Natural Numbers 1 Natural Numbers The set N

Natural Numbers 1

Natural Numbers � The set N = {1, 2, 3, 4, ……. . } is known as natural numbers or the set of positive integers � The natural numbers are used mainly for : � counting � ordering and � defining other concepts like generating pseudorandom numbers, assigning memory location to files, encrypting & decrypting messages, etc. 2

Basic Properties for Integers � Associative law (a + b) + c = a + (b + c) (a * b) * c = a * (b * c) � Commutative law a + b = b + c a * b = b * a � Distributive Law a * (b + c) = a * b + a * c � Additive identity a + 0 = 0 + a = a � Multiplicative identity a * 1 = 1 * a = a � Additive inverse a + (-a) = (-a) + a = 0 3

Prime A +ve integer greater than 1 that has no +ve divisor 1 and the number itself. Composite A +ve integer that has atleast one +ve divisor other than 1 & the number itself or which is not prime Fundamental Theorem of Arithmetic Every +ve integer n > 1 can uniquely be written as product of prime numbers. Find the prime factorization of 100, 999, 1024 4

Representation of Integers Let b be a +ve integer > 1. Then any +ve integer n can be uniquely expressed as n = ak bk + ak-1 bk-1 + …. . + a 1 b + a 0 where k is a non-negative integer, a 0, a 1, a 2, …, ak are nonnegative integers b, and ak ≠ 0 If n = 351, b = 2 351 = 1. 28 + 0. 27 + 1. 26 + 0. 25 + 1. 24 + 1. 23 + 1. 22 + 1. 21 + 1. 20 (351)10 = (101011111)2 5

Modular Arithmetic If a and b are integers and m is a +ve integer, then a is congruent to b modulo m a b (mod m) if m divides a - b � a b (mod m) iff a mod m = b mod m Determine whether 17 is congruent to 5 modulo 6 and whether 24 and 14 are congruent modulo 6. 2) List five integers that are congruent to 4 modulo 12 3) Decide whether each of the integers 80, 103, -29, -122 is congruent to 5 modulo 17 1) 6

Hashing Function � For direct access in file handling, the program is supplied with a key. Using this key, the program has to locate the required record of information. � Let K be set of keys and A be set of physical addresses. A function h : K A is called hash function if h(k) = k mod m where k K and m is the number of memory locations. �A hashing function h assigns memory location h(k) to the record that has k as its key. 7

Example � If m=111, � h(037149212) = 037149212 mod 111 = 65 � h(064212848) = 064212848 mod 111 = 14 � h(107405723) = 107405723 mod 111 = 14 This situation creates a collision. To remove collision, there are two methods: � Linear Probe � Chaining 8

Mathematical Induction is a method of mathematical proof typically used to establish a given statement for all natural numbers. Principle of Mathematical Induction Let P(n) be a statement about a natural number n N that is either true or false. The purpose of induction is to show that P(n) is true for all n N. 9

Steps of Induction �Basis Step is to prove the given statement for the first natural number. �Inductive Step is to prove the given statement for any one natural number implies the given statement for the next natural number. 10

Mathematical Induction �Basis Step: P(1) is true �Inductive Step: Assume that P(k) is true for any k N then prove that P(k+1) is true By principle of mathematical induction, P(n) is true for all n N 11

Example: Use mathematical induction to prove Sn = 2 + 4 + 6 + 8 +. . . + 2 n = n(n + 1) for every positive integer n. 1. Show that the formula is true when n = 1. S 1 = n(n + 1) = 1(1 + 1) = 2 True 2. Assume the formula is valid for some integer k. Use this assumption to prove the formula is valid for the next integer, k + 1 and show that the formula Sk + 1 = (k + 1)(k + 2) is true. Sk = 2 + 4 + 6 + 8 +. . . + 2 k = k(k + 1) Assumption 12

� Example continued: � Sk + 1 = 2 + 4 + 6 + 8 +. . . + 2 k + [2(k + 1)] = 2 + 4 + 6 + 8 +. . . + 2 k + (2 k + 2) = Sk + (2 k + 2) Group terms to form Sk. = k(k + 1) + (2 k + 2) Replace Sk by k(k + 1). = k 2 + k + 2 Simplify. = k 2 + 3 k + 2 = (k + 1)(k + 2) = (k + 1)((k + 1)+1) � The formula Sn = n(n + 1) is valid for all positive integer values of n. 13

Example: Use mathematical induction to prove for all positive integers n, True Assumption Group terms to form Sk. Replace Sk by k(k + 1). 14

Example continued: Simplify. � The formula is valid for all positive integer values of n. 15

Prove by Induction (Sums of Powers of Integers) : 16

� Second Variants : Suppose that b N and that we can prove these two statements: Base Case: P(k) is true for 0≤k≤b. Inductive Step: If P(k) is true for some k≥b, then P(k+1) is also true. Then, P(n) is true for all n N. � Third Variants (Strong Induction): Suppose that b N and we can prove two statements: Base Case: P(k) is true for 0≤k≤b. Inductive Step: If k≥b and P(i) is true for all i≤k, then P(k+1) is also true. Then, P(n) is true for all n N. 17

Principle of Strong Mathematical Induction �Let P(n) be statement involving positive integer n=1, 2, 3, … then �Step 1: Verify P(1) is true. (Basis Step) �Step 2: Assume that P(1), P(2), …, P(k) is true (Strong Inductive Hypothesis) �Step 3: Verify that P(k+1) is true using strong inductive hypothesis. (Inductive Step) 18

�A chocolate bar consists of a number of squares (say, n>0) arranged in a rectangular pattern. You split the bar into small squares always breaking along the lines between the squares. Prove that minimum number of breaks it takes is n-1. � Let P(n) denote the number of breaks needed to split a bar with n squares. � Base Step: P(1)=0 is true. � Inductive Step: Assume that P(k) is true for 2≤k≤n. � To prove that P(k+1)=k under the above assumption. 19

� Break the bar into two pieces of sizes n 1 and n 2, so that n 1+n 2=k+1. � By inductive hypothesis P(n 1) = n 1 -1 P(n 2) = n 2 -1 � Hence the total number of breaks is 1+(n 1 -1)+(n 2 -1) = k � Hence P(n) holds for all n>0. 20

Induction with Nonzero Base Cases � Sometimes we want to prove that some property P holds for all integers n ≥ b. � Inductive Argument: P(b) : Show that property P is true for b P(k) ⇨ P(k+1) : Show that if property P is true for k, then it’s true for k+1. � We can conclude that P(n) holds for all n ≥ b. � We don’t care about n < b. 21

� Example: Prove using strong induction that every amount of postage of 8 cents or more can be formed using just 3 -cent and 5 -cent stamps. � Let P(n) be the proposition that postage of n>8 cents can be formed using 3 -cent and 5 -cent stamps. � BASIS STEP: 8=3+5 � P(8) uses one 3 -cent and one 5 -cent stamp. � P(8) is true. 22

� INDUCTIVE STEP: Suppose it’s true for k. � There are two cases: � (1) If used a 5 -cent stamp to make k, replace it by two 3 -cent stamps. Get k+1. � (2) If did not use a 5 -cent stamp to make k, must have used at least three 3 -cent stamps. Replace three 3 -cent stamps by two 5 -cent stamps. Get k+1. � Hence, P(n) holds for all n ≥ 8. 23

� Example: Use strong mathematical induction to show that if n is an integer > 1, then n can be written as the product of primes. � Let P(n) be the proposition that n can be written as a product of primes. � BASIS STEP: P(2) is true since 2 itself is prime. � INDUCTIVE STEP: The inductive hypothesis P(j) is true for all integers j with 2 ≤ j ≤ k. � To show that P(k + 1) must be true under this assumption, two cases arise � Case– 1: If k + 1 is prime, then P(k + 1) is true. 24

� Case-2: If k + 1 is composite � k+1 = a. b where a & b are +ve integers with 2≤a≤ b<k+1 By the inductive hypothesis, a & b can be written as the product of primes k + 1 can also be written as the product of those primes. � Hence, by strong mathematical induction, every integer > 1 can be written as the product of primes. 25