MCA 1st sem /MCS-013/Solved Assignment/ Discrete Mathematics /2017-2018 New

Question.1.(A)

Ans a . Logical Connectivity:- 

In logic, a logical connective (also called a logical operator) is a symbol or word used to connect two or more sentences (of either a formal or a natural language) in a grammatically valid way, such that the sense of the compound sentence produced depends only on the original sentences. The most common logical connectives are binary connectives (also called dyadic connectives) which join two sentences which can be thought of as the function's operands. Also commonly, negation is considered to be a unary connective.

Different Types of Logical Connectivity:-

 Disjunction :-

Definition: The disjunction of two propositions p and q is the compound statement p or q, denoted by p ∨ q.

For example, ‘Zarina has written a book or Singh has written a book.’ Is the

disjunction of p and q, where

p : Zarina has written a book, and

q : Singh has written a book.

Similarly, if p denotes ‘ 2 > 0’ and q denotes ‘2 < 5’, then p ∨ q denotes the statement

‘2 is greater than 0 or 2 is less than 5.’.

Let us consider an example.

Example 1: Obtain the truth value of the disjunction of ‘The earth is flat’.

and ‘3 + 5 = 2’.

Solution: Let p denote ‘The earth is flat,’ and q denote ‘3 + 5 = 2’. Then we know

that the truth values of both p and q are F. Therefore, the truth value of p ∨ q is F.

Conjunction:-

As in ordinary language, we use ‘and’ to combine simple propositions to make

compound ones. For instance, ‘ 1 + 4 ≠ 5 and Prof. Rao teaches Chemistry.’ is

formed by joining ‘1 + 4 ≠ 5’ and ‘Prof. Rao teaches Chemistry’ by ‘and’. Let us

define the formal terminology for such a compound statement.

Definition: 

We call the compound statement ‘p and q’ the conjunction of the

statements p and q. We denote this by p ∧ q.

For instance, ‘3 + 1 ≠ 7 ∧ 2 > 0’ is the conjunction of ‘3 + 1 ≠ 7’ and ‘2 > 0’.

Similarly, ‘2 + 1 = 3 ∧ 3 = 5’ is the conjunction of ‘2 + 1 = 3’ and ‘3 = 5’.

Now, when would p ∧ q be true? Do you agree that this could happen only when both

p and q are true, and not otherwise? For instance, ‘2 + 1 = 3 ∧ 3 = 5’ is not true

because ‘3 = 5’ is false.

So, the truth table for conjunction.

Negation:-

Definition: The negation of a proposition p is ‘not p’, denoted by ~p.

For example, if p is ‘Dolly is at the study center.’, then ~ p is ‘Dolly is not at the study

center’. Similarly, if p is ‘No person can live without oxygen.’, ~ p is ‘At least one

person can live without oxygen.’.

Conditional Connectives:-

Consider the proposition ‘If Shruti gets 75% or more in the examination, then she

will get an A grade for the course.’. We can write this statement as ‘If p, and q’,

where

p: Shruti gets 75% or more in the examination, and

q: Shruti will get an A grade for the course.

This compound statement is an example of the implication of q by p.

Definition: Given any two propositions p and q, we denote the statement ‘If p, then

q’ by p → q. We also read this as ‘p implies q’. or ‘p is sufficient for q’, or ‘p only if

q’. We also call p the hypothesis and q the conclusion. Further, a statement of the

form p → q is called a conditional statement or a conditional proposition.

Precedence Rule

While dealing with operations on numbers, you would have realized the need for

applying the BODMAS rule. According to this rule, when calculating the value of an

arithmetic expression, we first calculate the value of the Bracketed portion, then apply

Of, Division, Multiplication, Addition and Subtraction, in this order. While

calculating the truth value of compound propositions involving more than one

connective, we have a similar convention which tells us which connective to apply

first.

Why do we need such a convention? Suppose we didn’t have an order of preference,

and want to find the truth of, say ~ p ∨ q. Some of us may consider the value of ( ~

p) ∨ q, and some may consider ~ ( p ∨ q). The truth values can be different in

these cases. For instance, if p and q are both true, then ( ~ p) ∨ q is true, but ~ (

p ∨ q) is false. So, for the purpose of unambiguity, we agree to such an order or

rule.

The rule of precedence: The order of preference in which the connectives are

applied in a formula of propositions that has no brackets is

:-

i) ~

ii) ∧

iii) ∨ and ⊕

iv) → and ↔

Q.1.(b)

A.1(b)

i)  

  

  

ii)

Q.1.(c)

A.1.(c)

(i)

(ii)

Q.1.(d)

A.1.(d) Logical Equivalence:-

Logical equivalence is a type of relationship between two statements or sentences in propositional logic or Boolean algebra. The relation translates verbally into "if and only if" and is symbolized by a double-lined, double arrow pointing to the left and right ( ). If A and B represent statements, then A B means "A if and only if B."

The statement A B is exactly the same as

(A B) * (B A)

where the asterisk (*) represents the logical AND operation, and the right-pointing, double-lined arrow ( ) represents logical implication

Question.2.(A)

Ans a . 

(i) (∃x) ( ∀y) ( ∃z) P 

(ii) ( ∀x) ( ∃y) (∃ z) P 

(B)

Ans b.

 (i)  Some students can pass in exam. 

∀x(C(x)∧P(x))

Then you say that everything in the domain is a student in the course and passed the exam.

However, presumably there are things in the domain other than students in the course. For example, that book that the student didn't read, or the exam that they took. So if they are in the domain, you would get that the book passed the exam, or that the exam is a student in the course ... which is not what you want. What you want is that if something is a student in the course, then they passed the exam, i.e. you want the conditional.

Of course, it is possible that the domain is simply all the students in the course, but then why would you need a predicate C(x)C(x)? If the domain was all students in the course, you could simply use ∃x¬B(x)∃x¬B(x) for the first premise, and ∀xP(x)∀xP(x) for the second.

So, either way, ∀(C(x)∧P(x))

∀(C(x)∧P(x)) is not what you want.

(ii) Everything is having life. 

∀x(L(x)∧ E(x))

We can say that Everything is having Life. 

and In another way , we can say that, 

~ (∀x(C(x)∧P(x))), that 's mean Nothing is having Life.

(C)

Ans c.

Definition Of Indirect Proof

Indirect proof is a type of proof in which a statement to be proved is assumed false and if the assumption leads to an impossibility, then the statement assumed false has been proved to be true.

Example of Indirect Proof

Sum of 2n even numbers is even, where n > 0. Prove the statement using an indirect proof. 

The first step of an indirect proof is to assume that 'Sum of even integers is odd.' 

That is, 2 + 4 + 6 + 8 + . . . .+ 2n = an odd number

⇒ 2(1 + 2 + 3 + 4 + . . . + n) = an odd number ⇒ 2 × = an odd number

⇒ n(n + 1) = an odd number, a contradiction, because n(n + 1) is always an even number. 

Thus, the statement is proved using an indirect proof.

(D)

Ans d.

An equivalence relation on a set S, is a relation on S which is reflexive, symmetric and transitive. Examples: Let S = ℤ and define R = {(x,y) | x and y have the same parity} i.e., x and y are either both even or both odd. The parity relation is an equivalence relation. 1. For any x ∈ ℤ, x has the same parity as itself, so (x,x) ∈ R. 2. If (x,y) ∈ R, x and y have the same parity, so (y,x) ∈ R. 3. If (x,y) ∈ R, and (y,z) ∈ R, then x and z have the same parity as y, so they have the same parity as each other (if y is odd, both x and z are odd; if y is even, both x and z are even), thus (x,z) ∈ R.

FOR EXAMPLE

Let S = ℝ and define the "square" relation R = {(x,y) | x 2 = y 2}. The square relation is an equivalence relation. 1. For all x ∈ ℝ, x 2 = x 2 , so (x,x) ∈ R. 2. If (x,y) ∈ R, x 2 = y 2 , so y 2 = x 2 and (y,x) ∈ R. 3. If (x,y) ∈ R and (y,z) ∈ R then x 2 = y 2 = z2 , so (x,z) ∈ R. For any set S, the identity relation on S, I S = {(x,x) | x ∈ S}. This is an equivalence relation for rather trivial reasons. 1. obvious 2. If (x,y) ∈ R then y = x, so (y,x) = (x,x) ∈ R. 3. If (x,y) ∈ R and (y,z) ∈ R then x = y = z, so (x,z) = (x,x) ∈ R.

Question.3.

a)

b)

ANS:- 

c)Explain De Morgan’s laws in relation to Boolean Algebra.

 Ans .DeMorgan’s Theorem is mainly used to solve the various Boolean algebra expressions.The Demorgan’s theorem defines the uniformity between the gate with same inverted input and output. It is used for implementing the basic gate operation likes NAND gate and NOR gate. The Demorgan’s theorem mostly used in digital programming and for making digital circuit diagrams. There are two DeMorgan’s Theorems.

Theorem 1

o    The left hand side (LHS) of this theorem represents a NAND gate with inputs A and B, whereas the right hand side (RHS) of the theorem represents an OR gate with inverted inputs.

o    This OR gate is called as Bubbled OR.

Table showing verification of the De Morgan's first theorem −

Theorem 2

o    The LHS of this theorem represents a NOR gate with inputs A and B, whereas the RHS represents an AND gate with inverted inputs.

o    This AND gate is called as Bubbled AND.

Table showing verification of the De Morgan's second theorem −

(d) What is principle of mathematical induction? Explain with the help of an example.

Ans :- Mathematical induction is a mathematical proof technique used to prove a given statement about any well-ordered set. Most commonly, it is used to establish statements for the set of all natural numbers.

Mathematical induction is a form of direct proof, usually done in two steps. When trying to prove a given statement for a set of natural numbers, the first step, known as the base case, is to prove the given statement for the first natural number. The second step, known as the inductive step, is to prove that, if the statement is assumed to be true for any one natural number, then it must be true for the next natural number as well. Having proved these two steps, the rule of inference establishes the statement to be true for all natural numbers. In common terminology, using the stated approach is referred to as using the Principle of mathematical induction.

Example: 'The sum of the natural numbers from 1 to n equals n(n+1)/2'.

This is a property which depends on n. We write this fact as E(n). E(3) means in our example :

' The sum of the natural numbers from 1 to 3 equals 3(3+1)/2'.

A proof by induction is a common method to prove such a property. We show how the method works and why it is correct.

Let V = { n | E(n) is true }. First we prove E(1). Then we know that 1 is in V.

Then we prove that: If E(k) is true, then E(k+1) is true.

If that is proved, we know that: if k is in V then (k+1) is in V.

But 1 is in V, so 2 is in V, so 3 is in V, etc...

Then V = { 1,2,3,4,....}

We have shown that the property E(n) is true for all natural numbers starting with 1.

In our example is E(n) is 1+2+3+4+5+...+ n = (1+n)n/2

First step:

We'll show the truth of E(1). We have to show: 1 = (1+1).1/2

We see immediately that both sides are equal. So, E(1) is true.

V contains '1'.

Second step:

Assume for a moment that E (k) is true.

Given : 1+2+3+4+5+...+k = (1+k)k/2

We'll show the truth of E(k+1).

Proof:

    left side

  = (1+2+3+4+5+...+k)+ (k+1)

  = (1+k).k/2 + (k+1)

  = (k+1)(k+2)/2

    right side

  = (1+(k+1)).(k+1)/2

  = (2+k)(k+1)/2

We see that the left side is equal to the right side.

Thus if E (k) is true then E(k+1) is true.

If k is in V, then k+1 is in V.

But 1 is in V, so 2 is in V, so 3 is in V, etc..

So, V = { 1,2,3,4,....}

Conclusion: The property is proved for all natural numbers starting from 1.

Question 4

(a)  How many different committees can be formed of 12 professionals, each containing at least 2 Professors, at least 3 Lecturers and 3 Administrative Officers from a set of 5 Professors, 8 Lectures and 5 Administrative Officers.

(b)   

PROFESSOR                                         2

 Lecturers                                                 3 

Administrative Officers                          3                                                                  

SOLUTION IS

C(5,2)  + C(8,3)  +  C(5,3)

( 5X4)/(2X1) + (8X7X6)/(3X2X1) + (5X4X3)/(3X2X1)

10+56+10

76

 (b) There are two mutually exclusive events A and B with P(A) =0.7 and P(B) = 0.6. Find the probability of followings:

 i) A and B both occur

P(A) = 1-P(A)= 1-0.7=0.3
P(B) = 1-P(B)= 1-0.6=0.4

 ii) Both A and B does not occur 

P(A’)=1-P(A)=1-0.7=0.3

P(B’)=1-P(B)=1-0.6=0.4

iii) Either A or B does not occur

P(A’)=1-P(A)=1-0.7=0.3

P(B’)=1-P(B)=1-0.6=0.4

P(A’ OR B’) = P(A’) + P(B’) - P(A’ AND B’) =0.3 + 0.4 - 0.3 = 0.4

(c) What is set? Explain the basic properties of sets.

First we specify a common property among "things" (we define this word later) and then we gather up all the "things" that have this common property.

For example, the items you wear: shoes, socks, hat, shirt, pants, and so on.

I'm sure you could come up with at least a hundred.

This is known as a set.

Or another example is types of fingers. This set includes index, middle, ring, and pinky.

So it is just things grouped together with a certain property in common.

Sets are the fundamental property of mathematics. Now as a word of warning, sets, by themselves, seem pretty pointless. But it's only when we apply sets in different situations do they become the powerful building block of mathematics that they are.

Question 5

(a) How many words can be formed using letter of UNIVERSITY using each letter at most once?

 i) If each letter must be used,

{U,N,I,V,E,R,S,T,Y}

n=9

NO’S OF WORD IS = 9!

9!= 7X6X5X4X3X2X1= 363880 WORD ( in the UNIVERSITY letter I two time be we count once time)

 ii) If some or all the letters may be omitted. 

=P(9,0) + p(9,1)+P(9,2) + p(9,3)+P(9,4) + p(9,5)+P(9,6) + p(9,7)

= 1 + 9 + (9x8) + (9x8x7) + (9x8x7x6) +(9x8x7x6x5) +(9x8x7x6x5x4) +(9x8x7x6x5x4x3)

=1+7+72 +504+3024 +15120 + 60480+181440

= 260648

b)Show using truth table that: ( p ® q) ® q Þ p Ú q

(c) Explain whether (p q) (q r) is a tautology or not.

(d) Explain addition theorem in probability

The addition theorem in the Probability concept is the process of determination of the probability that either event ‘A’ or event ‘B’ occurs or both occur. The notation between two events ‘A’ and ‘B’ the addition is denoted as '∪' and pronounced as Union.

The result of this addition theorem generally written using Set notation,

P (A ∪ B) = P(A) + P(B) – P(A ∩ B),

Where,

P (A) = probability of occurrence of event ‘A’

P (B) = probability of occurrence of event ‘B’

P (A ∪ B) = probability of occurrence of event ‘A’ or event ‘B’.

P (A ∩ B) = probability of occurrence of event ‘A’ or event ‘B’.

Addition theorem probability can be defined and proved as follows:

Let ‘A’ and ‘B’ are Subsets of a finite non empty set ‘S’ then according to the addition rule

P (A ∪ B) = P (A) + P (B) – P (A). P (B),

On dividing both sides by P(S), we get

P (A ∪ B) / P(S) = P (A) / P(S) + P (B) / P(S) – P (A ∩ B) / P(S) (1).

If the events ‘A’ and ‘B’ correspond to the two events ‘A’ and ‘B’ of a random experiment and if the set ‘S’ corresponds to the Sample Space ‘S’ of the experiment then the equation (1) becomes

P (A ∪ B) = P (A) + P (B) – P (A). P (B),

This equation is known as the addition theorem in probability.

Here the event A ∪ B refers to the meaning that either event ‘A’ or event ‘B’ occurs or both may occur simultaneously.

If two events A and B are Mutually Exclusive Events then A ∩ B = ф,

Therefore

P (A ∪ B) = P (A) + P(B) [since P(A ∩ B) = 0],

In language of set theory A ∩ B̅ is same as A / B.

(e) Prove that the inverse of one-one onto mapping is unique.

A function is onto if and only if for every yy in the codomain, there is an xx in the domain such that f(x)=yf(x)=y.

So in the example you give, f:R→R,f(x)=5x+2f:R→R,f(x)=5x+2, the domain and codomain are the same set: R.R. Since, for every real number y∈R,y∈R, there is an x∈Rx∈R such that f(x)=yf(x)=y, the function is onto. The example you include shows an explicit way to determine which xx maps to a particular yy, by solving for xx in terms of y.y. That way, we can pick any yy, solve for f′(y)=xf′(y)=x, and know the value of xx which the original function maps to that yy.

Side note:

Note that f′(y)=f−1(x)f′(y)=f−1(x) when we swap variables. We are guaranteed that every function ff that is onto and one-to-one has an inverse f−1f−1, a function such that f(f−1(x))=f−1(f(x))=xf(f−1(x))=f−1(f(x))=x.

Question 6

(a) How many ways are there to distribute 15 district objects into 5 distinct boxes with

 i) At least three empty box.

C(5,3) X P(15,2)X2¹³  +  C(5,4) X P(15,1)X1¹⁴  + C(5,5) X P(15,0)X0¹⁵

ii) No empty box. 

C(15,5) x 5⁵ x5!

(b) Explain principle of multiplication with an example. 

ANS:-Multiplication Principle states: If an event occurs in m ways and another event occurs independently in n ways, then the two events can occur in m × n ways.

Examples of Multiplication Principle

A pizza corner sells pizza in 3 sizes with 3 different toppings. If Bob wants to pick one pizza with one topping, there is a possibility of 9 combinations as the total number of outcomes is equal to Number of sizes of pizza × Number of different toppings.

(c)  Set A,B and C are: A = {1, 2, 3,5, 8, 11 12,13}, B = { 1,2, 3 ,4, 5,6 } and C={ 7,8,12, 13}. Find A B C , A B C, A B C and (B~C) 

A∩ B U C ={1,2,3,5,7,8,12,13}

A U B U C={ 1,2,3,4,5,6,7,8,11,12,13}

A U B ∩ C= {8,12,13}

(B~C)= { 1,2, 3 ,4, 5,6 }

(d) In a class of 40 students; 30 have taken science; 20 have taken mathematics and 8 has neither taken mathematic nor science. Find how many students have taken:

 i) both subjects.

Ans :-  

(i)

 Science(s) = 30

 Math(m) = 20

 and 8 student has neither taken mathematic nor science

Then ,

n(S) +n(M) - n(Total student) 

= 30+20 - 40

=10

ii) exactly one subject

Ans :- 20+10 = 30

Question 7 

(a) What is power set? Write power set of set A={1,2,5,6,7,9}.  

Ans (a)  {0}, {1}, {2}, {3}, {0,1}, {0,2}, {0,3} , {1,2}, {1,3}, {2,3}, {0,1,2}, {0,1,3},{1,2,3}, {0,2,3}, {0,1,2,3}

 P = {{}, {0}, {1}, {2}, {3}, {0,1}, {0,2}, {0,3} , {1,2}, {1,3}, {2,3}, {0,1,2}, {0,1,3},{1,2,3}, {0,2,3}, {0,1,2,3}}...{1,2,5,6,7},{2,5,6,7,9}........

Like that, the possible Power Set A = 64

(b) Draw truth table for (P®Q) n (Q®P) and explain whether it is a tautology or not

Ans (b) Same as above example .. Q5(c)

(c) What is a function? Explain domain and range in context of function, with the help of example 

Ans (c) 

Domain

The domain of a function is the complete set of possible values of the independent variable.

In plain English, this definition means:

The domain is the set of all possible x-values which will make the function "work", and will output real y-values.

When finding the domain, remember:

The denominator (bottom) of a fraction cannot be zero

The number under a square root sign must be positive in this section

Example 1 Here is the graph of \displaystyle{y}=\sqrt{{{x}+{4}}}

y= x+4:

(d) State and prove the Pigeonhole principle.

Ans (d) 

Suppose that a flock of 20 pigeons flies into a set of 19 pigeonholes to roost. Because there are
20 pigeons but only 19 pigeonholes, a least one of these 19 pigeonholes must have at least two
pigeons in it. To see why this is true, note that if each pigeonhole had at most one pigeon in it,
at most 19 pigeons, one per hole, could be accommodated. This illustrates a general principle
called the pigeonhole principle, which states that if there are more pigeons than pigeonholes,
then there must be at least one pigeonhole with at least two pigeons in it.

Theorem:

I) If “A” is the average number of pigeons per hole, where A is not an integer then

• At least one pigeon hole contains ceil[A] (smallest integer greater than or equal to A) pigeons
• Remaining pigeon holes contains at most floor[A] (largest integer less than or equal to A) pigeons

Or

II) We can say as, if n + 1 objects are put into n boxes, then at least one box contains two or more objects.
The abstract formulation of the principle: Let X and Y be finite sets and let f: X –> Y be a function.

• If X has more elements than Y, then f is not one-to-one.
• If X and Y have the same number of elements and f is onto, then f is one-to-one.
• If X and Y have the same number of elements and f is one-to-one, then f is onto.

Pigeonhole principle is one of the simplest but most useful ideas in mathematics. We will see more applications that proof of this theorem.

Example 1: If (Kn+1) pigeons are kept in n pigeon holes where K is a positive integer, what is the average no. of pigeons per pigeon hole?

Solution: average number of pigeons per hole = (Kn+1)/n
= K + 1/n
Therefore at least a pigeonholes contains (K+1) pigeons i.e., ceil[K +1/n] and remaining contain at most K i.e., floor[k+1/n] pigeons.
i.e., the minimum number of pigeons required to ensure that at least one pigeon hole contains (K+1) pigeons is (Kn+1).

Question 8 

(a)Find inverse of the following functions.

Ans (a)