STUDY MATERIAL FOR BSC MATHEMATICS DISCRETE …

STUDY MATERIAL FOR BSC MATHEMATICS DISCRETE MATHEMATICS

SEMESTER - V, ACADEMIC YEAR 2020 - 21

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UNIT CONTENT PAGE Nr

I MATHEMATICS LOGIC STATEMENT AND NOTATION 02

II NORMAL FORMS 31

III GROUP AND MONOIDS 39

IV LATTICES AND BOOLEAN ALGEBRA 76

V BINARY NUMBER SYSTEM 95



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UNIT - I MATHEMATICS LOGIC STATEMENT AND NOTATION

Definition: (Statement) A statement is a declarative sentence that is either true or false but not both. Example

1. Chennai is the capital of Tamil Nadu---It is a statement

2. ---It is a statement

3. ---it is a statement

4. Delhi is in America---=It is a statement

5. Canada is a country---It is a statement

6. This statement is false---It is not a statement

7. 1+101=110---It is a statement

8. Closed the door ---It is not a statement

Notation:

1. The statements will be denoted by distinct symbols selected from the capital letter

A,B,C,….

2. T is used to denote true statement

3. F is used to denote false statement

Atomic Statement:

Statements which cannot be further spit into simpler sentence are called atomic statements

(Primary statements or Primitive statements)

Example

Rama is a boy

Connectives

Five basic connectives

Sr English Language

usage

Logical connectives Types of

operator

Symbols

1. and Conjunction Binary ∧

2. or Disjunction Binary ∨

3. not Negation or denial unary ¬

4. if … then Implication or Binary



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conditional

5. if and only if Dis conditional Binary

Note:

Atomic statement do not contains connectives

Definition: (Truth Table)

A table, given a truth values of a compound statement interms of it is component parts is

called a truth table.

Compound statement is a statement which can be formed from atomic statements through

the use of connectives such as and, but, or, etc.

Definition: (Negation ¬)

The negation of a statement is generally formed by introducing the word ‘not’ at a proper

place in the statement.

Notation:

If P denote a statement, then the negation of P is written as ‘¬P’ and read as ‘not P’

Truth table:

P ¬P

T F

F T

Definition: (Conjunction) – (∧, Cap, And)

The conjunction of two statements P and Q is the statement P∧Q which is read ad (‘P and

Q’)

Truth table:

P Q P∧Q

T T T

T F F

F T F

F F F

The statement P∧Q has the truth value T whenever both P&Q have the truth value T;



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otherwise it has truth value F.

Definition: (Disjunction) – (∨, Cup, Or)

The disjunction of two statements P and Q is the statement P∨Q which is read as ‘P or Q’

Truth table:

P Q P∨Q

T T T

T F T

F T T

F F F

the statement P∨Q has the truth value F when ever both P and Q have the truth value F;

Otherwise it has truth value T.

Problem 1.

Let P : I went to my school yesterday write the statement ¬P

Solution:

Given that P : I went to my school yesterday

Then ¬P : I did not go to my school yesterday.

Problem 2.

Let P: London is a city. What is ¬P?

Solution

Given P : London is a city

Then ¬P : London is not a city or

¬P : It is not the case that London is a city

Problem 3:

Let P : Today is Monday

Solution:

Given P : Today is Monday

Then ¬P : to is not Monday

Problem 4:

Let

Solution:



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Given that

Then

Problem 5:

From the conjunction of P: It is raining today Q: There are 20tables in this classroom

Solution:

: It is raining today and there are 20 tables in this classroom

Problem 6:

P: It is snowing Q: I am cold

Solution:

: It is snowing and I am cold

Problem 7:

Solution:

: and

Problem 8:

Solution:

: and

Problem 9:

4

Solution:

: and 4

Problem 10:

Let P : 2 is a positive integer

And Q: is a rational integer

What is ?

Solution:

is a positive integer or is a rational integer

Problem 11.:

Let P:

Solution:



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or

Problem 12:

London is the capital of India

Solution:

or London is the capital of India

Exercise problems:

Problem 1:

Using the statement

R: Mark is rich

H: Mark is happy

Write the following statement in symbolic form

a) Mark is poor / but happy

b) Mark is rich or unhappy

c) Mark is neither rich or happy

d) Mark is poor or he is both rich and unhappy

Solution

a) In symbolic form, the given statement is ¬R∧H

b) R∨(¬H)

c) ¬R∧¬H

d) (¬R)∨(R∧(¬H))

Problem 2:

Construct the truth table for the statement formula

i) P∨¬Q ii) P∧¬P iii) (P∨Q)∨( ¬P)

Solution:

i) P∨¬Q

P Q ¬Q P∨¬Q

T T F T

T F T T



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F T F F

F F T T

ii) P∧¬P

P ¬P P∧¬P

T F F

F T F

iii) (P∨Q)∨( ¬P)

P Q P∨Q ¬P (P∨Q)∨( ¬P)

T T T F T

T F T F T

F T T T T

F F F T T

Exercise Problem:

Construct the truth table for the following formula.

a) ¬(¬P∨¬Q) b) ¬(¬P∧¬Q) c) P∧(P∨Q) d) P∧(Q∨P)

e) (¬P∧(¬Q∧R))∨(Q∧R)∨(P∧R) f) (P∧Q)∨( ¬P∧Q)∨(P∧¬Q)∨(¬P∧¬Q)

a) Truth table for ¬(¬P∨¬Q)

P Q ¬P ¬Q ¬P ∨¬Q ¬(¬P∨¬Q)

T T F F F T

T F F T T F

F T T F T F

F F T T T F

b) Truth table for ¬(¬P∧¬Q)

P Q ¬P ¬Q ¬P ∧¬Q ¬(¬P∧¬Q)

T T F F F T



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T F F T F T

F T T F F T

F F T T T F

c) Truth table for P∧(P∨Q)

P Q P∨Q P∧(P∨Q)

T T T T

T F T T

F T T F

F F F F

d) Truth table for P∧(Q∨P)

P Q Q∨P P∧( Q∨P)

T T T T

T F F T

F T F F

F F F F

e) Truth table for (¬P∧(¬Q∧R))∨(Q∧R)∨(P∧R)

P Q R ¬P ¬Q ¬Q∧R ¬P∧(¬Q∧R) Q∧R P∧R (Q∧R)∨(P∧R) (¬P∧(¬Q∧R))∨

(Q∧R)∨(P∧R)

T T T F F F F T T T T

T T F F F F F F F F F

T F T F T T F F T F T

T F F F T F F F F F F

F T T T F F F T F T T

F T F T F F F F F F F

F F T T T T T F F T T

F F F T T F F F F F F



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f) Truth table for (P∧Q)∨( ¬P∧Q)∨(P∧¬Q)∨(¬P∧¬Q)

P Q ¬P ¬Q P

∧Q

¬P

∧Q

P

∧¬Q

¬P

∧¬Q

(P∧Q)∨(

¬P∧Q)

(P∧¬Q)∨(¬P∧¬Q) (P∧Q)∨( ¬P∧Q)∨

(P∧¬Q)∨(¬P∧¬Q)

T T F F T F F F T F T

T F F T F F T F F T T

F T T F F T F F T F T

F F T T F F F T F T T

Example

Given the truth values of P and Q as T and those of R and S as F, find the truth values of the

following

a) P∨(Q∧R)

b) (P∧(Q∧R))∨ ¬((P∨Q)∧(R∨S))

c) (¬(P∧Q)∨ ¬R)∨(((¬P∧Q)∨¬R)∧S

Solution:

Given that the truth value of P is T

The truth value of Q is T

The truth value of R is F

The truth value of S is F

a) To find the truth value of P∨(Q∧R)

P Q R S Q∧R P∨(Q∧R) P∧(Q∧R) P∨Q R∨S (P∨Q)∧

(R∨S)=A

¬A (P∧(Q∧R))∨

¬A

T T F F F T F T F F T T

P Q R S ¬P ¬R P∧Q ¬(P∧Q

)

(¬P∧Q) (¬P∧Q)∨

¬R=B

((¬P∧Q)∨

¬R)=C

C∧S B∨

(C∧S)

T T F F F T T F F T T F T

a) The truth value of P∨(Q∧R) is T



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b) The truth value of (P∧(Q∧R))∨ ¬((P∨Q)∧(R∨S)) is T

c) The truth value of (¬(P∧Q)∨ ¬R)∨(((¬P∧Q)∨¬R)∧S is T

Conditional Statement:

If P and Q are any two statements, then the statement which read as “If P, then Q” is

called a conditional statement

The statement has a truth value F when Q has the truth value F and P has the truth

value T, otherwise it has the truth value T.

The truth table for is given below

P Q

T T T

T F F

F T T

F F T

Example 1:

Express in English the statement where P: The sun is shining today

Q: 2+7>4

Solution:

Given that P: The sun is shining today

Q: 2+7>4

: If the sun is shining today, then 2+7>4

Example 2:

Write the following statement in the symbolic form

P: If either Jerry takes calculus or Ken takes Sociology, then Larry will take English

Solution:

Let J: Jerry takes calculus

K: Ken takes Sociology

L: Larry will take English

Then, the given statement can be written as (J∨K) L is symbolic form

Example 3:

Write in symbolic form the statement the crop will be destroyed if there is a flood



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Solution:

Let F: There is a flood

C: The crop will be destroyed

Then, the given statement can be written as F C in symbolic form

Note 1:

In P Q

i) The statement P is called the antecedent

ii) The statement Q is called the consequent

Note 2:

i) Q is necessary for P

ii) P is sufficient for Q

iii) Q if P

iv) P only if Q

v) P implies Q – are represented by P Q

Construct a truth table for

(P Q)∧(Q P)

Problem :

Write the following statement in symbolic form

You cannot ride the riller coaster if you are under four feet tall unless you are older than 16

years old

Solution:

Let P: You cannot ride the riller coaster

Q: you are under four feet tall

R: you are older than 16

Then the given statement can be written as (Q∨¬R) P

Bi-conditional statement ( if and only if)

Let P and Q be any two statements. Then the statement P Q which is read as “P if and only

if Q” and abbreviated as “P if Q” is called biconditional statements

The statement P Q has the truth value T whenever both P and Q have identical truth

values



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The truth table for is given below

P Q

T T T

T F F

F T F

F F T

Example 1:

Let P: You can take the flight

Q: You buy a ticket

Find

Solution:

Given that P : You can take the flight

Q: You buy a ticket

: You can take the flight if and only if you buy a ticket

Example 2:

Let P: You cannot take the flight

Q: You do not buy a ticket

: You cannot take the flight if and only if you do not buy a ticket

Problem 1:

Construct the truth table for and ∧

Solution:

P Q ∧

T T T T T T

T F F F T F

F T F T F F

F F T T T T

Hence the truth values of and ∧ are identical

Problem 2:

Construct a truth table for ¬(P∧Q) ( ¬P∨¬Q)



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P Q P ∧Q ¬ (P ∧Q) ¬P ¬Q ¬P ∨¬Q ¬(P∧Q)↔( ¬P∨¬Q)

T T T F F F F T

T F F T F T T T

F T F T T F T T

F F F T T T T T

Exercise problem 1:

Show that the truth values of the following formulas are independent of their components

a) (P∧(P Q)) Q

P Q P ∧ (P∧(P→Q))→Q

T T T T T

T F F F T

F T T F T

F F T F T

The given formula is independent of their components

b) (P Q) (¬P∨Q)

P Q ¬P (¬P∨Q) (P→Q) ↔(¬P∨Q)

T T T F T T

T F F F F T

F T T T T T

F F T T T T


c) ((P Q)∧(Q R)) (P R)

P Q R P→Q Q→R P→R ((P→Q)∧(Q→R)) ((P→Q)∧(Q→R)) →(P→R)

T T T T T T T T

T T F T F F F T



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T F T F T T F T

T F F F T F F T

F T T T T T T T

F T F T F T F T

F F T T T T T T

F F F T T T T T


d) (P Q) ((P∧Q)∨(¬P∧¬Q))

P Q ¬P ¬Q P ∧Q ¬P ∧¬Q (P∧Q)∨(¬P∧¬Q) (P↔Q)

↔((P∧Q)∨(¬P∧¬Q))

T T F F T F T T T

T F F T F F F F T

F T T F F F F F T

F F T T F T T T T

Construct the truth tables for the following formulas

a) (Q∧(P Q)) P

P Q Q∧(P→Q) (Q∧(P→Q)) →P

T T T T T

T F F F T

F T T T F

F F T F T

b) ¬(P∨(Q∧R)) ((P∨Q)∧(P∨R))

P Q R Q∧R P∨(Q∧R) ¬(P∨(Q∧R)) (P∨Q) (P∨R) (P∨Q)∧

(P∨R)

¬(P∨(Q∧R))

↔((P∨Q)∧(P∨R))



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T T T T T F T T T F

T T F F T F T T T F

T F T F T F T T T F

T F F F T F T T T F

F T T T T F T T T F

F T F F F T T F F F

F F T F F T F T F F

F F F F F T F F F F

Problem 3:

A connective denoted by is defined by the following table find a formula using P/Q and

the connectives ∧/∨ and ¬ whose truth values are identical to the truth values of P▽Q

Solution:

P Q P▽Q

T T F

T F T

F T T

F F F

P Q P▽Q P ∧Q P ∨Q ¬

T T F T T T T F

T F T F T F F T

F T T F T T F T

F F F F F T T F

The formula ¬ id identical with P▽Q

Given the truth values of P,Q are ¬ and these R and S as F. Find the truth values of the

following

a) ¬[(P∧Q)∨ ¬R]∨[(Q ¬P) (R∨¬S)]



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A=¬[(P∧Q)∨ ¬R]

B=[(Q ¬P) (R∨¬S)]

P Q R S R∧Q ¬R (P∧Q)∨

¬R

A (Q ¬P) ¬S R∨¬S B A∨B

T T F F T T T F F T T T T

b) (P R)∧(¬Q S)

P Q R S P↔R ¬Q ¬Q→S (P↔R)∧(¬Q→S)

T T F F F F T F

c) [P∨(Q (R∧¬P))] (Q∨¬S)

Where A = [P∨(Q (R∧¬P))]

B = (Q∨¬S)

P Q R S R∧¬P Q→(R∧¬P) A Q∨¬S A B

T T F F F T T T T

Well – formed formula (w.f.f)

A well formed formula can be generated by the following rules

1. A statement variables standing alone is a well formed formula

2. If A is a w.f.f, then ¬A is a w.f.f

3. If A and B are w.f.f, then (A∨B), (A∧B), (A B), law (A B) are w.f.f

Example 1:

¬(P∧Q), ¬(P∧Q), (P (P∧Q)), (P (P∨Q)), (((P Q)∧(Q R)) (P R)) are w.f.formulas the

following are nor w.f.f

¬P∧Q is not a w.f.f but ¬(P∧Q) is a w.f.f and is a w.f.f

(P Q) (∧Q) is not a w.f.f since ∧ is a binary operator but a unary operator, (∧Q is

not valid)

P Q - it is not a w.f.f because closing parathesis is missing



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(P∧Q) Q - it is not a w.f.f because closing parathesis is missing in the beginning

Tautologies

A statement that is true for all possible values of its propositional variables is called a

tautology or a logical truth

Contradiction

A statement that is always false is called a contradiction

Note:

i) The negation of a contradiction is a tautology

ii) A propositional function that is neither a tautology nor a contradiction is called a

contingency

iii)

Tautology Contradiction Fallacy

In the result column all the

entries are true (T)

In the result column, all the

entries are false (F)

In the result column, any one

entry is false (F)

Problem 1:

Show that

i) P∨¬P is a tautology

ii) P∧¬P is a contradiction

P ¬P P∨¬P P∧¬P

T F T F

F T T F

Therefore P∨¬P is a tautology & P∧¬P is a contradiction

Problem 2:

Show that Q∨(P∧¬Q)∨( ¬P∧¬Q) is a tautology

Proof:

Let us form a truth table

P Q ¬P ¬Q P ∧¬Q ¬P ∧¬Q Q∨(P∧¬Q) Q∨(P∧¬Q)∨( ¬P∧¬Q)

T T F F F F T T



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T F F T T F T T

F T T F F F T T

F F T T F T F T

The given proposition in the result column, all the value are true and so the given

proposition is a tautology

Problem 3:

Using the truth table, verify that whether the proposition (P∧Q)∧¬(P∨Q) is a tautology or

not

Proof:


P Q P∧Q P∨Q ¬ (P∨Q) (P∧Q)∧¬(P∨Q)

T T T T F F

T F F T F F

F T F T F F

F F F F T F

In the result column, all the values are false and so the given proposition is not a tautology.

It is a contradiction.

Problem 4:

Show that the proposition (P∨Q) (Q∨P) is a tautology

Proof:

P Q P∨Q Q∨P (P∨Q)↔(Q∨P)

T T T T T

T F T T T

F T T T T

F F T F T

In the result column, all the values are true and so (P∨Q) (Q∨P) is a tautology

Exercise Problem

From the formulas given below select those whice are well formed and indicate which ones



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are tautologies or contradiction

a) (P (P∨Q)) b) ((P ¬P)) ¬P) c) ((¬Q∧P)∧Q)

d) (P (Q R)) ((P Q) )

e) ((¬P Q) (Q

f) ((P∧Q) P)

Solution :

a) Let us form a truth table

P Q P∨Q (P→(P∨Q))

T T T T

T F T T

F T T T

F F T T

In the result column all the values are true and so that the given proposition is a tautologies

b) Let us form a truth table

P ¬P P→(¬P) ((P→(¬P))→¬P)

T F F T

F T T T


c) Let us form a truth table

P Q ¬Q ¬Q ∧P ((¬Q∧P)∧Q)

T T F F F

T F T T F

F T F F F

F F T F F



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In the result column, all the values are false and so the given proposition is not a tautology.

It is a contradiction

d) Let us form a truth table

P Q R Q→R P→Q P→R P→(Q→R) (P→Q)

→(P→R)

P (Q R)) ((P Q) )

T T T T T T T T T

T T F F T F F F T

T F T T F T T T T

T F F T F F T T T

F T T T T T T T T

F T F F T T T T T

F F T T T T T T T

F F F T T T T T T


e) ((¬P Q) (Q

P Q ¬P ¬P→Q Q→P ((¬P→Q) →(Q→P)

T T F T T T

T F F T T T

F T T T F F

F F T F T T

In the result is well formed and so neither tautologies nor contradiction

f) ((P∧Q) P)

P Q P∧Q ((P∧Q) P)

T T T T

T F F F

F T F T



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F F F T

In the result is well formed and so neither tautologies nor contradiction

Equivalence of Formulas

Definition

Let A and B be two statement formulas and let denote all the variables occering

both A and B. Consider an assignment of truth values to and the resulting truth

values of A&B. If the truth value of A is equal to the truth value of B one of the possible

set of truth values assigned to then A&B are said to be equivalent

Example

Show that P is equivalent P∧P, P∨P and ¬¬P to and by using truth table

Solution


P ¬P ¬¬P P∨P P∧P

T F T T T

F T F F F

Then we can say that P is equivalent to P∧P

we can say that P is equivalent to P∨P

we can say that P is equivalent to ¬¬P

Note:

Above ¬¬P is equivalent to P∧P

¬¬P is equivalent to P∨P

P∨P is equivalent to P∧P

Show that

i) (P∧¬P)∨Q is equivalent to Q

ii) (P∨¬P) is equivalent to Q∨¬Q

Proof

Let us form the truth table

P Q ¬P P∧¬P (P∧¬P)∨Q

T T F F T



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T F F F F

F T T F T

F F T F F

Q and (P∧¬P)∨Q have identical truth values & so they are equal

P Q ¬P P∨¬P ¬Q Q∨¬Q

T T F T F T

T F F T T T

F T T T F T

F F T T T T

(P∨¬P) and Q∨¬Q have identical truth values & so they are equal

Problem 1;

Show that P is equivalent to

i) P∨(P∧Q) ii) P∧(P∨Q) iii) (P∧Q)∨ (P∧¬Q) iv) (P∨Q)∧(P∨¬Q)

Proof

i) P∨(P∧Q)


P Q P∧Q P∨(P∧Q)

T T T T

T F F T

F T F F

F F F F

P and P∨(P∧Q) have identical truth value so they are equal

ii) P∧(P∨Q)


P Q P∨Q P∧(P∨Q)

T T T T



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T F T T

F T T F

F F F F

iii) (P∧Q)∨ (P∧¬Q)


P Q ¬Q P∨Q P∨¬Q P∧Q P∧¬Q (P∧Q)∨ (P∧¬Q)

T T F T F T F T

T F T T T F T T

F T F T F F F F

F F T F F F F F

P and (P∧Q)∨ (P∧¬Q) have identical truth values and so they are equal

iv) (P∨Q)∧(P∨¬Q)

P Q ¬Q P∧Q P∧¬Q (P∨Q)∧(P∨¬Q)

T T F T F T

T F T F T T

F T F F F F

F F T F F F

Problem 2.

Show that P Q and ¬P∨Q are logically equivalent

Proof:

Let us form a truth table P Q ⇔ ¬P∨Q

P Q ¬P ¬Q ¬P∨Q

T T F F T T

T F F T F F

F T T F T T

F F T T T T



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P Q and ¬P∨Q have identical truth value and so they are equal

Exercise Problem

Show that following equivalence

a) ¬(P∧Q)⇔ ¬P∨¬Q

b) ¬(P∨Q)⇔ ¬P∧¬Q

c) ¬(P Q)⇔ P∧¬Q

d) ¬(P Q)⇔ (P∧¬Q)∨(¬P∧Q)

Solution

a) ¬(P∧Q)⇔ ¬P∨¬Q


P Q ¬P ¬Q P∧Q ¬ ¬P∨¬Q

T T F F T F F

T F F T F T T

F T T F F T T

F F T T F T T

¬(P∧Q) and ¬P∨¬Q have identical values and so they are equal

b) ¬(P∨Q)⇔ ¬P∧¬Q


P Q ¬P ¬Q P∨Q ¬ ¬P∧¬Q

T T F F T F F

T F F T T F F

F T T F T F F

F F T T F T T

¬(P∨Q) and ¬P∧¬Q have identical values and so they are equal

c) ¬(P Q)⇔ P∧¬Q




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P Q ¬Q P Q ¬ P∧¬Q

T T F T F F

T F T F T T

F T F T F F

F F T T F F

¬(P Q) and P∧¬Q have identical values and so they are equal

d) ¬(P Q)⇔ (P∧¬Q)∨(¬P∧Q)


P Q ¬P ¬Q P∧¬Q ¬P∧Q P Q ¬ (P∧¬Q)∨(¬P∧Q)

T T F F F F T F F

T F F T T F F T T

F T T F F T F T T

F F T T F F T F F

¬(P Q) and (P∧¬Q)∨(¬P∧Q) have identical values and so they are equal

Equivalent formulas:

Idempotent Laws:

i) P∧P ⇔ P

ii) P∨P ⇔P

Associative Laws:

i) (P∨Q)∨R ⇔ P∨(Q∨R)

ii) (P∧Q)∧R ⇔ P∧(Q∧R)

Commutative Laws:

i) P∨Q ⇔ Q∨P

ii) P∧Q ⇔ Q∧P

Distributive Laws:

i) P∨(Q∧R) ⇔ (P∨Q)∧(P∨R)

ii) P∧(Q∨R) ⇔ (P∧Q)∨(P∧R)



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Domination Laws:

i) P∨F⇔P

ii) P∧T ⇔ P

iii) P∨T ⇔ T

iv) P∧F ⇔ F

v) P∨¬P ⇔T

vi) P∧¬P ⇔ F

Absorption Laws:

i) P∧(P∨Q) ⇔ P

ii) P∨(P∧Q) ⇔ P

De Morgan’s Laws:

i) ¬(P∧Q) ⇔¬P∨¬Q

ii) ¬(P∨Q) ⇔¬P∧¬Q

Problem

Show that P (Q R)⇔P (¬Q∨R)⇔(P∧Q) R without constructing truth table

Proof:

P (Q R)⇔P (¬Q∨R) (by Q R⇔¬Q∨R)

⇔¬P (¬Q∨R) (by P Q⇔¬P∨Q)

⇔(¬P ¬Q)∨R (by associative law)

⇔¬(P Q)∨R (by Demorgan’s law)

⇔(P∧Q) R (by P Q⇔¬P∨Q)

Hence P (Q R)⇔P (¬Q∨R)⇔(P∧Q) R

Duality Law

True formulas A and A* are said to be duals of each other if either one can be obtained from

the other by replacing ∧ by ∨ and ∨ by ∧

The connectives ∧ and ∨ are called duate of each other

If the formula A containing the variables T or F, then A*, its dual is obtained by replacing T

by F and F by T in addition to the above mentioned interchanges

Example

Write the duals of a) (P∨Q)∧R b) (P∧Q)∨T c) ¬(P∨Q)∧(P∨¬(Q∧¬S))



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Solution:

a) The dual of (P∨Q)∧R is (P∧Q)∨R

b) The dual of (P∧Q)∨T is (P∨Q)∧F

c) The dual of ¬(P∨Q)∧(P∨¬(Q∧¬S)) is ¬(P∧Q)∨(P∧¬(Q∨¬S))

Problem

Show that (¬P∧(¬Q∧R))∨(Q∧R)∨(P∧R)⇔R with out constructing truth table

Solution:

(¬P∧(¬Q∧R))∨(Q∧R)∨(P∧R)

⇔((¬P∧¬Q)∧R)∨((Q∨P)∧R) (by associative and distributive law)

⇔[(¬P∧¬Q)∨(P∨Q)]∧R (by distributive law)

⇔[¬(P∨Q)∨(P∨Q)]∧R (by De morgan’s law)

⇔T∧R

⇔R

Tautological implications

For any statement formula P Q the statement formula i) Q P called its converse ii)

¬P ¬Q is called its inverse iii) ¬Q ¬P is called its contrapositive

Definition:

A statement A is said to tautologically imply a statement B if and only id A B is a tautologh.

This idea will be denoted by A⇒B

Example 1:

Show that P Q⇔¬Q ¬P

Proof


P Q P Q ¬P ¬Q ¬Q→¬P

T T T F F T

T F F F T F

F T T T F T

F F T T T T

P Q⇔¬Q ¬P

Example 2:



Page 28 of 119

Show that Q P⇔¬P ¬Q

Proof


P Q Q P ¬P ¬Q ¬P→¬Q

T T T F F T

T F T F T T

F T F T F F

F F T T T T

Q P⇔¬P ¬Q

Show that following implecation

a) (P∧Q)⇒(P Q)

b) P⇒ (Q P)

c) (P (Q P))

Solution:

a) E.T.P (P∧Q) (P Q) is a tautologe

P Q P∧Q P Q (P∧Q) → (P→Q)

T T T T T

T F F F T

F T F T T

F F F T T

In the last column, all the truth values are T and so (P∧Q) (P Q) is a tautology

Alter

Now (P∧Q) (P Q)

⇔(P∧Q) (¬P∨Q) (⸪P Q⇔¬P∨Q)

⇔¬(P∧Q)∨ (¬P∨Q) (⸪P Q⇔¬P∨Q)

⇔(¬P∧¬Q)∨ (¬P∨Q) (by De morgan’s Law)

⇔¬P∨(¬Q∨Q) (by associative law)



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⇔¬P∨T

⇔T

Duality Principle Theorem:

Let A and A* be dual formulas and let be all the automic variables that occur in

A and A* (i.e. A=A( and A*=A*( ). Then

¬A( )⇔A*(¬ ) (i.e. The negation of a formula is equivalent to its

dual in which every variable is replaced by its negation) also, A¬( ⇔

¬A* (

Theorem

If any two formulas are equivalent, then their duals are also equivalent to each other

i.e, If A⇔B, then A*⇔B*

Proof:

Let ( be all the atomic variable appearing in the formulas A and B

Let A⇔B (T.P A*⇔B*)

Then A B is a tautology

(i.e) A( B is a tautology and

A(¬ ) B(¬ ) is a tautology … (1)

(1)⇒ ¬A* ( ¬B* ( is a tautology

⇒¬A* ¬B*

⇒A* B*

Problem:

Verify equivalence ¬A( ⇔A*(¬ ) if A(P,Q,R) is P∧ (Q∨R)

Solution:

Given that A(P,Q,R) = P∧ (Q∨R) …(1)

Now A(P,Q,R) = P∧ (Q∨R))

A(P,Q,R) = P∧(Q∨R) …(2)

Now A*(P,Q,R)= P∨ (Q∧R) (replacing ∧ by ∨ and ∨ by ∧ in (1))

A*( P, Q, R)= P)∨ ( Q∧ R)

=P∨ (Q∨R)

= P∨(Q∨R) ….(3)



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From (1) and (2) we set A(P,Q,R)⇔ A*( P, Q, R)

Hence the equivalence is verified.

Problem

Show that i) ¬(P∧Q) ( ¬P∨(¬P∨Q)⇔( ¬P∨Q) ii) (P∨Q)∧(¬P∧(¬P∧Q))⇔ ¬P∧Q

Proof:

i) ¬(P∧Q) ( ¬P∨(¬P∨Q)

⇔¬(¬(P∧Q))∨((¬P∨¬P)∨Q) (⸪P Q⇔¬P∨Q & by associative law)

⇔ (P∧Q)∨(¬P∨Q) (¬P ∨¬P ⇔¬P)

⇔ (P∨(¬P∨Q))∧(Q∨(¬P∨Q)) (by distributive law)

⇔ ((P∨¬P)∨Q)∧(Q∨¬P) (by associative law)

⇔ (T∨Q)∧(Q∨¬P) (⸪P ∨¬P⇔T)

⇔T∧(Q∨¬P) (⸪T∨Q⇔T)

⇔¬P∨Q (⸪T∨P⇔P)

⇔¬P∨Q

ii) (P∨Q)∧(¬P∧(¬P∧Q))⇔ ¬P∧Q

⇔ (P∨Q)∧(¬P∧¬P)∧Q) (by associative law)

⇔ (P∨Q)∧(¬P∧Q) (by De Margan’s Law)

⇔ (P∨Q)∧(¬P∧Q)

⇔ (P∧(¬P∧Q))∨(Q∧(¬P∧Q)) (by distributive law)

⇔ ((P∧¬P)∧Q)∨(¬P∧Q))

⇔ (F∧Q)∨(¬P∧Q) (⸪P ∧¬P=F)

⇔F∨(¬P∧Q) (⸪F∧Q=F)

⇔ (¬P∧Q) (⸪F∨(¬P∧Q)= (¬P∧Q))

(ii) is proved.



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UNIT - II NORMAL FORMS

Disjunctive Normal forms:

Definition: (Elementary Product)

The product of the variables and their negation in a formula is called an Elementary product

Definition: (Elementary sum)

The sum of the variables and their negation in a formula is called an Elementary sum.

Note:

Sum means disjunction and product means conjunction

Example:

Let P and Q be two statement tautology then

i) P∨¬P is an elementary sum

ii) P∨Q is an E.S

iii) P∨¬Q is an E.S

iv) ¬P∨Q is an E.S

v) ¬P∨¬Q is an E.S

vi) P∧¬P is an E.P

Disjunctive normal form (DNF)

Definition

A formula which is equivalent to a given formula and which consists of a sum of elementary

products is called a disjunctive normal form of the given formula

Procedure to obtain DNF

1. An equivalent formula can be obtained by replacing and with ∧, ∨ and ¬

2. Apply negation to the formula or to a part of the formula and not to the variables

3. Using Demorgan’s laws, apply negation to variables

4. Repeated application of distributive laws will give +ve

Remark 2:

A given formula is false if every elementary product in DNF is identically false .

DNF = (elementary product)∨ (elementary product)∨….(..)

Problem 1:

Obtain disjunctive normal form of P∧(P Q)



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Solution

P∧(P Q)⇔P∧(¬P∨Q)

⇔ (P∧¬P)∨(P∧Q)

Which is a sum of elementary product

∴DNF is (P∧¬P)∨(P∧Q)

Problem 2:

Obtain disjunctive normal form of ¬(P∨Q) (P∧Q) is (P∧¬Q)∨(Q∧¬P)

Solution:

We know that P Q (P∧Q)∨( ¬P∧¬Q) …(1)

Now, ¬(P∨Q) (P∧Q)

⇔[¬(P∨Q)∧(P∧Q)]∨[ (P∨Q)∧(P∧Q)] …(by (1))

⇔ [(¬P∨¬Q)∧(P∧Q)]∨[ (P∨Q)∧( ¬P∧¬Q)]

⇔ [(¬P∨P)∧(¬Q∧Q)]∨[P∧(¬P∨¬Q)∨(Q∧(¬P∧¬Q))]

⇔ [F∧F]∨[(P∧¬P)∨(P∧¬Q)]∨[(Q∧¬P)∨(Q∧¬Q)]

⇔F∨[(F∨(P∧¬Q))∨((Q∧¬P)∨F)]

⇔F∨[(P∧¬Q))∨(Q∧¬P)]

⇔ (P∧¬Q))∨(Q∧¬P)

The disjunctive normal form of ¬(P∨Q) (P∧Q) is (P∧¬Q)∨(Q∧¬P)

Solution:

We know that P Q (P∧Q)∨( ¬P∧¬Q) …(1)

⇔[¬(P∨Q)∧(P∧Q)]∨[(P∨Q)∧(P∧Q)] …(by (1))

⇔ [(¬P∨¬Q)∧(P∧Q)]∨[(P∨Q)∧(¬P∨¬Q)] (by Demorgan’s Law)

⇔ [(¬P∨¬Q)∧ (P∧Q)]∨[(P∧(¬P∨¬Q)∨(Q∧(¬P∧¬Q)]

⇔ [(P∨Q)∧(¬P∨¬Q)]∨[(P∧¬P)∨(P∧¬Q)]∨[(Q∧¬P)∧(Q∧¬Q)]

⇔ [P∨Q∧¬P∨Q]∨[P∧¬P∨P∧¬Q]∨[Q∧¬P∧Q∧¬Q]

Which is a sum of elementary product

Obtain a disjunctive normal form of P ((P Q)∧ ¬(¬Q∨¬P))

Solution:

⇔P ((¬P∨Q)∧ ¬¬(Q∨P))

⇔P ((¬P∨Q)∧ (Q∨P))

⇔¬P∨((¬P∨Q)∧ (Q∨P))



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⇔ [¬P∨(¬P∨Q)]∧[¬P∨(Q∨P)]

⇔ [(¬P∨Q)]∧[(¬P∨Q)∧(¬P∨P)]

⇔ [(¬P∨Q)]∧[(¬P∨Q)∧T]

⇔ (¬P∨Q)∧(¬P∨Q)

⇔¬P∨Q

Alter

⇔[¬P∧ (¬P∨Q)]∨[Q∧ (¬P∨Q)]

⇔[(¬P∧¬P)∨ (¬P∧Q)]∨[(Q∧¬P)∨ (Q∧Q)]

⇔[¬P∨(¬P∧Q)]∨[(Q∧¬P)∨Q]

Principal disjunctive normal forms:

Let P and Q be two statement variables. Then, we can write formulas P∧Q, P∧¬Q, ¬P∧Q

and ¬P∧¬Q

These formulas are called minter or Boolean conjunctions of P and Q

A formula which is equivalent to a given formula and which consists of sum of its minterms

is called “principal disjunctive normal form” (PDNF)

Construction of PDNF without truth table

To replace conditionals and bi-conditionals by their equivalent formula involving ∧,

∨, ¬ only

To use Demorgan’s law and distributive law

To drap any elementary product which is a contradiction

To obtain min-terms in the disfunction by introducing missing factors

To delete identical minterms keeping only one, that appear in the disjunction

Find the minterms of P, Q, R

Solution:

There minterms. They are P∧Q∧R, P∧Q∧¬R, P∧¬Q∧R, ¬P∧Q∧R, ¬P∧¬Q∧R, ¬P∧Q∧¬R,

P∧¬Q∧¬R, ¬P∧¬Q∧¬R

Note:

i) P∧Q or Q∧P is include but not both

ii) P∧¬P and Q∧¬Q are not allowed

iii) No two minterms are equivalent



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iv) Each minterm has the truth value T. for every exactly one combination of the truth

values of the varivbles Pand Q

Note 2:

In general for a given n-number of variables, there will be minterms

Minterms of P and Q

P Q ¬P ¬Q P∧Q ¬P∧Q P∧¬Q ¬P∧¬Q

T T F F T F F F

T F F T F F T F

F T T F F T F F

F F T T F F F T

Problem 1:

Obtain the PDNF for ¬P∧Q

Solution:

¬P∧Q

⇔[¬P∧(Q∨¬Q)]∨[Q∧(P∨¬P)] (⸪Q∨¬Q⇒T)

⇔ [(¬P∧Q)∨( ¬P∧¬Q)]∨[(Q∧P)∨(Q∧¬P)] (by distributive law)

⇔ (¬P∧Q)∨( ¬P∧¬Q)∨(Q∧P)∨(Q∧¬P) (⸪(P∧Q)∨(P∧Q)⇔P∧Q)

Which is a product of minterms

The PDNF is (¬P∧Q)∨( ¬P∧¬Q)∨(Q∧P)

Problem 2:

Obtain PDNF for (P∧Q)∨( ¬P∧R)∨(Q∧R)

Solution:

(P∧Q)∨(¬P∧R)∨(Q∧R)

⇔[(P∧Q)∧(R∨¬R)]∨[(¬P∧R)∧(Q∨¬Q)]∨[(Q∧R)∧(P∨¬P)] (⸪P∧T=T)

⇔[((P∧Q)∧R)∨((P∧Q)∧¬R)]∨[((¬P∧R)∧Q)∨((¬P∧R)∧¬Q)]∨[((Q∧R)∧P)∨((Q∧R)∧¬P)]

⇔[(P∧Q∧R)∨(P∧Q∧¬R)]∨[(¬P∧R∧Q)∨(¬P∧R∧¬Q)]∨[(Q∧R∧P)∨(Q∧R∧¬P)]

⇔(P∧Q∧R)∨(¬P∧R∧Q)∨(P∧Q∧¬R)∨(¬P∧R∧¬Q)

Which is a product of P,Q

Problem 3:

Obtain PDNF for P [(P Q)∧¬(¬Q∨¬P)]



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Solution:

P [(P Q)∧¬(¬Q∨¬P)]

P [(¬P∨Q)∧ (Q∧P)]

¬P∨[(¬P∨Q)∧ (Q∧P)]

¬P∨[(¬P∨Q)∧(Q∧P)]

[¬P∧(Q∨¬Q)]∨[(¬P∨Q) ∧ (Q∧P)]

[(¬P∧Q)∨(¬P∧¬Q)]∨[¬P∧(Q∧P)∨Q∧(Q∧P)]

(¬P∧Q)∨(¬P∧¬Q)∨(¬P∧(Q∧P))∨(Q∧P)

∴ The PDNF is (¬P∧Q)∨(¬P∧¬Q)∨(¬P∧(Q∧P))∨(Q∧P)

Conjunctive normal form

A formula which is equivalent to a given formula and which consists of a product of

elementary sum is called a conjunctive normal form of the given formulas

Remark

CNF = product of elementary sums

= (elementary sum)∧ (elementary sum)∧….. (elementary sum)

Principal conjunctive normal form

Let P and Q be two statement variables. Then we can write formula P∨Q, P∨¬Q, ¬P∨Q,

¬P∨¬Q.

These formulas are called maxterms of Boolean disjunction of P and Q

A formula which is equivalent to a given formula and which consists of product of its

maxterm is called “Principal of cinjunctive normal form” (PCNF)

Problem 1:

Obtain a conjunctive normal form of P∧(P Q)

Proof

P∧(P Q)⇔ P∧(¬P∨Q)

Which is a product of elementary sums

∴ The CNF of P∧(P Q)is P∧(¬P∨Q)

Problem 2:

Obtain a conjunctive form [Q∨(P∧R)]∧¬[(P∨R)∧Q]

Solution:

[Q∨(P∧R)]∧¬[(P∨R)∧Q]



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[(Q∨P)∧(Q∨R)]∧[¬(P∨R)∨¬Q]

[(Q∨P)∧(Q∨R)]∧[(¬P∧¬R)∨¬Q]

(Q∨P)∧(Q∨R)∧(¬P∨¬Q)∧(¬R∨¬Q)

Product of elementary sums

The CNF is (Q∨P)∧(Q∨R)∧(¬P∨¬Q)∧(¬R∨¬Q)

Maxterms

For a given number of variables the max terms consists of conjunctions in which each

variable or its negation, but not both appears only once

Remark

1. The maxterms are duals of minterms

2. Each of the maxterms has the truth value F for exactly one combination of the truth

values of the variables.

3. Different max terms have the truth table F for different combinations of the truth

values of the variable

P Q ¬P ¬Q P∨Q ¬P∨Q P∨¬Q ¬P∨¬Q

T T F F T T T F

T F F T T F T T

F T T F T T F T

F F T T F T T T

Obtain the principal conjunctive normal form of the formula S given by (¬P R)∧(Q P)

Solution:

(¬P R)∧(Q P)

(¬P R)∧[(Q∧R)∨(¬Q∧¬P)]

[¬ (¬P)∨R]∧[(Q∧R)∨(¬Q∧¬P)]

(P∨R)∧[(Q∨(¬P∧¬P)∧P∨(¬Q∧¬P)]

(P∨R)∧[ (Q∨¬P)∧(Q∨¬P)∧(P∨¬Q)∧(P∨¬P)]

(P∨R)∧(Q∨¬P)∧(Q∨¬P)∧(P∨¬Q)∧T

(P∨R)∧(Q∨¬P)∧(P∨¬Q)

Alter

(¬P R)∧[(Q P)∧(P Q)]



Page 37 of 119

[¬ (¬P)∨R]∧[(¬Q∨P)∧(¬P∨Q)]

(P∨R)∧(¬Q∨P)∧(¬P∨Q)

[(P∨R)∧(Q∨¬P)]∧(¬Q∨P)∨(R∧¬R)]∧[(¬P∨Q)∨(R∧¬R)]

(P∨R∨Q)∧(P∨R∨¬Q)∧(¬P∨R∨Q)∧(¬P∨R∨¬Q)∧(¬P∨R∨Q)∧(¬P∨¬R∨Q)

Which is a product of max.terms



Page 38 of 119



Page 39 of 119

UNIT - III GROUP AND MONOIDS

Definition: (Group) A non-empty set G together with a binary operation is said to be a group if it satisfied the following

i) is associative

(i.e.)

ii) Identity property

There exists an elements such that (e is called

identity element of G)

iii) Inverse property

, there exists an element such ( is called the

inverse of a)

Definition (Monoid) :

A non-empty set M together with a binary operation said to be a monoid if it

satisfied the following

i) is associative

(i.e.)

ii) Identity property

There exists an elements such that (e is called

identity element of M)

Semi group:

A non-empty set b together with a binary operation said to be a semi group if

its satisfying the following condition

(i) Closed

(ii) Associative

Example 1:

Let be the set of natural numbers then (i) is a semi group but not a

monoid (ii) is a semi group and also a monoid

Solution:

Let



Page 40 of 119

To prove is a semi group, but not a monoid

Closure Property

Let (To prove

Then clearly (⸪addition of two natural numbers is again)

∴Closure property is true a natural number

Associative property

Let (To Prove )

Then, clearly

∴Associative property in true

Let

Then,

∴ has no identity element and hence

is a semi group but not a monoid

To Prove is a semi group and also a monoid

Closure Property

Let (To prove

Then clearly (⸪Product of two natural numbers is again natural number)



Let (To Prove )

Then, clearly

∴ Associative property in true

Identity

Let . Then such that so 1 is the identity element of N

∴ is a semi group and also a monoid

Example 2

Let then (i) is a monoid and (ii) is a monoid

Solution:

Closure Property

Let (To prove

Then clearly (⸪addition of two positive integers is again positive integer)



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∴Closure property is true


Let (To Prove )

Then, clearly


Identity:

Let

Then, such that and so 0 is the identity element of

∴ has no identity element and hence



Closure Property

Let (To prove

Then clearly (⸪Product of two natural numbers is again natural number)



Let (To Prove )

Then, clearly


Identity



Example 3

Let be the set of all even numbers. Then (i) is a semi group but not

monoid and (ii) is a semi group but not a monoid

Solution:

To prove is a semi group but not monoid

Closure Property

Let (To prove

Then clearly (⸪addition of two even numbers is again even number)




Page 42 of 119


Let (To Prove )

Then, clearly


Identity:

Let

Then, such that But ∉ E

∴ E has no identity element



Closure Property

Let (To prove

Then clearly (⸪Product of two even numbers is again even number)



Let (To Prove )

Then, clearly


Identity



Example 4

Let be the set of all even numbers. Then (i) (ii) monoid

Solution:

To prove are monoid

Closure Property

Let

Then clearly (⸪Union of any two subsets of is again a subset of )



Let (To Prove )



Page 43 of 119

Then, clearly )


Identity:

Let

Then,

∴ Clearly has no identity element



Closure Property

Let

Then clearly (⸪Intersection of any two subsets of is again a subset of

)



Let (To Prove )

Then, clearly )


Identity:

Let

Then,

∴ Clearly has no identity element


Example 5:

Let be the set of all even numbers. Then is a monoid

Solution:

To prove is a monoid

Closure Property

Let (To prove

Then clearly (⸪Sum of two rational numbers is again rational number)





Page 44 of 119

Let (To Prove )

Then, clearly


Identity:

Let

Then,

∴ clearly 0 is the identity element

Hence is a semi group but not a monoid

Example 6:

Show that the set is a semi group under the operation is

a monoid

Proof

In the binary operation is defined by where

To prove is a monoid

Closure Property

Let (To prove

Then

clearly



Let (To Prove )

Then, L.H.S =

=

=

…(1)

R.H.S =

=

…(2)

L.H.S=R.H.S (by (1) and (2))



Page 45 of 119


is a semi group

Identity:

Let

Then such that

Now,

Then it is a monoid

Example 7:

Let X be a non-empty set and be the set of all mappings from X to X. Let * denote the

operation of composition of these mappings (i.e.) for , is given by

. Then the algebra is a monoid.

Solution:

Let X be a non-empty set

/ f

e the binary operation composition of function is defined by

Closure Property

Let (To prove

Then and are mappings

Clearly, is a map and so



Let (To Prove &

, and

Let

Then L.H.S =

… (1)

R.H.S =

… (2)



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L.H.S=R.H.S (by (1) and (2))


Identity:

Let

Then such that

And

∴

Identity property is true

Then is a monoid

Definition:

Let and be any two monoids. A mapping is said to be a

monoid homomorphism, if (i)

(ii)

Definition:

Let be a monoid homomorphism then f is said to be a

(i) Monoid monomorphism if f is 1-1

(ii) Monoid epimorphism of f is onto

(iii) Monoid isomorphism of f is 1-1 and onto

Prove that monoid homomorphism preserves (i) associative property (ii) identity property

(iii) commutative property (iv) invertibility

Proof:

Let and be any two monoids. A mapping is said to be a

monoid homomorphism, if (i) …(1)

(ii) …(2)


Let

Then (To Prove )

Now, (by (1))



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(by (1)

(by (2)

)

Theorem associative property is satisfied

Identity:

Given that is the identity element of M, is the identity element of T

∴ …(3)

To Prove

Let

Then, (by (2))

(by (1))

And

(by (1))

And so

Then identity property is satisfied

Commutative property :

Let such that … (4)

To prove

Now,

Then commutative property is satisfied

Inversibility:

Let and be the inverse of

Then, …(5)

To prove is the inverse of (i.e.

Let



Page 48 of 119

Then (by(1))

(by (5))

(by (2))

Similarly we can prove that

Theorem

For any commutative monoid the set of idempotent elements of M forms a

submonoid

Proof:

Let be a commutative monoid

Closure Property

Let

Then and … (1) (by the definition of T)

(⸪

(⸪ M is commutative)

(by(1))

And so

Claim

Clearly

∴

Hence T is a submonoid

Example:

Let be a monoid and let and be a monoid with the operation

given by the following table

* e 0 1

e e 0 1

0 0 0 0

1 1 0 1

A map is defined by

and for …(1)



Page 49 of 119

Then g is not monoid homomorphism because

Example:

Let I be the set of integers and let be the set of all equivalence “conqruence moduls M”

for any the integer M then, and are monoids

Proof:

Let

In we define and as following

Consider the composition table for and

0 1 2 3 4

0 0 1 2 3 4

1 1 2 3 4 0

2 2 3 4 0 1

3 3 4 0 1 2

4 4 0 1 2 3

0 1 2 3 4

0 0 0 0 0 0

1 0 1 2 3 4

2 0 2 4 1 3

3 0 3 1 4 2

4 0 4 3 2 1

Clearly closure property are true in and

Identity element in is 0

and identity element in is 1

and are monoids

Similarly we can form a table for and

Let

Then closure property and associative property is true. Also 0 is the identity element of



Page 50 of 119

and 1 is the identity element of

and are monoids by definition (1)

Example

Let be the given set. Let S denote the set of all mappings from A to A then, we

have mappings available

Proof:

Let where

There 1S is a monoid under the composition of function 0

Let us form a table for composition of function

0

Closure Property

Let

Then, clearly


Let where

Then, Clearly

Identity Property

Let

Then, such that

S has an identity element I

Hence is a monoid

Algebraic Structures



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Binary and N-ary Operation

Let S be a non-empty set and be a mapping. Then, f is called a binary operation

on S.

In general, is called an n-ary operation and n is called

the order of the operation

For , is called a unary operation similarly for , binary

operation

Definition: (Algebraic System)

A non-empty set G together with one or more n-ary operation say * is called an algebraic

system or algebraic structure or algebra and It is denoted by

Note: etc. are same of binary operations

Notation:

We shall denote an algebraic system by where S is a non-empty set and

are operations of S

Example:

Let I be set of Integers. Consider the algebraic system of where are the

operation of addition and multiplication on I. List on important property of this operation

will be given below

: for any

(Associative)

: for any

(Commutative)

a distinguished element such that here, is the identity element write

addition.

: such that

(inverse)

: for any

(Associative)

: for any

(Commutative)

a distinguished element such that



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here, is the identity element write multiplication.

For any

(Distributive)

The operation distributes over

For any and

(cancellation law)

Exercise 1:

Which of the following system satisfy the properties of which are designed by

to , to , D and C?

a)All odd integers b)All even integers c)All Positive integers d)All non-negative

e) f)

a) All odd integers

Let 0 be the set of all odd integers for any

(Associative)

: for any

(Commutative)


(Identity)

here, is the identity element under addition.

: such that

(inverse)

: for any

(Associative)

: for any

(Commutative)


here, is the identity element with respect to multiplication.

For any

(Distributive)



Page 53 of 119


For any and

(cancellation law)

b) All even integers

Let E be the set of all even integers for any

(Associative)

: for any

(Commutative)


(Identity)


: such that

(inverse)

: for any

(Associative)

: for any

(Commutative)



For any

(Distributive)


For any and

c) All even integers

Let P be the set of all Positive integers for any

(Associative)

: for any



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(Commutative)


(Identity)


: such that

(inverse)

: for any

(Associative)

: for any

(Commutative)



For any

(Distributive)


For any and

The following examples are algebraic systems with the binary operation which share more

of the properties of

Example 2:

Let R be the set of real numbers. Let and be the operation on R. Then the algebraic

system satisfies all the properties given for the system

Example 3:

In the algebraic system where N is the set of natural numbers and the operation

and have their usual meanings, all the properties listed for are satisfied except

Solution:

Let

Then it satisfied the following property



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Associative

Let

Associative property is true

: Commutative

Let

Clearly Commutative property is true

Let

Then such that

is the addition identity in N.

:

Let

Then, no such that

In N no element has an additive inverse

: for any


:

Let

Commutative Property is true

Let

Then such that

is the multiplication identity element in N.

P



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Let

Distributive property is true


C

Let

Such that

Then,

Clearly all the properties are satisfied except by N

Example 4:

Let S be a non-empty set and be its power set for any sets defined the

operation and of P(S) as

(X is not a cartesion product) then the algebraic system satisfied all the

properties listed except (C)

Solution:

Associative

Let


: Commutative

Let

Clearly Commutative property is true

identity

Let

Then there exist such that

is the addition identity in .



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: Inverse

Let

Then,

∴ A has an additive inverse

: Associative

Let

Clearly Associative property is true

: Commutative

Let

Clearly Commutative Property is true

Identity

Let

Then such that

is the multiplication identity

D

Let

Distributive property is true

C

Let

To prove



Page 58 of 119

Now

Is not satisfied

Example 6:

Consider the set and the operation and on B given by the following table

0 1

0 0 1

1 1 0

0 1

0 0 0

1 1 1

Then B is the algebraic system and satisfying the all the above properties

Some simple algebraic system and general properties

In the section we give examples of algebraic systems. Conceding of a single unary or binary

operation

Example 1:

Let and r be a unary operation on M given by

Then algebra is called the clock algebra

This can be illstrated as follows

Note:

Every element of M can be generated from the element by respected application for

the operation ‘r’

1 can be called a generator of the algebraic system

Example 2:

Let and S denote the set of all mapping from X to X let us write



Page 59 of 119

where

Then is an algebraic system where the operation * is a left composition of function

also * is associative

Solution:

Let and

Let us take where

The composition table for the operation * is given below

*

Clearly it is an algebraic system in which the operation * is associative

Note 1:

is the identity element with respect to the operation *

Note 2:

Not all the elements are S are invertiable

Example 2:

Let and given by let

identity mapping X be denoted by . If we form the composite functions.

, , and so given then

Let us denote by and consider the set

Then, F is called the operation composition then clearly it is an algebraic system

Definition:

Let and be two algebraic system of the same type in the sense that both ○

and * are binary (n-ary) operations. A mapplings is called a

homomorphism or simple morphism, if

Note 1:



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Note 2:

is a homomorphism image of

Definition:

A homomorphism is called a

i) Epimorphism if g is onto

ii) Monomorphism if g is 1-1

iii) Isomorphism of g is 1-1 and onto

Definition :

Let and be two algebraic system. If an isomorphism mapping

exists, then and are said to be isomorphic

Definition:

Let and be two algebraic system such that . A homomorphism

in such case and called an endomorphism

Example 1:

If is an abelian group, then for all show that

Proof:

Given that is an abelian group

To prove

Let

Then

( by associative law)

(⸪ G is abelian)

Example 2:

In a symmetric group the binary operation ◇ is defined in the

following table

◇



Page 61 of 119

Where

Find all these elements a and b

Such that (i) (ii) (iii)

Solution:

Let

Then

This condition is satisfied

Let is the identity element



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∴ P is satisfied this condition

Definition:

Let S be a non-empty set. A bijection is called a permutation of S

Example 1:

Let

Then, and are permutations of S

Example 2:

Let

Then there are 24 permutations of S

Definition:

The order of a group (G,*) denoted by , is the number of elements of G, when G is finite

Definition:

A group (G,*) is said to be an abelian group if the binary operation is commutative

If

Example 1:

Let I be the set of integers. Then the algebra is an abelian group

Example 2:

The set of all rational numbers excluding zero is any abelian group under multiplication

Note 1:

A group of order one has only the identity element (ie.) .

A group of order two has one more element besides the identity element (i.e.)

* e a

e e a

a a e



Page 63 of 119

Let us consider then G is a group under the binary operation * and it is defined

in the following table

* e a B

E e a B

A a b E

B b e A

Note :

The above three group are abelian and so the groups of order 2 &3 that all such groups are

abelian

In fact the groups of order 4 and 5 are also abelian

Remark

Groups of order 6 are necessarily abelian

Problem

is an abelian group, then for all . Show that

Solution:

Given that is an abelian group

To prove

Let .

LHS (n times)

(n times) (G is abelian)

(n times) (n times)

To prove this result by using induction on n

Let

Then

The result is true for



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Induction hypothesis

Let us assume that result is true for then we have …(1)

Now we have to prove it for n

Now

(by (1))

Problem 2:

Show that if every element in a group is its own inverse, then the group must be abelian

Proof:

The inverse of

Given that every element is inverse of itself

Let

Then

G is an abelian group

Problem 3:

Let the composition take the for and where

0 1 2 3 4 5 6

0 0 1 2 3 4 5 6

1 1 2 3 4 5 6 0

2 2 3 4 5 6 0 1

3 3 4 5 6 0 1 2

4 4 5 6 0 1 2 3

5 5 6 0 1 2 3 4

6 6 0 1 2 3 4 5



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1 2 3 4 5 6

1 1 2 3 4 5 6

2 2 4 6 1 3 5

3 3 6 2 5 1 4

4 4 1 5 2 6 3

5 5 3 1 6 4 2

6 6 5 4 3 2 1

Theorem

Every row and column in the composition table of a group is a permutation of the

elements of G.

Proof:

Claim 1:

No row or column in the composition table can have the element of G more than once

For 1

Let us assume that the row corresponding to an element has two entries are both k,

Then where …(

(by cancellation law)

Which is contradiction to

Our assumption is wrong and hence the claim (1)

Claim 2:

Every element of G appears in each row and column of the table of composition

For 2

Let us consider the row corresponding to the element and

Since , b must appear in the row corresponding to the element

Every row of the composition table is obtained by a permutations of the element of G and

each row is a distinct permutation

Example

Let and the operation on G is given by

Then is not associative



Page 66 of 119

Solution:

Let

L.H.S

R.H.S

Clearly is associative

Definition: (Group)

A non-empty set G together with a binary operation is said to be a group if

following conditions are satisfied

i) is associative

(i.e.)

ii) There exists an elements such that (e is called

identity element of G)

iii) For any element a in G there exists an element such ( is

called the inverse of a)

Definition : (Permutation)

Let A be a finite set. A bijection from A to itself is called a permutation of A

Example :

If given by , is permutation of A

Subgroups:



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Definition:

Let G be set with a binary operation definition on it. Let . If for each

, we say that S is closed with respect to the binary operation

Order of the group:

Let G be a group and let . The least positive integer n such that is called the

order of a. If there is no positive integer n such that , then the order of a is said to be

infinite

Definition (Normal Subgroup)

A subgroup H of G is called a normal subgroup of G if for all

Definition : (Coset)

Let H be a cubgroup of a group G. Let . Then the set is called the

left coset of H defined by a in G

Similarly is called the right coset of H

Theorem :

Let H be a subgroup of G then

Any two left cosets of H are either identical or disjoint

Union of all the left cosets of H is G

The number of elements in any left coset is the same as the number of elements

in H

Proof:

Let and two left cosets

Suppose and are not disjoint

We prove that

Since and are not disjoint,

∴ there exists an element

∴ and

∴ and

∴

ii)Let then

∴ Every element of G belongs to a left coset of H



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∴ The union of all the left cosets of H is G

iii)The map defined by defined by is clearly a bijection.

Hence every left soccer has the same number of element S as H

Lagrange’s Theorem

Let G be a finite group of order n and H be any subgroup of G. then the order of H dinides

the order of G.

Proof:

Let and

Then the number of distinct left cosets of H in G is r

By above theorem, these r left cosets are mutually disjoint, they have the same number of

elements namely M and their union is G

∴

Hence m divides n

Caylay’s Theorem:

Any finite group is isomorphic to a group of permutations

Proof:

We shall first find a set of permutations. Then we prove that is a group of

permutations and finally we exhibit an isomorphism

Step 1:

Let G be a finite group of order n. let define by

Now, is 1-1, since =

is onto

Since, if then

Thus is bijection

Since G has n elements, is just permutation on n symbols

Let

Step 2:

We prove is a group

Let



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Hence

Hence is closed under composition of mapping is the identity element

The inverse of is G is

Step 3:

We prove

Define by

Hence is 1-1 obviously is onto

Also

Hence is an isomorphism

Abelian group:

A Group G is said to be abelian if the for all

A group which is no abelian is called a non-abelian group

Kernel of f:

Let be a homomorphism let . Then K is called the kernel

of f and is denoted by ker f

Theorem

Let be a homomorphism then the kernel K of f is a normal subgroup of G

Proof:

Since

And hence

Now let . Then

∴

Thus . Hence k is a subgroup of G

Now, let and

Then



Page 70 of 119

∴

Hence k is a normal subgroup of G

Problem

is an abelian group

Proof:

Closure Property

Let

Then clearly



Let

Then,

Thus the operation + is Associative in Z

Identity

The element is the identity

Since

The identity property is satisfied

Inverse

Let , then there exists

Such that

Hence is group

Commutative

Let

Then

∴ is an abelian group

Definition

Let be a group and define and

Above define

Above we have where

Exercise



Page 71 of 119

Let the permutation of the elements of be given by

. Find

and , also find

Solution:

Let

To find

Group codes

Coding theory has developed technics for introducing extra information intrasmitted data

which help in detaceing and some times in correct in errors

Some of these technics make use of group theory

A communication system will consist of three essential part

i) Transmitter (source)

ii) Channer (storage medium)

iii) Receiver

Transmitter Channel Receiver

In actual practice the communication channel will be subject to a lot of disturbances like

noise due to whether interference electrical problems and so once.

In order to the dated and to correct the errors due to noise in communication system we

use what is no know as encoder and decoder



Page 72 of 119

Transmitter Encoder Channel Decoder Receiver Noise

Definition (Encoder)

A device which is used to improve the efficence of the communication channel is called

encoder

It transforms the incoming messages in such a way that the presence of noise on the

transformed messages is detected

A decoder is used to transform the encoder messages into their original form that is estable

to the receiver

The basic unit of information called a message is a finite sequence of characters from a

finite alfaber

A word consists of no of symbols (0’s and 1’s) in it

A code is a collection of words that are to be used to represent distinct message

A word in code is also called code word

A block code is a code consisting of words that are of the same length

Definition

A code which is a group under the binary operation + is called a group code

+ 0 1

0 0 1

1 1 0

Where + is defined by

Encodeing function

Let where be the set of all m digit sequences

Let where and be a 1-1 function. Then the function e is called

an encoding function which represents every word in as a word in

If then is called the code word representing b

if then the number of 1’s in x is called the weight of x and is denoted by

Example 1:

Find the weight of each of the following words in (a) (b)

(c) (d)

Solution :



Page 73 of 119

Given that where

We know that if , then the weight of number of 1’s in x

a) Let

Then the number of 1’s in

b) Let


c) Let


d) Let


Let us consider the following 8 symbols each are this can be

represented by a sequence of three binary digits

Due to disturbances or noise the receiver can make an uncorrectable and undetectable

mistake because of the interchange of 0 or 1 in a particular code

For example

If the code 000 for A is changed by noise to 100 or 010 or 001 which gives B or C or D

To avoid this ‘Hamming’ developed a code by introducing extra digits called as party choices

In a message which is ‘n’ digit long the first ‘m’ digits are used for the information

and the remaining digits are used for the detaction and correction of the later

digits are known as parity check code

Page 74 of 119

Even parity check:

Let us consider a code word of length ‘n’ in which the first digits we used for the

informantion part of the message and the last digit is either 0 or 1, show that the number of

ones (1’s) is even or odd. Thus encoding processare is called an even parity check

Let us define formally as follows

The encoding function is called the parity check code if

Define where

Example :

Consider the encoding function and define if

and where

Original code Code with extra digit

000

001

010

011

100

101

110

111

0000

0011

0101

0110

1001

1010

1100

1111

If we consider a word 000 then

If an error occers in a single digit then 0000 becomes 0001 or 0010 or 0100 or 1000

But this combination is not available for any of the other code

By supplying extra digits we are in a position to detect and correct a single error appearing in

one of the codes

Note:

If more than one error occurs in a code, then we will have to supply more than one extra digit

Example :

Consider the following encoding function if defing

(i.e.) the encoding function and repeat each word of three times

Page 75 of 119

Solution

Let

Then

Assume that the transmission channel makes an error in the under line digit and that we

receive the word

This is a code word so we have detect the error

Hamming distance

Let X and Y be word in the hamming distance is the weight of

The distance between and is the number of positions in which and

differs

Minimum distance

Minimum distance of an encoding function for the minimum of the distance

between all distinct pairs of code words

Example:

Find the distance x and y where a) b)

Solution :

a) Number of position differs from

b) Number of position differs from

Page 76 of 119

UNIT - IV LATTICES AND BOOLEAN ALGEBRA

Definition: (Partial Ordering) A binary relation R in a set P is called a partial order relation or partial ordering in P iff R is reflexive, antisymmetric and transitive Note: If is a partial ordering on P, then the ordered pair is called a partially order set or poset Definition: (Lattice) A lattice is a partial ordered set in which every pair of elements has a greatest lower bound and a last upper bound Notation:

The g.l.b of a subset will be donated by and

The l.u.b of a subset will be donated by and

(i.e) the meet or product of a & b

and the join or sum of a & b

Example 1:

Let S be any set and be its power set. The partially ordered set is a lattice in

which the meet the and join are the same as the operations ∩ and U respectively

Solution:

Given that S is an set and is its powerset

To prove is a lattice

Enough to prove every pair of elements has a g.l.b and l.u.b

Let

Then, g.l.b of and

Then, l.u.b of and

is a lattice

Note:

When S has a single element the corresponding lattice is a chain containing two element

Note:

When S has two elements, draw the diagram for the corresponding lattice

For let

Then is a lattice (by the above exam) where

Page 77 of 119

{a,c} {a,b

} {c} {b}

{a}

{b,c}

Note:

When S has three elements, draw the diagram for the corresponding lattice for

Let

Then is a lattice (by the above example)

Where

The diagram of the lattice is

Example 2:

Let be the set of all positive integers and D denote the relation of division in such that for

any if a divides b. Then is a lattice in which the join of a and b is given by

the least common of and the meet of a and b is given by the greatest common divisor of

a and b (i.e.) and

Solution :

Let

GCD of and

To prove is a lattice

Enough to prove

Every pair of elements has a g.l.b and a l.u.b

Let

Then

And

Therefore is a lattice

{a,b}

{b}

{a}

{a,b,c}

Page 78 of 119

Example 3:

Let n be a positive integer and be the set of all divisors of n. Let P denote the relation of

division such that . Then is a lattice. Draw the diagram of lattices

and

The lattice

Here

Hasse Diagram

A pictorial representation of a poset is called a Hasse diagrame

Example

Let and the relation R defined on X by . Draw Hasse

diagrame for

Solution:

Let

.

Then

Hasse diagrame for is

6

3 2

1

36

6

24

36

6

6

12

36

6

6

6

36

6

6

2

36

6

6

3

36

6

6

36

6

24

6

12

6 6

6

2

6

3

6

Page 79 of 119

Example 2:

Let find lub for the poset where

Solution:

Let

The Hasse Diagrame for is

The table of lub and glb

Example 3:

Determine whether the posets

i)

ii)

Solution:

12

6

6

6

4

6

1

6

2

6

3

6

Page 80 of 119

Let

Then

The Hasse diagrame is

Here, glb=1

Lub does not exist

So X is not a lattice

Let

Then

The Hasse diagrame is

Here

And

X is a lattice

Some properties of Lattices

Let L be a lattice

Idempotent Law

4 5

2

1

3

4

8

2

1

16

4

8

2

1

16

Page 81 of 119

Commutative Law

Associative Law

Absorption Law

Proof of

Let

To prove

Then ---(1) (by the definition

And also

∴ (by the definition

---(2)

From (1) and (2) we get

to prove

Let

Then ----(3)

And (by

(i.e) ---(4)


Similarly we can prove the other problem to and to

Theorem:

Let be a lattice in which and denote the operations of meet and join respectively.

Then, for any

Proof:

Given that L is a lattice such that

and for all

Let

Claim 1:

For 1

Let

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Then

(by ---(1)

By the definition ---(2)clearly

(1) And (2) ⇒

Conversely, let ---(i) (to prove (Enough to prove (by(i))

Clearly

∴

And so

Claim 2:

For 2 let

Then

---(3)

By definition ---(4) clearly

(3)&(4)⇒

Conversely let ---(B)

(To prove (Enough to prove (by (B))

Then clearly

∴ (by (B))

And so

Theorem:

Let be a lattice for any . The following properties called isotonicity hold

and

Proof:

Let be a lattice for any such that ---(1)

(i)To prove

Enough to prove

(by associative law)

(by commutative law)


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(by idompotant law)

(⸪

Hence (i) is proved

(ii)To prove

Enough to prove


(by commutative law)


(by idompotant law)

(⸪

Hence (ii) is proved

Theorem:

Let be a lattice for any . The following ineaualities called the distributive

inequalities hold

i)

ii)

Proof:

Let be a lattice for any .

i) To prove

We have ---(1) and

---(2)

∴ (by using (1) and (2))

∴ ---(3) (by idempotent law)

Also ---(3) and

---(4)

(3& (4)⇒

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(by idempotent)

Hence (i) is proved

ii) To prove

We have ---(5) and

---(6)

∴ (by using (1) and (2))

∴ ---(7) (by idempotent law)

Also and

---(8)

(7)& (8)⇒

(by idempotent)

Hence (ii) is proved

Theorem: Modular inequality

Let be a lattice for any . The following holds

Proof:

Let be a lattice for any

Let To prove

Then and ---(1)

We know that the distributive inequality is

(by using (1))

Hence the proof.

Other forms of modular inequality

1. if and

2. if

Proof:

We know that the distributive inequalities are

i. ---(1)

ii. ---(2)

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Given that

and

i. To prove

Put in (2) we get

(i.e.)

ii.

Put in (1) we get

Problem:

Show that in a lattice, if , then and

Proof:

Let be a lattice for any such that

Then , , , , ,

To prove

By the given condition and so

To prove

Now,

And

Hence the proof

Problem:

In a lattice, show that (i)

(ii)

Proof:

(by distributive inequality)

And so ---(1)

Page 86 of 119

(by (1))

Some special lattices

Recall that in a lattice, every pair of elements has a least upper bound and a greatest lower

bound.

Result:

Every finite subset of a lattice has a least upper bound and a greatest lower bound

Infinite subset of a lattice does not have least upper bound

For 1

Let L be a lattice and be a subset of L

To prove S has lub and a glb

Define

∴ S has a GLB and LUB

For example

Let be the set of positive integers. Then has no LUB

Definition :

A lattice L is called complete if each of its non-empty subsets has 0 least upper bound and a

greatest lower bound

Note 1:

Every finite lattice must be complete also every complete lattice must have a least element and

greatest element

Note 2:

The least and greatest element of a lattice if they exist are called the bound of the lattice and

are denoted by zero and one respectively

Bounded lattice

A lattice said to be a bounded lattice, if it has both elements 0 and 1 consider the lattice

where the and

Remark

The bounded 0 and 1 of a lattice satisfied the following identity

and

and

Page 87 of 119

Remark:

Obviously 0 is the identy of the operation ⊕ and 1 is the identity of the operation

In a bounded lattice one and zero are duals each other

Definition

Let be a bounded lattice. An element is called an complement of an element

if and

(i.e.) If ‘a’ is a complement of ‘b’ then ‘b’ is a complement of ‘a’ and vicoversal

Note:

and

0 and 1 are complement of each other

Result:

1 is the only complement of 0

For 1

Let where be the complement of zero ---(1) but we have

Which is a contradiction

1 is the only complement of 1

Definition

A lattice is said to be a complement lattice if every element of L has atleasy on

complement

Show that the following lattices are not distributive

Let

Then

1

0

1

0

Page 88 of 119

This is true

But

∴

This is true

Now

∴ The distributive law is not true

And so it is not a distributive lattice

Similarly

So that it is not distributive

Problem:

Show that in a complemented distributive lattice

Proof:

Assume that To prove

Then, ---(1) and ---(2)

Now, (by (1))

(by associative)

∴

Conversely assume that

Let to prove

Then

Conversely

to prove

Then

(iii)⇒(iv)

Page 89 of 119

Assume that to prove

(iv)⇒(i)

Assume that to prove

Now and

Applying demargon’s law

Show that Demorgan’s laws given by and hold in a

complemented distributive lattice

Solution :

To prove enough to prove

Now

And

Now

∴

Similarly we can prove

Page 90 of 119

Boolean algebra:

Definition:

A Boolean algebra is complemented distributive lattice and is denoted bu

Notation :

is a lattice under the peration and ⊕

It is bounded lattice with bounds 0 and 1

(0 is the least and 1 is the greatest)

Every element in B has unique complement

Properties:

A Boolean algebra satisfies the following properties

is lattice in which the operation and ⊕ satisfy the following identities

Problem

is a distributive lattice satisfies the following identity

for all

and

Problem:

is a bounded lattice in which for any the following hold

Problem:

is a uniquely complemented lattice in which the complement of any element

is denoted by by satisfied the following identities

for all

Page 91 of 119

Problem:

There exists a particle ordering relation on B such that

State the properties of Boolean algebra

Show that is a Boolean

Example

Let be a set the operations and on B are tabulated below then B is a Boolean

algebra

0 1

0 0 0

1 0 1

Prove the following Boolean identities

a)

b)

c)

d)

Proof:

a) To prove

(by distributive)

(

(

b)

0 1

0 0 1

1 1 0

0 1

1 0

Page 92 of 119

(by distributive law)

hence proved b

c)


(by absorption and distributive law)

(by absorption law)

d)


(by absorption law)

Problem 1:

In any Boolean algebra show that

a)

b)

c)

d)

e)

Problem 2:

Let be a Boolean algebra define the operations and on the elements of B by

Show that

a)

b)

Page 93 of 119

c)

d)

Proof:

Let to prove

Conversely:

Let To prove

b) let …(2) To prove

(

Conversly :

Let To Prove

c) Let

…(3)

…(4)


Page 94 of 119

Proof:

Given that

To prove

Page 95 of 119

UNIT - V BINARY NUMBER SYSTEM

The binary number system is a system that uses only the digits 0 and 1 as codes Reset and carry The units wheel has reset to 0 and sent a carry to the tens wheel. This action is called as reset and carry Binary odometers A binary odometer is a device which has only two digits 0 and 1 then each wheel truns, it displays 0, then 1, then back to 0 and the cycle repeats. A four digit binary odometer starts with 0000 After 1 mile it indicates 0001 After 2 mile it indicates 0010 After 3 mile it indicates 0011 After 4 mile it indicates 0100 After 5 mile it indicates 0101 After 6 mile it indicates 0110 After 7 mile it indicates 0111 After 8 mile it indicates 1000 After 9 mile it indicates 1001 After 10 mile it indicates 1010 After 11 mile it indicates 1011 After 12 mile it indicates 1100 After 13 mile it indicates 1101 After 14 mile it indicates 1110 After 15 mile it indicates 1111 Note:

1. The word bit is the abbreviations for binary digit 2. A list of 4 bits numbers from 0000 to 1111 equivalent to decimal 0 to 15 3. When a binary has 4 bits it is called a nibble 4. A binary with 8 bits is know as a byte

Binary to decimal conversion Method:

1. Write the binary number 2. Directly under the binary number write 1,2,4,8,16,…. Working from right to left 3. If a zero appears in a digits position cross out the decimal weight for that position 4. Add the remaining weight to the obtained the decimal equivalent

Example 1:

Binary

1 0 1

4 2 1 ⇒ 4+1=5

Page 96 of 119

For fractions the weight of digits positions to the right of the binary point are given by

and so

Example 2:

0.101

For mixed number

The weights for mixed numbers

Example 3:

1 1 0. 0 0 1

110.001=6.125

Double Dabble method

In this method you progressively divide the decimal number by 2 writing down the remainder for each division The remainder taken in reverse from the binary number Example: Converse the decimal 13 to 15 binary Solution: 2 13

2 6 – 1

2 3 - 0

1 - 1

13 = 1101

Fractions: For fractions you multiple by 2 and record a carry in integer position. The carries read

downward are the binary fraction Example: 0.85 convert to binary Solution:

with a carry of 1 with a carry of 1 with a carry of 1 with a carry of 1 with a carry of 1

Page 97 of 119

with a carry of 1 Note:

1. In this case, we stopped the conversion process after gathering six binary digits

2. you can find its decimal equivalent with this formula decimal

Example:

1111 has 4 bits equivalent decimal =

1

11

111

1111

11111

111111

1111111

Convert decimal 23.6 to a binary number

Solution:

Solution:

2 23

2 11 – 1

2 5 - 1

2 2 - 0

1 - 0

23 = 10111

0.6 2 = 1.2 = 0.2 with a carry of 1

0.2 2 = 0.4 = 0.4 with a carry of 0

0.4 2 = 0.8 = 0.8 with a carry of 0

0.8 2 = 1.6 = 0.6 with a carry of 1

0.6 2 = 1.2 = 0.2 with a carry of 1

∴ 0.6 = 0.10011

Answer 23.60=10111.100110 A digit computer process 32 bits long binary number if a 32 bit number has all 1’s what is its decimal equivalent

Page 98 of 119

Solution: The decimal equivalent = Where n = 32

=

= 4294967295

Octal odometer In octal odometer each display wheel contains only eight digits, number 0 to 7 when

wheel turns from 7 back to zero, it sents a carry to the next higher level initially an octal odometer shows

After 1 km 0000

2 km 0001

3 km 0002

4 km 0003

5 km 0004

6 km 0005

7 km 0006

8 km 0007

9 km 0010

10 km 0012

11 km 0013

12 km 0014

13 km 0015

14 km 0016

15 km 0017

16 km 0020

17 km 0021

18 km 0022

19 km 0023

20 km 0024

21 km 0025

22 km 0026

23 km 0027

24 km 0030

Page 99 of 119

Octal to decimal conversion: In the octal number system each digits position corresponds to a power of 8 as follows

To convert from octal to decimal multiply each octal digit by its weight and add the resulting product Example 1: Convert octal 23 to decimal Solution: 0 0 2 3

3+16=19

Equivalent decimal = 19

∴ The decimal equivalent to 23 is 19

Example 2:

Convert octal 257 to decimal

Solution:

0 2 5 7

7

7+40+128=175 Equivalent decimal = 175 ∴ The decimal equivalent to 257 is 175 Decimal to octal conversation:

Octal dabble method similar to double dabble method is used with octal number. In this method we deived by eight writing down the reminder after each devision. The remider in reverse order form the octal number Fraction: With decimal fraction multiply by 8 writing the carry into the integer position Example 1: Convert the decimal 175 to octal

Page 100 of 119

Solution:

8 175

8 21 – 7

2 - 5

175 0257

Example 2: Convert the 0.23 to octal Solution: 0.23 8 = 1.84 = 0.84 with a carry of 1 0.84 8 = 6.72 = 0.72 with a carry of 6 0.72 8 = 5.76 = 0.76 with a carry of 5 0.76 8 = 6.08 = 0.08 with a carry of 6 0.08 8 = 0.64 = 0.64 with a carry of 0 The octal number of 0.23 is 0.16560 Problem 1: Octal to decimal 271 Solution: 2 7 1

1+56+128=185

The decimal number is 185 Problem 2: Octal to decimal 356 Solution: 3 5 6

6+40+192=238

The decimal number is 238

Problem 3:

10101101

Page 101 of 119

1 0 1 0 1 1 0 1

128 64 32 16 8 4 2 1 ⇒ 128+32+8+4+1=173


Problem 4:

111100111

1 1 1 1 0 0 1 1 1

256 128 64 32 16 8 4 2 1⇒ 256+128+64+32+4+2+1=487


Problem 5:

2 369

2 184 – 1

2 92 - 0

2 46 - 0

2 23 - 0

2 11 - 1

2 5 - 1

2 2 - 1

1 - 0

The binary number is = 101110001 Number system and codes Decimal numbers The number system contains 0,1,2,… are called decimal numbers Binary numbers: The binary number system that uses only the 0 and 1 as codes are other digits thrown away. To represent decimal numbers and letters of the alphabet with binary code you have to use different things of binary digits for each number or letter The word with is the abbreviation for binary digit when a binary number has four bits then it is called nibite. A binary number with 8 bits is known as Byte Binary to decimal conversion

1. Write the binary number 2. Directly under the binary number write 1,2,4,8,16,…. Working from right to left 3. If a zero appears in a digits position cross out the decimal weight for that position 4. Add the remaining weight to obtained the decimal equivalence

Example 1:

10010

Page 102 of 119

Solution :

1 0 0 1 0

16 8 4 2 1 ⇒ 16+2=18

18=10010

Example 2:

10101

Solution :

1 0 1 0 1

16 8 4 2 1 ⇒ 16+4+1=21

21=10101

Example 3:

1101

Solution :

1 1 0 1

8 4 2 1 ⇒ 8+4+1=13

13=1101

Fractions :

Example 1:

0.101

Solution :

0. 1 0 1

0.5 0.25 0.125 ⇒ 0.5+0.125=0.625

Example 2:

110.0101

Solution :

1 1 0

4 2 1 ⇒ 4+2=6

0 1 0 1

0.5 0.25 0.125 0.625 ⇒ 0.25+0.625=0.3125

110.0101=6.3125

Example 3:

1011.11

Page 103 of 119

Solution :

1 0 1 1

8 4 2 1 ⇒ 8+2+1=11

1 1

0.5 0.25 ⇒ 0.5+0.25=0.75 1011.11=11.75 Example 4: 1011.001 Solution : 1 0 1 1

8 4 2 1 ⇒ 8+2+1=11

0 0 1

0.5 0.25 0.125 ⇒ 0.125

1011.11=11.125

Decimal to Binary Conversion: Example 1: 18 Solution: 2 18

2 9 – 0

2 4 - 1

2 2 - 0

1 - 0

18 = 10010

Example 2:

21

Solution:

2 21

2 10 – 1

2 5 - 0

2 2 - 1

1 - 0

21 = 10101

Example 3:

Page 104 of 119

13

Solution:

2 13

2 6 – 1

2 3 - 0

1 - 1

13 = 1101

Fractions:

Example 1: 0.6 Solution: 0.6 2 = 1.2 = 0.2 1 0.2 2 = 0.4 = 0.4 0 0.4 2 = 0.8 = 0.8 0 0.8 2 = 1.6 = 0.6 1 0.6 2 = 1.2 = 0.2 1 ∴ 0.6 = 0.10011 Example 1: 23.6 Solution: 2 23

2 11 – 1

2 5 - 1

2 2 - 1

1 - 0

23 = 10111 0.6 = 0.10011 ∴ 23.6 = 10111.10011

1. Convert decimal 28.86 to a binary number 2. What is the binary number for decimal 255 3. Let convert 0.96 to a binary number 4. Convert 1110.1010 to a decimal number

Page 105 of 119

Answer

28.86

2 28

2 14 – 0

2 7 - 0

2 3 - 1

1 - 1

28 = 11100

0.86 2 = 1.72 = 0.72 1

0.72 2 = 1.44 = 0.44 1

0.44 2 = 0.88 = 0.88 0

0.88 2 = 1.76 = 0.76 1

0.76 2 = 1.52 = 0.52 1

0.52 2 = 1.04 = 0.04 1

0.04 2 = 0.08 = 0.08 0

0.08 2 = 0.16 = 0.16 0

0.16 2 = 0.32 = 0.32 0

0.32 2 = 0.64 = 0.64 0

0.64 2 = 1.28 = 0.28 1

0.28 2 = 0.56 = 0.56 0

0.56 2 = 1.12 = 0.12 0

0.12 2 = 0.24 = 0.24 0

0.24 2 = 0.48 = 0.48 0

0.48 2 = 0.96 = 0.96 0

0.96 2 = 1.92 = 0.92 1

0.92 2 = 1.84 = 0.84 1

0.84 2 = 1.68 = 0.68 1

0.68 2 = 1.36 = 0.36 1

0.36 2 = 0.72 = 0.72 0

0.72 2 = 1.44 = 0.44 1

28.86 = 11100.1101110000001000111001

Page 106 of 119

Answer 2:

255

2 255

2 127 – 1

2 63 - 1

2 31 - 1

2 15 - 1

2 7 - 1

2 3 - 1

1 - 1

255 = 11111111

Answer

0.96

0.96 2 = 1.92 = 0.92 1

0.92 2 = 1.84 = 0.84 1

0.84 2 = 1.68 = 0.68 1

0.68 2 = 1.36 = 0.36 1

0.36 2 = 0.72 = 0.72 0

0.72 2 = 1.44 = 0.44 1

0.44 2 = 0.88 = 0.88 0

0.88 2 = 1.76 = 0.76 1

0.76 2 = 1.52 = 0.52 1

0.52 2 = 1.04 = 0.04 1

0.04 2 = 0.08 = 0.08 0

0.08 2 = 0.16 = 0.16 0

0.16 2 = 0.32 = 0.32 0

0.32 2 = 0.64 = 0.64 0

0.64 2 = 1.28 = 0.28 1

0.28 2 = 0.56 = 0.56 0

0.56 2 = 1.12 = 0.12 0

0.12 2 = 0.24 = 0.24 0

0.24 2 = 0.48 = 0.48 0

0.48 2 = 0.96 = 0.96 0

Page 107 of 119

0.96 2 = 1.92 = 0.92 1

∴0.96 = 111101011100001010001

Answer

1 1 1 0. 1 0 1 0

8 4 2 1 0.5 0.25 0.125 0.625⇒ 8+4+2. 0.5+0.125 = 14.625

∴1110.1010 = 14.625

Octal number The decimal number system has a base of 10, because it uses the digits 0 to 9 the binary

system has a base of 2 because it uses only the digits 0 and 1. Then the octal number system has a base of 8 although we can use any 8 digits to use the first 8 decimal digits then the no 8 or 9 in the octal number code then the digits 0 to 7 have exactly the same physical meaning as decimal symbol. Octal to decimal conversion: In the octal number system each digits position corresponds to a power of 8 as follows

Example 23

2 3

⇒

Problem 1: Convert the octal number 257 to a decimal number 2 5 7

⇒

Problem 2: Convert the octal number 27.26 to a decimal number 2 7 . 2 6

.

. +

23.

23.3437

Decimal to Octal conversion: Problem 1: Convert a decimal number 175 to an octal Solution: 8 175

8 21 – 7

Page 108 of 119

2 - 5

175 257

Problem 2: Convert a decimal number 363 to an octal Solution: 8 363

8 45 – 3

5-5

363 553

Fraction: (Decimal Octal) Problem 1: 0.23

0.23 8 = 1.84 = 0.84 1

0.84 8 = 6.72 = 0.72 6

0.72 8 = 5.76 = 0.76 5

0.76 8 = 6.08 = 0.08 6

0.08 8 = 0.64 = 0.64 0

0.23 = 0.16560

Problem 2: Convert 34.56 to an octal. 8 363

4 – 2

34 42

0.56 8 = 4.48 = 0.48 4

0.48 8 = 3.84 = 0.84 3

0.84 8 = 6.72 = 0.72 6

0.72 8 = 5.76 = 0.76 5

0.76 8 = 6.08 = 0.08 6

0.56 = 0.43656

∴34.56 = 42.43656

Octal Binary: Problem 1: Change octal 23 to its binary equivalent 2 3

Page 109 of 119

010 011

∴23 010011

Problem 2: Change octal 562 to its binary equivalent 5 6 2

101 110 010

∴562 101110010

Problem 3: Change octal 332 to its binary equivalent 3 3 2

011 011 010

∴332 011011010

Binary octal conversion Conversion from binary to octal is a reversal procedures simply remember to group the

bits in threes. Starting at the binary point. Then convert each group of three to its octal equivalent Example 1: 1011.01101 001 011 . 011 010

1 3 . 3 2

1011.01101 13.32

Example 2: Convert the Binary number 110011.01101 into an octal 110 011 . 011 010

6 3 . 3 2

110011.01101 63.32

Example 3: Convert the Binary number 1101.110110 into an octal

001 101 . 110 110

1 5 . 6 6

1101.110110 15.66

Hexadecimal number Hexadecimal number system has base of 16, although any 16 digits may be used,

everyone uses 0 to 9 and A to F in other words. After reaching 9 in the hexadecimal system continue as follows A,B,C,D,E,F.

Page 110 of 119

Decimal Binary Hexadecimal

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

0000

0001

0010

0011

0100

0101

0110

0111

1000

1001

1010

1011

1100

1101

1110

1111

0

1

2

3

4

5

6

7

8

9

A

B

C

D

E

F

Hexadecimal to Binary conversion: Problem 1: 9AF

9 A F

1001 1010 1111

∴ 9AF 100110101111

Problem 2:

C5E2

C 5 E 2

1100 0101 1110 0010

∴ C5E2 1100101011100010

Binary to Hexadecimal conversion

Problem 1:

10001100

1000 1100

Page 111 of 119

8 12

10001100 8C

Problem 2:

110111010001101

0110 1110 1000 1101

6 14 8 12

110111010001101 6E8C

Fraction:

Problem 1:

0110001111111.0011110

0000 1100 0111 1111 . 0011 1100

0 12 6 15 . 0.1875 0.75

C6F.9375

Hexadecimal to decimal conversion In the hexadecimal number system each digit position corresponds to a power of 16.

The weight of the digit position in the hexadecimal number are as follows

To convert from hexadecimal to decimal multiply each hexadecimal digit by its weight and add the resulting products note that Example 1: F8E6.39 convert Hexadecimal number to decimal F 8 E 6 . 3 9

.

61440+2048+224+6.22266

63718.2227

Example 2: 9AF.26 convert Hexadecimal number to decimal 9 A F . 2 6

.

2304+160+15.(0.125+0.0234)

2479.1484

Decimal to hexadecimal conversion: Problem 1: Convert a decimal number 2479 to Hexadecimal Solution:

Page 112 of 119

16 2479

16 154 – 15

9 - 10

2479 9AF

Problem 2:

Convert a decimal number 65535 to Hexadecimal

Solution:

16 65535

16 4095 – 15

16 255 - 15

15 - 15

65535 FFFF

Octal to binary (1332)

1 3 3 2

001 011 011 010

1332 001011011010

65535 FFFF

ASCII Code: To get information into and out of a computer we need to use some kind of alpha

numeric code (1 for letters numbers other symbols). An input, output conde known as American standard code for information Interchange (ASCII). This code allows to manufacture to standardize computer forward such as keyboard, printer and video displays The ASCII code is a 7 bit code whose format is where each X is a zero or one. The ASCII code for the upper case and lower case letters of the alphabet and some of the most commonly used symbols. Example 1: The capital letter A has of 100 and of 0001. Then letter a is called as 1100001 b 1100010

B 1000010

EBCDIC as alpha numeric code. The abbreviation of EBCDIC is extended binary coded decimal interchange code. It is an 8 bit code. And primarily used in IBM made devices. The bit assignments used in IBM made devices. The bit assignments of EBCDIC are different from the ASCII. But the character symbols are the same. The excess – 3 code is an important 4 bit code sometimes used with binary coded decimal numbers. To convert and decimal number into its excess – 3 code from add 3 to each decimal digit and then convert the sum to a BCD number

Page 113 of 119

Example :

2 3

+3 +3 5 6

0101 0110

Four cases remember The four cases are (i) 0 + 0 = 0 (ii) 0 + 1 = 1 (iii) 1 + 0 = 1 (iv) 1 + 1 = 10 Example 1: Add following the binary number 11100 and 11010 Solution:

11100

11010

110110

8 bit arthemetic

In this binary addition we can a column by column to find the sum of two binary numbers regard less of how long they may be addition is done an 8 bits numbers such as

Example 1: Add these 8 bits numbers 01010111 and 00110101 then show the same number in hexadecimal notation Solution: 0101 0111

0011 0101

1000 1100

0 1 0 1

⇒ 4+1 = 5

0 1 1 1

⇒ 4+2+1 = 7

0 0 1 1

⇒ 2+1 = 3

0 1 0 1

⇒ 4+1 = 5

5 7 H

3 5 H

Page 114 of 119

8 C H

8CH

The hexadecimal number is 8CH Example 2: Add these 16 bits numbers 0000 1111 1010 1100 and 0011 1000 0111 1111 then show the same number in hexadecimal and decimal Solution: 0000 1111 1010 1100 0011 1000 0111 1111

0100 1000 0110 1011

0 0 0 0 ⇒ 0

1 1 1 1

⇒ 8+4+2+1 = 15

1 0 1 0

⇒ 8+2=10

1 1 0 0

⇒ 8+4=12

0 0 1 1

⇒ 2+1=3

1 0 0 0

⇒ 8

0 1 1 1

⇒ 4+2+1=7

1 1 1 1

⇒ 8+4+2+1 = 15

⇒ 0100 1000 0010 1011

4 8 2 11

482B

0 F A C H

3 8 7 F H

4 8 2 B H

The hexadecimal number is 482BH Example 3: What does the meaning of and ?

Page 115 of 119

Solution : It has the base value 2

It is a binary number Then has a base value 10 ∴ It is a decimal number Binary Subtraction: We have four basic cases of binary subtraction (i) 0 - 0 = 0 (ii) 0 - 1 = 1 (iii) 10 - 1 = 1 (iv) 1 - 1 = 0 Example 1: 1101 – 1010 Solution:

1 1 0 1

1 0 1 0

0 0 1 1

Example 2:

0111-0101

Solution:

0 1 1 1

0 1 0 1

0 0 1 0

Example 3:

0101-0011

Solution:

0 1 0 1

0 0 1 1

0 0 1 0

Example 4: Solve the binary subtraction of 1100 1000 – 0111 1101 Solution: 1100 1000 0111 1101 0100 1011

Example 5:Show how to subtract 8510 from 15010 with unsigned 8 bit number

150

2 150

2 75 – 0

2 37 - 1

2 18 - 1

Page 116 of 119

2 9 - 0

2 4 - 1

2 2 - 0

1 - 0

2 85

2 42 – 1

2 21 - 0

2 10 - 1

2 5 - 0

2 2 - 1

1-0

15010 binary number : 1001 0110 8510 binary number : 0101 0101 0100 0001

Hexadecimal:

1 0 0 1

⇒ 8+1=9

0 1 1 0

⇒ 4+2=6

0 1 0 1

⇒ 4+1=5

9 6 H

5 5 H

4 1 H

The hexadecimal in 41H Example 5: Show how to add 8510 from 15010 with unsigned 8 bit number 15010 binary number : 1001 0110 8510 binary number : 0101 0101 1110 1011

9 6 H

5 5 H

14 11 H ⇒ EBH

∴ The hexadecimal is EBH Binary Multiplication:

Page 117 of 119

In order words multiplication is done with addition inspection and division with subtraction instruction. An added – subtracter is all that is need for addition subtraction multiplication division (Ex.) Multipication is equivalent to repeated addition given to problem such that as The first no is called multiplication and the second no is multipier. Multiplying 8 by 4 is the same as addition 8 four times 8+8+8+=?. One way to multiply 8 by 4 is programming a computer add 8 untill the total of 4 “8’s have been added this approach is known as program multiplication by repeated addition

1 0 0 0

1 0 0 0

10 0 0 0

10 0 0 0

10 0 0 0 0

Problem 1: Give the sum each of the following 38+78=? The equivalent binary value of 3 is 011 The equivalent binary value of 7 is 111 0 1 1 1 1 1 0 1 0 1

Problem 2: Give the sum each of the following 816+F16=? The equivalent binary value of 8 is 1000 The equivalent binary value of 15 is 1111 1 0 0 0

1 1 1 1

10 1 1 1

The binary addition of 816+F16=10111 Problem 3: Give the sum each of the following 58+68=? The equivalent binary value of 5 is 101 The equivalent binary value of 6 is 110 1 0 1

1 1 0

10 1 1

The binary addition of 58+68=1001

Page 118 of 119

Problem 4: Give the sum each of the following 416+C16=? The equivalent binary value of 4 is 0100 The equivalent binary value of 12 is 1100 0 1 0 0

1 1 0 0

10 0 0 0

The binary addition of 416+C16=10000 Workout each of these binary sums: Problem 1: 0000 1111 + 0011 0111 0000 1111 0011 0111 0100 0110

Problem 2: 0001 0100 + 0010 1001 0001 0100 0010 1001 0011 1101 Problem 3: 0001 1000 1111 0110 + 0000 1111 0000 1000 0001 1000 1111 0110 0000 1111 0000 1000 0010 0111 1111 1110 The binary addition is 0010 0111 1111 1110 Subtract the following 0100 1111 – 0000 0101 0100 1111 0000 0101 0100 1010 Problem 1: Show the binary addition of 75010+53810 using the 16 bits number The equivalent binary value of 750 is 0000 0010 1110 1110 The equivalent binary value of 538 is 0000 0010 0001 1010 0000 0010 1110 1110

0000 0010 0001 1010

0000 0101 0000 1000

The hexadecimal is

02EFH

021AH

Page 119 of 119

0508H

Problem 2:

Show the subtraction binary form 4710+2310 = ?

2 47

2 23 – 1

2 11 - 1

2 5 - 1

2 2 - 1

1 - 0

2 23

2 11 – 1

2 5 - 1

2 2 - 1

1 - 0

The equivalent binary value of 47 is 0010 1111 The equivalent binary value of 23 is 0001 0111 0010 1111

0001 0111

0001 1000

The binary subtraction is

0001 1000

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