Math.3336: Discrete Mathematics Mathematical Induction Instructor: Dr. Blerina Xhabli Department of Mathematics, University of Houston https://www.math.uh.edu/ blerina Email: blerina@math.uh.edu Fall 2018 Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 1/37
Assignments to work on Homework #6 due Monday, 10/15, 11:59pm No credit unless turned in by 11:59pm on due date Late submissions not allowed, but lowest homework score dropped when calculating grades Homework will be submitted online in your CASA accounts. You can find the instructions on how to upload your homework in our class webpage. Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 2/37
Fermat s Little Theorem Fermat s Little Theorem: If p is prime and gcd(a, p) = 1, then a p 1 1 (mod p). Furthermore, for every a Z, a p a (mod p). Fermat s little theorem is useful in computing the remainders modulo p of large powers of integers. Find 7 222 (mod 11). By Fermat s little theorem, 7 10 1 (mod 11). Therefore, (7 10 ) k 1 (mod 11), k Z. It follows that 7 222 = 7 10 22+2 = (7 1 0) 2 2 7 2 (1) 2 2 49 5 (mod 11). As a result, 7 222 5 (mod 11) Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 3/37
Proof of Fermat s Little Theorem Proof: It is clear that for any relatively prime integers a, p, if a p 1 1 (mod p), then a p = a a p 1 a 1 = a (mod p), i.e. a p a (mod p). Note all integers 1, 2, 3,..., p 1 are relatively prime to p. Furthermore, since gcd(a, p) = 1, then all integers a, 2a, 3a,..., (p 1)a are relatively prime to p. Let i, j {1, 2, 3,..., p 1}, then one can show that ia ja (mod p) iff i = j. This shows that each of a, 2a, 3a,..., (p 1)a is distinctly equivalent to one of 1, 2, 3,..., p 1. As a result, by properties of congruence we get: a 2a 3a (p 1)a 1 2 3 (p 1) (mod p) Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 4/37
Proof of Fermat s Little Theorem As a result, by properties of congruence we get: a 2a 3a (p 1)a 1 2 3 (p 1) (mod p) It follows that (p 1)!a p 1 (p 1)! (mod p) Since gcd(p, (p 1)!) = 1, then we deduce that a p 1 1 (mod p). Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 5/37
Euler s Totient Function The Euler s Totient Function at a given positive integer n, denoted by φ(n), gives the number of all positive integers smaller than n relatively prime to n. If p is prime, then φ(p) = p 1. Because p is relatively prime to any of 1, 2, 3,..., p 1. If n = pq, then φ(n) = (p 1)(q 1). Because all positive integers less equal than n that have a factor of p or q are p, 2p, 3p,..., qp and q, 2q, 3q,..., pq. The rest of them are relatively prime to n = pq, i.e. φ(n) = pq p q + 1 = (p 1)(q 1) Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 6/37
Euler s Totient Function in Fermat s Little Theorem By Fermat s Little Theorem, if gcd(a, p) = gcd(a, q) = 1, then a p 1 1 (mod p) a q 1 1 (mod q) Furthermore, we have a (p 1)(q 1) 1 = (a p 1 1)(a q 1 + a q 2 + + a 2 + a + 1) = (a q 1 1)(a p 1 + a p 2 + + a 2 + a + 1) Since p (a p 1 1) and q (a q 1 1), it follows that p (a (p 1)(q 1) 1) and q (a (p 1)(q 1) 1), i.e. a (p 1)(q 1) 1 (mod p) a (p 1)(q 1) 1 (mod q) Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 7/37
Euler s Totient Function in Fermat s Little Theorem Thus, we have a (p 1)(q 1) 1 (mod p) a (p 1)(q 1) 1 (mod q) A quick application of Chinese Remainder Theorem implies that a (p 1)(q 1) 1 (mod pq). To generalize, given that gcd(a, n) = 1 we have a φ(n) 1 (mod n). Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 8/37
High Level Math Behind RSA In the RSA system, private key consists of two very large prime numbers p, q Public key consists of a number n, which is the product of p, q and another number e e is a number relatively prime with (p 1)(q 1) = φ(n) Encrypt messages using n, e, but to decrypt, must know p, q In theory, can extract p, q from n using prime factorization, but this is intractable for very large numbers Security of RSA relies on inherent computational difficulty of prime factorization Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 9/37
Encryption in RSA To send message to Bob, Alice first represents message as a sequence of numbers Call this number representing message M Alice then uses Bob s public key n, e to perform encryption as: C is called the ciphertext C = M e (mod n) Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 10/37
Encryption Example Encrypt message STOP using RSA with n = 2537, e = 13 First convert each letter to a number in [0, 25]: S = 18, T = 19, O = 14, P = 15 Group sequence into blocks of 4 digits: M = 1819 1415 Now encrypt each block as C = M 13 (mod 2537) For first block, 1819 13 (mod 2537) = 2081; for second block 1415 13 (mod 2537) = 2182 Ciphertext: 2081 2182 Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 11/37
RSA Decryption How do we decrypt cipher text using private keys p, q? We know that e is relatively prime to (p 1)(q 1). Decryption key d is the inverse of e modulo (p 1)(q 1): d e 1 (mod φ(n) = (p 1)(q 1)) As we saw earlier, d can be computed reasonably efficiently if we know (p 1)(q 1) However, since adversaries do not know p, q, they cannot compute d with reasonable computational effort! Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 12/37
RSA Decryption, cont. We know that where p, q are primes. a (p 1)(q 1) 1 (mod pq), Our goal is to find the decryption key d, which is the inverse of e modulo (p 1)(q 1), i.e. need d such that Then we have: d e + t (p 1)(q 1) = 1. (M e ) d = M ed = M 1+(p 1)(q 1) ( t) M (mod n = pq) Since the ciphertext C is just M e, C d (mod n) allows decrypting the message Since Bob can compute d using p, q, Bob can easily decrypt message, but no one else can! Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 13/37
Decryption Example A message M is encrypted through RSA cipher with p = 41, q = 61 and e = 17, and the following cipher text is obtained: C = 2133 2330 1235 0655 1403 To decrypt this cipher text, we need to compute d, the inverse of e modulo (p 1)(q 1). Here, (p 1)(q 1) = 2400, thus solve: 17x 1 (mod 2400) Apply extended Euclidean algorithm to compute s, t such that: We get s = 847, t = 6 17s + 2400t = 1 Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 14/37
Example, cont. Decrypt C = 2133 2330 1235 0655 1403 using p = 41, q = 61, n = 2501, and e = 17. The inverse of e = 17 modulo (p 1)(q 1) = 2400 is d = 847 1553 (mod 2400) 2133 1553 (mod 2501) = 1108; 2330 1553 (mod 2501) = 0504 1235 1553 (mod 2501) = 0818; 0655 1553 (mod 2501) = 0614 1403 1553 (mod 2501) = 1403 Thus, decrypted message is M = 1108 0504 0818 0614 1403, or in English, Life is Good. Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 15/37
Chapter 5 Induction and Recursion Chapter 5 Overview Mathematical Induction Section 5.1 Strong Induction Section 5.2 Well-Ordering Section 5.2 Recursive Definitions Section 5.3 Structural Induction Section 5.3 Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 16/37
Section 5.1 Introduction to Mathematical Induction Many mathematical theorems assert that a property holds for all natural numbers, odd positive integers, etc. Mathematical induction: very important proof technique for proving such universally quantified statements Induction will come up over and over again in other classes: algorithms, programming languages, automata theory,... Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 17/37
Analogy Suppose we have an infinite ladder, and we know two things: 1 We can reach the first rung of the ladder 2 If we reach a particular rung, then we can also reach the next rung From these two facts, can we conclude we can reach every step of the infinite ladder? Answer is yes, and mathematical induction allows us to make arguments like this Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 18/37
Mathematical Induction Used to prove statements of the form x Z + P(x) An inductive proof has two steps: 1 Base case: Prove that P(1) is true 2 Inductive step: Prove n Z + P(n) P(n + 1) Induction says if you can prove (1) and (2), you can conclude: x Z + P(x) Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 19/37
Inductive Hypothesis In the inductive step, need to show: n Z + P(n) P(n + 1) To prove this, we assume P(n) holds, and based on this assumption, prove P(n + 1). The assumption that P(n) holds is called the inductive hypothesis Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 20/37
Example 1 Prove the following statement by induction: n n Z + (n)(n + 1), i = 2 i=1 Base case: n = 1. In this case, 1 the base case holds. i=1 i = 1 and (1)(1+1) 2 = 1; thus, Inductive step: By the inductive hypothesis, we assume P(k): k i = i=1 Now, we want to show P(k + 1): k+1 i = i=1 (k)(k + 1) 2 (k + 1)(k + 2) 2 Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 21/37
Example 1, cont. First, observe: k+1 i = i=1 k i + (k + 1) i=1 By the inductive hypothesis, k i=1 i = (k)(k+1) 2 ; thus: k+1 i = i=1 (k)(k + 1) 2 + (k + 1) Rewrite left hand side as: k+1 i = k 2 + 3k + 2 2 i=1 = (k + 1)(k + 2) 2 Since we proved both base case and inductive step, property holds. Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 22/37
Example 2 Prove the following statement for all non-negative integers n: n 2 i = 2 n+1 1 i=0 Since need to show for all n 0, base case is P(0), not P(1)! Base case (n = 0): 2 0 = 1 = 2 1 1 Inductive step: By the inductive hypothesis, we assume P(k): k 2 i = 2 k+1 1 i=0 Now, we want to show P(k + 1): k+1 2 i = 2 (k+1)+1 1 = 2 k+2 1 i=0 Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 23/37
Example 2, cont. Thus, we have: k+1 k 2 i = 2 i + 2 k+1 i=0 i=0 By the inductive hypothesis, we have: k+1 2 i = 2 k+1 1 + 2 k+1 i=0 Rewrite as: k+1 2 i = 2 2 k+1 1+ = 2 k+2 1 i=0 Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 24/37
Example 3 Prove that n < 2 n for all nonnegative integers n Base case (n=0): 0 < 2 0 Inductive step: Need to show k + 1 < 2 k+1 assuming k < 2 k From the inductive hypothesis, we know k + 1 < 2 k + 1 Since 1 2 k for k Z +, this implies k + 1 < 2 k + 2 k Thus, k + 1 < 2 2 k = 2 k+1 Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 25/37
Example 4 Prove that 2 n < n! for all integers n 4 Base case (n=4): 2 4 = 16 < 24 = 4! Inductive step: By the inductive hypothesis, we know 2 k < k!, and we need to show that 2 k+1 < (k + 1)! By induction hypothesis, we have 2 k+1 = 2 2 k < 2 k! Since k 4, 2 < k + 1, therefore: 2 k+1 < 2 k! < (k + 1)k! = (k + 1)! By induction, we conclude that 2 n < n!, n 4. Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 26/37
Example 5 Prove that 3 (n 3 n) for all positive integers n. Base case (n=1): 3 (0 = 1 3 1) Inductive step: Show 3 (k + 1) 3 (k + 1) assuming 3 k 3 k First, rewrite (k + 1) 3 (k + 1) as k 3 + 3k 2 + 3k + 1 k 1 = (k 3 k) + 3(k 2 + k) By the inductive hypothesis, 3 (k 3 k); and also 3 3(k 2 + k) By Thm from earlier, it follows that 3 (k 3 k + 3(k 2 + k)) Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 27/37
Example 6 We can also use induction to prove results about sets. Use induction to prove generalized DeMorgan s law for sets: n n A j = A j for n 2 j =1 j =1 We ll prove this by induction on the number of sets Base case (n=2): A 1 A 2 = A 1 A 2 We already proved this earlier! Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 28/37
Example 6, cont. Inductive step: P(k) P(k + 1) k+1 j =1 ( k ) A j = A j A k+1 j =1 Now, using De Morgan s law, rewrite left hand side: k+1 j =1 ( k ) A j = A j A k+1 j =1 From inductive hypothesis, we have ( k j =1 A j ) = k j =1 A j, thus: k+1 j =1 A j = k A j A k+1 = j =1 k+1 j =1 A j Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 29/37
Correctness of Induction Why is induction a valid proof technique? Suppose we can prove the base case and inductive step, but n P(n) does not hold for positive integers. There must be a least element k for which P(k) doesn t hold. Two possibilities: Either (i) k = 1 or (ii) k 2 (i) k cannot be 1 because we proved P(1) in base case (ii) Since k is the least element, we know P(k 1) holds But, in the inductive step we proved P(k 1) P(k); thus, P(k) must also hold! Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 30/37
Section 5.2 Strong Induction Slight variation on the inductive proof technique is strong induction Regular and strong induction only differ in the inductive step Regular induction: assume P(k) holds and prove P(k + 1) Strong induction: assume P(1), P(2),.., P(k); prove P(k + 1) Regular induction and strong induction are equivalent, but strong induction can sometimes make proofs easier Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 31/37
Motivation for Strong Induction Prove that if n is an integer greater than 1, then it is either a prime or can be written as the product of primes. Let s first try to prove the property using regular induction. Base case (n=2): Since 2 is a prime number, P(2) holds. Inductive step: Assume k is either a prime or the product of primes. But this doesn t really help us prove the property about k + 1! Claim is proven much easier using strong induction! Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 32/37
Proof Using Strong Induction Prove that if n is an integer greater than 1, then it is either a prime or can be written as the product of primes. Base case: same as before. Inductive step: Assume each of 2, 3,..., k is either prime or product of primes. Now, we want to prove the same thing about k + 1 Two cases: k is either (i) prime or (ii) composite If it is prime, property holds. Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 33/37
Proof, cont. If composite, k + 1 can be written as pq where 2 p, q k By the IH, p, q are either primes or product of primes. Thus, k + 1 can also be written as product of primes Observe: Much easier to prove this property using strong induction! Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 34/37
A Word about Base Cases In all examples so far, we had only one base case i.e., only proved the base case for one integer In some inductive proofs, there may be multiple base cases i.e., prove base case for the first k numbers In the latter case, inductive step only needs to consider numbers greater than k Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 35/37
Example Prove that every integer n 12 can be written as n = 4a + 5b for some non-negative integers a, b. Proof by strong induction on n and consider 4 base cases Base case 1 (n=12): 12 = 3 4 + 0 5 Base case 2 (n=13): 13 = 2 4 + 1 5 Base case 3 (n=14): 14 = 1 4 + 2 5 Base case 4 (n=15): 15 = 0 4 + 3 5 Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 36/37
Example, cont. Prove that every integer n 12 can be written as n = 4a + 5b for some non-negative integers a, b. Inductive hypothesis: Suppose every 12 i k can be written as i = 4a + 5b. Inductive step: We want to show k + 1 can also be written this way for k + 1 16 Observe: k + 1 = (k 3) + 4 By the inductive hypothesis, k 3 = 4a + 5b for some a, b because k 3 12 But then, k + 1 can be written as 4(a + 1) + 5b Instructor: Dr. Blerina Xhabli, University of Houston Math.3336: Discrete Mathematics Mathematical Induction 37/37