Natural Numbers

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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.

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Natural Numbers

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

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Basic Properties for Integers

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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

Let b be a +ve integer > 1. Then any +ve integer n can be uniquely expressed as

n = ak bk + ak-1 bk-1 + ….. + a1 b + a0

where k is a non-negative integer, a0, a1, a2,…, ak are nonnegative integers b, and ak ≠ 0

If n = 351, b = 2351 = 1 . 28 + 0 . 27 + 1 . 26 + 0 . 25 + 1 . 24 + 1 . 23 +

1 . 22 + 1 . 21 + 1 . 20 (351)10 = (101011111)2

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Representation of Integers

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

1) 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 123) Decide whether each of the integers 80, 103, -29, -122 is congruent to 5 modulo 17

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Modular Arithmetic

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 : KA 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.

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Hashing Function

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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

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.

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Mathematical 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.

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Steps of 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

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Mathematical Induction

Example:

Use mathematical induction to prove

Sn = 2 + 4 + 6 + 8 + . . . + 2n = n(n + 1)

for every positive integer n.

1.Show that the formula is true when n = 1.

S1 = 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 + . . . + 2k = k(k + 1) Assumption

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Example continued: Sk + 1 = 2 + 4 + 6 + 8 + . . . + 2k + [2(k + 1)]

= 2 + 4 + 6 + 8 + . . . + 2k + (2k + 2) = Sk + (2k + 2) Group terms to form Sk.

= k(k + 1) + (2k + 2) Replace Sk by k(k + 1).

= k2 + k + 2k + 2 Simplify. = k2 + 3k + 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.

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Example: Use mathematical induction to prove for all positive integers n,

Assumption

True

Group terms to form Sk.

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2 2 2 2 2 2

1

( 1)(2 1)1 2 3 4 .6

n

i

n n ni n

1( 1)(2( ) 1) 1(2)(2 1) 6 1

611

6 61S

2 2 2 2 2 ( 1)(2 1)1 2 3 46k kS k k k

2 2 21

2 2 21 2 3 4 ( 1)k k kS

2( 1)kS k 2 2 1kS k k

2( 1)(2 1) 2 16

k k k k k Replace Sk by k(k + 1).

The formula is valid for all positive integer values of n.

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( 1)(2 1)6n

n n nS

Example continued:

3 2 22 3 6 12 66 6

k k k k k 3 22 9 13 6

6k k k

2( 3 2)(2 3)6

k k k

( 1)( 2)(2 3)6

k k k

( )[( ) 1][2(1 ) ]6

1 11k k k

Simplify.

Prove by Induction (Sums of Powers of Integers) :

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1

( 1) 1 2 3 412

.n

i

n ni n

2 2 2 2 2 2

1

( 1)(2 1) 1 2 3 42.6

n

i

n n ni n

2 23 3 3 3 3 3

1

( 1) 1 2 3 43.4

n

i

n ni n

24 4 4 4 4 4

1

( 1)(2 1)(3 3 1) 1 2 3 44.30

n

i

n n n n ni n

2 2 25 5 5 5 5 5

1

( 1) (2 2 1) 1 2 3 41

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.n

i

n n n ni n

Second Variants : Suppose that bN 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 nN.

Third Variants (Strong Induction): Suppose that bN 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 nN.

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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)

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Principle of Strong Mathematical Induction

)1P()P(...)3P()2P()1P( kk

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.

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Break the bar into two pieces of sizes n1 and n2, so that n1+n2=k+1.

By inductive hypothesis P(n1) = n1-1 P(n2) = n2-1 Hence the total number of breaks is 1+(n1-1)+(n2-1) = k Hence P(n) holds for all n>0.

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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.

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Induction with Nonzero Base Cases

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.

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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.

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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.

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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.

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