Induction Proof

1. Induction Proof

2. Well-ordering A set S is well ordered if every subset has a least element. • [0, 1] is not well ordered since (0,1] has no least element. Examples: • N is well ordered (under the  relation). • Any coutably infinite set can be well ordered. • The least element in a subset is determined by a bijection (list) which exists from N to the countably infinite set. • Z can be well ordered but it is not well ordered under the  relation. • Z has no smallest element. • The set of finite strings over an alphabet using lexicographic ordering is well ordered.

3. Mathematical Induction Let P(x) be a predicate over a well ordered set S. In the case that S = N, the natural numbers, the principle has the following form. P(0) P(n)  P(n +1) x P(x) The hypotheses are • H1: P(0) (Basis Step) • H2: P(n) P(n +1) for n arbitrary. (Induction Step)

4. How It Works • First, prove that the predicate is true for the smallest element of the set S (0 if S = N). • Then, show if it is true for an element (n if S=N) implies it is true for the “next” element in the set (n + 1 if S=N). Meaning • Knowing it is true for the first element means it must be true for the element following the first or the second element • Knowing it is true for the second element implies it is true for the third and so forth. • Therefore, induction is equivalent to modus ponens applied an countable number of times!!

5. Outline of Induction Proof • State what P(n) is. • Basis: • Prove that P(0) is true. • Induction hypothesis: • Assume that P(n) is true, for an arbitrary n. • Induction step: • Prove that P(n+1) is true, using P(n). (Usually, direct proof)

6. Example Prove:ni= n(n +1)/2 i=0 In logical notation we wish to show n [ni = n(n +1)/2 ] i=0 Hence, P(n) is [ni] = n(n +1)/2 i=0 Basis: Prove H1: P(0): 0=0(0 +1)/2 Induction Hypotheses: Assume P(n) is true for n arbitrary. Now use this and anything else you know to establish that P(n+1)must be true.

7. Example Induction step: Write down the assertion P (n+1) P(n + 1)is the assertion n+1i=(n+1)((n +1)+1)/2 i=0 n+1i =ni +n+1 i=0 i=0 = n(n +1)/2 + n+1 (frominductionhypothesis) = (n(n +1) + 2(n+1) )/ 2 = (n+1)(n+2)/ 2 = (n+1)((n +1)+1)/2 Q.E.D.

8. More General Rule • Suppose we wish to prove for some specific integer kx [xk P(x)] • Now we merely change the basis step to P(k)and continue.

9. Example Prove 3n + 5 is in O(n2). Definition:f(n) is in O(g(n)) if there are constants c>0 and k such that f(n)cg(n) for n  k. Proof: We must find C and k such that 3n + 5 Cn2 for nk (or n > k-1). If we try C = 1, then the assertion is not true until k = 5. Hence we prove by induction that 3n + 5 n2 for all n 5. The assertion becomes n[n  5 3n + 5 n2 ] and the predicate P(n) is 3n + 5 n2.

10. Example Basis step: P(5): 35 + 5 = 20 £ (5)2 . Induction hypothesis: Assume P(n): 3n + 5 £ n2 is true for arbitrary n. Induction step: Prove P(n+1): 3(n +1)+ 5 £ (n +1)2 From 3n + 5 £ n2 , we have (3n + 5) + 3 £ n2 +3. Now we must show that n2 + 3 £ (n + 1)2 = n2 + 2n + 1 which is true iff 3 £ 2n + 1 which is true iff n³ 1. But we have already restricted n ³ 5 so n ³ 1 must hold. That is, "n[n ³ 5®3n + 5 £ n2 ], i.e. 3n + 5 is in O(n2). Q.E.D.

11. Double Qualifier • In doubly quantified assertions of the form "m"n[P(m,n)] • we often assume m (or n ) is arbitrary to eliminate a quantifier and prove the remaining result using induction.

12. Strong Induction H1: P(0) H2: P(0) P(1)  ... P(n)  P(n +1) xP(x) The two rules are equivalent but sometimes the second is easier to apply.

13. Example For any integer n>1, n can be written as a product of primes. Proof: Let P(n) be the predicate n can be written as a product of primes. Basis: P(2): 2 can be written as a product of 2, which is a prime. Induction hypothesis: For any integer kn, k can be written as a product of primes.

14. Example Induction step: Prove n+1 can be written as a product of primes. • If n+1 is a prime, then it can be written as a product of itself only (by def. of primes). • If n+1 is not a prime, then there exist integers p and q such that pq=n+1 (by def. of primes) and p and q are less than n+1. Since p and q are less than n+1, p and q can be written as products of primes (by induction hypothesis). Thus, n+1 can be written as a product of primes that make p and q. Q.E.D.

15. Recursive or Inductive Definitions • Basis step • For sets • State the basic building blocks (BBB's) of the set. • For functions • State the values of the function on the BBB’s. • Inductive or recursive step: • For sets • Show how to build new things from old with some construction rules. • For functions • Show how to compute the value of a function on the new things that can be built knowing the value on the old things.

16. Recursive or Inductive Definitions • Extremal clause: • For sets • If you can't build it with a finite number of applications of steps 1. and 2. then it isn't in the set. • For functions • A function defined on a recursively defined set does not require an extremal clause. • Often omitted. • To prove something is in the set you must show how to construct it with a finite number of applications of the basis and inductive steps. • To prove something is not in the set is often more difficult.

17. Example A recursive definition of N: • Basis: • 0 is in N (0 is the BBB). • Induction: • if n is in N then so is n + 1 (how to build new objects from old: “add one to an old object to get a new one”). • Extremal clause: • If you can't construct it with a finite numberof applications of the basis and induction, it is not in N.

18. Example Given the recursive definition of N, we can give recursive definitions of functions on N: • f(0)=1 • The initial condition or the value of the function on the BBB’s. • f(n+1)=(n+1)f(n) • The recurrence equation, how to define f on the new objects based on its value on old objects. • f is the factorial function: f(n) = n! • Note how it follows the recursive definition of N. • Proof of assertions about inductively defined objects usually involves a proof by induction.

19. Proof guideline • Prove the assertion is true for the BBBs in the basis step. • Prove that if the assertion is true for the old objects it must be true for the new objects you can build from the old objects. • Conclude the assertion must be true for all objects.

20. Example We define aninductively where n is in N. Basis: a0= 1 Induction: a(n+1) = ana Theorem: m n [aman= am+n] Proof: Since the powers of a have been defined inductively, we must use a proof by induction somewhere. Get rid of the first quantifier on m by Universal Instantiation: Assume m is arbitrary. Now prove, by induction, the remaining quantified assertion n [aman= am+n]

21. Example Basis step:Show it holds for n=0. The left side becomes ama0= am(1)= am. The right side becomes am+0= am. Hence, the two sides are equal to the same value. Induction hypothesis: Assume the assertion is true for n: aman= am+n. Induction step: Now show it is true for n+1 (aman+1=am+n+1). aman+1 =am(ana) (by inductive step in the definition of an) aman+1 = (aman)a (by associativity of multiplication) aman+1 = am+na (bythe induction hypothesis) aman+1 = am+n+1 (by inductive step in the definition of an)

22. Rooted Trees The set of rooted trees, where a rooted tree consists of a set of vertices containing a distinguished vertex called the root, and edges connecting these vertices, can be defined recursively as follows: Basis step: A single vertex r is a rooted tree. Recursive step: If T1, T2, …, Tn are rooted trees with roots r1, r2, …, rn , espectively, the graph form by starting with a root r, which is not in any of T1, T2, …, Tn , and adding an edge from r to r1, r2, …, rn is also a rooted tree.

23. Extended Binary Trees The set of extended binary trees can be defined recursively as follows: Basis step: The empty set is an extended binary tree. Recursive step: If T1 and T2 are extended binary trees, there is an extended binary tree, denoted by T1T2, consisting of a root r together with edges connecting the root to the root of the left subtree T1 and the right subtree T2, when these trees are not empty.

24. Full Binary Trees The set of full binary trees can be defined recursively as follows: Basis step: A single vertex r is a full binary tree. Recursive step: If T1 and T2 are full binary trees, there is an full binary tree, denoted by T1T2, consisting of a root r together with edges connecting the root to the root of the left subtree T1 and the right subtree T2.

25. 2 0 1 0 Height of Trees The height of a full binary tree T, denoted by h(T), can be defined as follows: Basis step: The height of a full binary tree T consisting of a single vertex r is h(T)=0. Recursive step: If T1 and T2 are full binary trees, the height of a full binary tree T = T1T2 is h(T)=1+max(h(T1), h(T2)).

26. Number of Nodes in a Full Binary Tree Theorem: If T is a full binary tree, the number of nodes in T, denoted by n(T), is not more than 2h(T)+1 -1. Proof: Basis step: For a full binary tree T of height 0, T is consisted of one vertex. Then, n(T)=1, and 2h(T)+1-1 =20+1-1=1. Thus, n(T) 2h(T)+1-1. Induction hypothesis: Assume n(T) 2h(T)+1-1 for any full binary tree T of the height less than k. Induction step: If T1 and T2 are full binary trees of the height less than k, the number of nodes in a full binary tree T = T1T2 is 1+n(T1)+n(T2).

27. Number of Nodes in a Full Binary Tree n(T) =1+n(T1)+n(T2) (from the construction of T)  1 + (2h(T1)+1-1) + (2h(T2)+1-1)(from induction hypothesis) 1 + (2h(T1)+1-1) + (2h(T2)+1-1) = 2h(T1)+1+ 2h(T2)+1-1  2max(2h(T1)+1, 2h(T2)+1) -1 2max(2h(T1)+1, 2h(T2)+1) -1 = 2(2 max(h(T1)+1,h(T2))+1) -1 = 22h(T)-1 = 2h(T)+1-1 That is, n(T)  2h(T)+1-1. Q.E.D.