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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
STUDY MATERIAL FOR BSC MATHEMATICS DISCRETE MATHEMATICS
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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:
STUDY MATERIAL FOR BSC MATHEMATICS DISCRETE MATHEMATICS
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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
The given formula is independent of their components
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
The given formula is independent of their components
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:
Let us form a truth table
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
In the result column all the values are true and so that the given proposition is a tautologies
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
In the result column all the values are true and so that the given proposition is a tautologies
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
Let us form a truth table
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)
Let us form a truth table
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)
Let us form a truth table
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)
Let us form a truth table
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
Let us form a truth table
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
Let us form a truth table
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
Let us form a truth table
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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)
Let us form a truth table
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
Let us form a truth table
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:
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Show that Q P⇔¬P ¬Q
Proof
Let us form a truth table
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)]
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[¬ (¬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
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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
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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)
∴Closure property is true a natural number
Associative property
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
Associative property
Let (To Prove )
Then, clearly
∴Associative property in true
Identity:
Let
Then, such that and so 0 is the identity element of
∴ 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)
∴Closure property is true a natural number
Associative property
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 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)
∴Closure property is true
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Associative property
Let (To Prove )
Then, clearly
∴Associative property in true
Identity:
Let
Then, such that But ∉ E
∴ E has no identity element
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 even numbers is again even number)
∴Closure property is true a natural number
Associative property
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 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 )
∴Closure property is true
Associative property
Let (To Prove )
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Then, clearly )
∴Associative property in true
Identity:
Let
Then,
∴ Clearly has no identity element
is a semi group but not a monoid
To Prove is a semi group and also a monoid
Closure Property
Let
Then clearly (⸪Intersection of any two subsets of is again a subset of
)
∴Closure property is true
Associative property
Let (To Prove )
Then, clearly )
∴Associative property in true
Identity:
Let
Then,
∴ Clearly has no identity element
is a semi group but not a monoid
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)
∴Closure property is true
Associative property
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Let (To Prove )
Then, clearly
∴Associative property in true
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
∴Closure property is true
Associative property
Let (To Prove )
Then, L.H.S =
=
=
…(1)
R.H.S =
=
…(2)
L.H.S=R.H.S (by (1) and (2))
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∴Associative property in true
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
∴Closure property is true
Associative property
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))
∴Associative property in true
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)
Associative property
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
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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)
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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
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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
Associative property
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)
a distinguished element such that
(Identity)
here, is the identity element under addition.
: such that
(inverse)
: for any
(Associative)
: for any
(Commutative)
a distinguished element such that
here, is the identity element with respect to multiplication.
For any
(Distributive)
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The operation distributes over
For any and
(cancellation law)
b) All even integers
Let E be the set of all even integers for any
(Associative)
: for any
(Commutative)
a distinguished element such that
(Identity)
here, is the identity element under addition.
: such that
(inverse)
: for any
(Associative)
: for any
(Commutative)
a distinguished element such that
here, is the identity element with respect to multiplication.
For any
(Distributive)
The operation distributes over
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)
a distinguished element such that
(Identity)
here, is the identity element under addition.
: such that
(inverse)
: for any
(Associative)
: for any
(Commutative)
a distinguished element such that
here, is the identity element with respect to multiplication.
For any
(Distributive)
The operation distributes over
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
Associative property is true
:
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
The operation distributes over
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
Associative property is true
: 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
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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
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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
◇
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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
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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
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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
STUDY MATERIAL FOR BSC MATHEMATICS DISCRETE MATHEMATICS
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Page 70 of 119
∴
Hence k is a normal subgroup of G
Problem
is an abelian group
Proof:
Closure Property
Let
Then clearly
∴Closure property is true
Associative property
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
STUDY MATERIAL FOR BSC MATHEMATICS DISCRETE MATHEMATICS
SEMESTER - V, ACADEMIC YEAR 2020 - 21
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
STUDY MATERIAL FOR BSC MATHEMATICS DISCRETE MATHEMATICS
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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 :
STUDY MATERIAL FOR BSC MATHEMATICS DISCRETE MATHEMATICS
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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
Then the number of 1’s in
c) Let
Then the number of 1’s in
d) Let
Then the number of 1’s in
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
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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
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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
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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
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{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}
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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
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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
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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
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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)
From (3) and (4) we get
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)
(by associative law)
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(by idompotant law)
(⸪
Hence (i) is proved
(ii)To prove
Enough to prove
(by associative law)
(by commutative law)
(by associative 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)
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(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
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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
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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)
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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
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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
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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
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(by distributive law)
hence proved b
c)
(by distributive law)
(by absorption and distributive law)
(by absorption law)
d)
(by distributive law)
(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)
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c)
d)
Proof:
Let to prove
Conversely:
Let To prove
b) let …(2) To prove
(
Conversly :
Let To Prove
c) Let
…(3)
…(4)
From (3) and (4) we get
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
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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
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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
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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
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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
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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
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1 0 1 0 1 1 0 1
128 64 32 16 8 4 2 1 ⇒ 128+32+8+4+1=173
The decimal number is 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
The decimal number is 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
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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
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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:
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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
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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
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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
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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
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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
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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.
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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
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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:
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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
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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
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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 ?
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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
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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:
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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
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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
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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