Lecture 3.2: Strong Induction

1. Lecture 3.2: Strong Induction CS 250, Discrete Structures, Fall 2011 Nitesh Saxena *Adopted from previous lectures by Cinda Heeren, Zeph Grunschlag Lecture 3.2 -- Strong Induction

2. Course Admin • HW2 due tomorrow at 11am • Mid-Term 1 • We are grading them now, and should have the results by next weekend • Solution has been posted • HW1 graded • Please pick up your graded answers, if you haven’t done so already • Any questions, please get in touch with the TA • Further questions, please contact me Lecture 3.2 -- Strong Induction

3. Outline • More practice: induction • Strong Induction Lecture 3.2 -- Strong Induction

4. Induction: some more exercises (Rosen) • Show that n < 2n , for n=0, 1, 2,… • Prove (generalized De Morgan’s) Lecture 3.2 -- Strong Induction

5. In our proofs, to show P(k+1), our inductive hypothesis assures that ALL of P(0), P(1), … P(k) are true, so we can use ANY of them to make the inference. Strong Mathematical Induction If • P(0) and • k0 (P(0)  P(1)  …  P(k))  P(k+1) Then • n0 P(n) Lecture 3.2 -- Strong Induction

6. Example Recall the Fibonacci sequence: {fn} = 0,1,1,2,3,5,8,13,21,34,55,… defined by f0 = 0, f1 = 1, and for n > 1 fn = fn-1+fn-2 . Notice that every third Fibonacci number is even: LEMMA: For all natural numbers n, 2|f3n. Lecture 3.2 -- Strong Induction

7. Example Proof. Base case n = 0. f3·0 = f0 =0 which is divisible by 2  Induction step, n > 0: Assume F3k is divisible by 2 F3k+3 = f3k+2 +f3k+1= (f3k+1 +f3k) +f3k+1 = 2f3k+1+f3k By hypothesis, 2|f3k therefore 2|(2f3k+1+f3k), so 2|f3k+3 , and the proof is complete. Lecture 3.2 -- Strong Induction

8. Strong Induction Example Sometimes a stronger version of induction is needed, one that allows us to go back to smaller values than just the previous value of n. E.g. consider the Fibonacci sequence vs. the sequence 2n: {fn} = 0, 1, 1, 2, 3, 5, 8, 13, 21, 34 {2n} = 1, 2, 4, 8, 16, 32, 64, 128, 256, 512 LEMMA: For all n, fn < 2n Lecture 3.2 -- Strong Induction

9. Strong Induction Example LEMMA: For all n, fn < 2n Proof. Base n = 0: f0 = 0 < 1 = 20 Induction n > 0: Assume fk< 2k Now, fk+1= fk +fk-1<2k+fk-1 Q: Now what? Lecture 3.2 -- Strong Induction

10. Strong Induction Example A: Would want to apply same formula to k-1. But strictly speaking, can’t because induction hypothesis only let us look at previous domino. This limitation on induction need not be so: If we could assume that the first k dominos falling implies that the k+1thdomino falls, would be able to reduce back to smaller values, as needed here. Strong induction formalizes this ability. Lecture 3.2 -- Strong Induction

11. Strong Induction Example So now, we can complete stuck proof: LEMMA: For all n, fn < 2n Proof. Base cases (both needed as can’t apply induction step on f1 since f-1 is undefined) n = 0: f0 = 0 < 1 = 20 n = 1: f1 = 1 < 2 = 21 Lecture 3.2 -- Strong Induction

12. Strong Induction Completing Example So now, we can complete stuck proof: LEMMA: For all n, fn < 2n Proof. Base cases (both needed as can’t apply induction step on f1 since f-1 is undefined) n = 0: f0 = 0 < 1 = 20 n = 1: f1 = 1 < 2 = 21 Induction n > 0: fk+1 = fk+fk-1<2k+ 2k-1applying both P (k) and P (k-1) which can be assumed by strong induction hypothesis. Doing more algebra: 2k + 2k-1= 2k(1 + ½) = 1.5*2k < 2*2k = 2k+1 Therefore, fk+1 < 2k+1 Lecture 3.2 -- Strong Induction

13. Another Example (Rosen) • Prove that every integer > 1 can be expressed as a product of prime numbers Lecture 3.2 -- Strong Induction

14. Today’s Reading • Rosen 5.2 Lecture 3.2 -- Strong Induction