Mathematical Induction 1

2 Suppose we have a sequence of propositions which we would like to prove: P (0), P (1), P (2), P (3), P (4), … P (n), … We can picture each proposition as a domino: P (i)

3 Mathematical Induction When the domino falls, the corresponding proposition is considered true: P (i)

4 Mathematical Induction So sequence of propositions is a sequence of dominos. … P (n+1)P (n) P (2)P (1)P (0)

5 Mathematical Induction When the domino falls (to right), the corresponding proposition is considered true: P (i) true

6 Mathematical Induction Suppose that the dominos satisfy two constraints. 1) Well-positioned: If any domino falls (to right), next domino (to right) must fall also. 2) First domino has fallen to right P (0) true P (i+1)P (i)

7 Mathematical Induction Suppose that the dominos satisfy two constraints. 1) Well-positioned: If any domino falls to right, the next domino to right must fall also. 2) First domino has fallen to right P (0) true P (i+1)P (i)

8 Mathematical Induction Suppose that the dominos satisfy two constraints. 1) Well-positioned: If any domino falls to right, the next domino to right must fall also. 2) First domino has fallen to right P (0) true P (i) true P (i+1) true

9 Mathematical Induction Then can conclude that all the dominos fall! … P (n+1)P (n) P (2)P (1)P (0)

10 Mathematical Induction Then can conclude that all the dominos fall! … P (n+1)P (n) P (2)P (1)P (0)

11 Mathematical Induction Then can conclude that all the dominos fall! …P (0) true P (n+1)P (n) P (2)P (1)

12 Mathematical Induction Then can conclude that all the dominos fall! …P (0) true P (1) true P (n+1)P (n) P (2)

13 Mathematical Induction Then can conclude that all the dominos fall! P (2) true …P (0) true P (1) true P (n+1)P (n)

L1414 Mathematical Induction Then can conclude that all the dominos fall! P (2) true …P (0) true P (1) true P (n+1)P (n)

15 Mathematical Induction Then can conclude that all the dominos fall! P (2) true …P (0) true P (1) true P (n) true P (n+1)

16 Mathematical Induction Then can conclude that all the dominos fall! P (2) true …P (0) true P (1) true P (n) true P (n+1) true

17 Mathematical Induction Principle of Mathematical Induction: If: 1) [basis] P (0) is true 2) [induction]  n P(n)=>P(n+1) is true Then:  n P(n) is true This formalizes what occurred to dominos. P (2) true …P (0) true P (1) true P (n) true P (n+1) true

Proof By Weak Induction Claim:S(n) is true for all n >= k Basis: Show formula is true when n = k (often k is taken equal to 1 or 0) Inductive hypothesis: Assume formula is true for an arbitrary n Step: Show that formula is then true for n+1 18

Weak Induction Suppose S(k) is true for fixed constant k Often k = 0 or k=1 Suppose S(n) is true S(n) => S(n+1) Then S(j) is true for all j >= k This means that if S(k) is true for fixed constant k and it follows from S(j) is true for j=n that S(j) is true for j=n+1, then S (j) is true for all j >= k 19

Induction Example: Arithmetic Progression Arithmetic Progression is a sequence of numbers such that the difference of any two successive members of the sequence is a constant: d is the difference of the progression 20

Induction Example 1: Arithmetic Progression Prove Basis: If n = 1, then Inductive hypothesis (assume true for n): Assume Step (show true for n+1): 21

Induction Example 2: Arithmetic Progression The sum of the first n members of the arithmetic progression is given by 22

Induction Example 2: Arithmetic Progression Prove … + n = n(n+1) / 2 Basis: If n = 0, then 0 = 0(0+1) / 2 Inductive hypothesis (assume true for n): Assume … + n = n(n+1) / 2 Step (show true for n+1): … + n + n+1 = ( …+ n) + +(n+1)= = n(n+1)/2 + n+1 = [n(n+1) + 2(n+1)]/2 = (n+1)(n+2)/2 = (n+1)((n+1) + 1) / 2 23

Strong Induction Strong induction means Basis: show S(0) Hypothesis: assume S(k) holds for arbitrary k <= n Step: Show S(n+1) follows from S(n-i),…,S(n-1),S(n) Another variation: Basis: show S(0), S(1) Hypothesis: assume S(n) and S(n+1) are true Step: show S(n+2) follows from S(n), S(n+1) 24

Homework pp (Section 2.5) 25