14 2 7 Induction Hypothesis Suppose k N and k 14 and suppose P 12 P 13 P 14 P k Notice that k 14 is an essential assumption here based on the way we set up our Base Cases We can t use the following argument to deduce P 13 say so we need to make this restriction Induction Step WWTS P k 1 Since k 14 k 1 15 so k 1 3 12 Thus P k 1 3 applies This tells us k 1 3 k 2 is achievable By adding another 3 Brendan note to the representation of k 2 we can achieve k 1 Huzzah Minimal Criminal argument The second condition in the statement of the Principle of Mathematical Induction is a conditional statement k N P 1 P 2 P k P k 1 Thus we can consider proving it by contrapositive instead k N P k 1 P 1 P 2 P k That is if we assume some instance of the claim P n fails we can deduce some previous instance fails as well This cannot descend forever because we already proved the base case does hold Thus if we can actually prove this contrapositive successfully then we have shown that the claim holds for all instances The minimal criminal idea comes from the fact that there is no first instance of the claim that fails Claim Prime factorizations are unique Proof Let P n be the proposition There is only one prime factorization of n We will prove P n holds for every natural n 2 by induction on n Base Case Obviously P 2 holds because 2 2 is the only factorization Induction Hypothesis Suppose k 2 and suppose P k 1 fails This means we can factor k 1 in two ways Induction Step Assuming k 1 is even then both factorizations must have a 2 in them By dividing out 1 that factor of 2 we obtain two factorizations of k2 Moreover they must be different factorizations since we only removed a factor common to both This shows that P k 2 holds Assuming k 1 is odd then we have two cases If k 1 itself is prime then it certainly only has one factorization so this case does not apply If k 1 is not prime then we know it has an odd prime factor let s call that factor p Note This appeals to an unproven as of yet result known as Euclid s Lemma We will prove it later in the course perhaps By removing that factor p from both factorizations of k 1 we again obtain two different factorizations of k 1 p This shows that P k 1 p holds and noting that p 3 we have shown the desired conditional statement Relations Definition and Examples Definition Let A B be sets A relation between A and B is a set of ordered pairs R A B Given elements a A and b B we say a and b are related if and only if a b R The set A is called the domain and the set B is called the codomain The set R is called the relation set and we say R is a relation between A and B When A B we say R is a relation on the set A It is also fairly common to write x R y to mean x y R A relation R is often defined by identifying a property of the elements of A and B and saying a b R P a b is true Example 0 1 Let W English words and L English letters Define the relation R by setting w R w begins with Then mathematics m R and golf g R but knowledge n R and you u R Also note that zyzyxyqy z R because zyzyxyqy W It is sometimes the case that A B so R defines a relation on pairs of elements from one set Example 0 2 Let A B Z and define x y R x and y have the same parity Then 2 8 R and 3 7 R and 99 99 R but 1 2 R and 0 3 R Example 0 3 Define a relation on R by setting x y R x y Note the difference in font between R and R Then 1 R and 0 100 R but 2 2 R and 1 R Notice that these are ordered pairs which we may forget about since A B R so the order of the elements matters Indeed knowing that x y R doesn t necessarily imply that y x R in general In this example that implication is always false even Example 0 4 Let S 3 and define a relation R on P S by x y R x y Thus 1 1 2 R and 3 R and 3 3 R but 1 2 1 R Example 0 5 Let S be the set of students in your class Define a relation R1 between S and N by saying s n R1 if person s S is n years old Write out a few elements of this relation set Now define a relation R2 on S by saying s t S if persons s and t are the same age in years Write out a few elements of this relation set How does this compare to the relation R1 Does R2 somehow encode the same information about the elements of the set S Why or why not Is there a way we can use R1 to determine R2 What about the other way around Properties Definition Let A be a set and let R be a relation on A i e R A A We say R is reflexive if x A x x R i e every element is related to itself 2 We say R is symmetric if x y A x y R y x R i e the order of the comparison doesn t matter We say R is transitive if x y z A x y R y z R x z R i e the relation always transitions through a middle man We say R is anti symmetric if x y A x y R y x R x y i e two different elements can be related in at most one way or not at all Write out the contrapositive of the line above to see why this characterization is true Note that anti symmetric is different than not symmetric Think about the relation it is both symmetric …