IGNOU BCA 2 SEM Assignment MCS-013

Course Code : MCS-013

Course Title : Discrete Mathematics

Assignment Number : MCA(1)/013/Assign/2012

Assignment Marks : 100

Last Date of Submission : 30th April,2012

There are eight questions in this assignment, which carries 80 marks. Rest 20 marks are

for viva-voce. Answer all the questions. You may use illustrations and diagrams to enhance

the explanations. Please go through the guidelines regarding assignments given in the

Programme Guide for the format of presentation.

Question 1:

Marks ( 4 + 4 +4)

a) Make truth table for

i) ~p→(q  ~ r)  ~p  q

Ans:

p q r ~p ~r (q V ~r) (~p ^ q) (q V ~r) ^ (~p ^ q) ~p→(q  ~ r)  ~p  q

0 0 0 1 1 1 0 0 0

0 0 1 1 0 0 0 0 0

0 1 0 1 1 1 1 1 1

0 1 1 1 0 1 1 1 1

1 0 0 0 1 1 0 0 1

1 0 1 0 0 0 0 0 1

1 1 0 0 1 1 0 0 1

1 1 1 0 0 1 0 0 1

i) ~p→r  ~q  ~p  ~r

p q r ~p ~q ~r (~q ^ ~p) [r V(~q ^ ~p)] [(r V (~q ^ ~p)) V ~r] ~p→r  ~q  ~p  ~r

0 0 0 1 1 1 1 1 1 1

0 0 1 1 1 0 1 1 1 1

0 1 0 1 0 1 0 0 1 1

0 1 1 1 0 0 0 1 1 1

1 0 0 0 1 1 0 0 1 0

1 0 1 0 1 0 0 1 1 0

1 1 0 0 0 1 0 0 1 0

1 1 1 0 0 0 0 1 1 0

b) What are conditional connectives? Explain use of conditional connectives with an example.

Ans:

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.

So, for example, in the conditional proposition ‘If m is in Z, then m belongs to Q.’ the hypothesis is ‘m ∈

Z’ and the conclusion is ‘m ∈ Q’.

Mathematically, we can write this statement as

m ∈ Z → m ∈ Q.

Let us analyse the statement p → q for its truth value. Do you agree with the truth table we’ve given

below (Table 3)? You may like to check it out while keeping an example from your surroundings in mind.

Table 3: Truth table for implication

p q p->q

T T T

T F F

F T T

F F T

Eg

‘If Ayesha 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: Ayesha gets 75% or more in the examination, and

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

therefore, P->Q

c) Write down suitable mathematical statement that can be represented by the following symbolic

properties.

i) (  x) (  y) (  z) P

ii)  (x) (  y) (  z) P

Question 2:

Marks (4 + 4)

a) What is proof? Explain method of direct proof with the help of one example.

Ans :

A proof of a proposition p is a mathematical argument consisting of a sequence

of statements p1 , p2 , · · · , pn from which p logically follows. So, p is the conclusion of this

argument.

The statement that is proved to be true is called a theorem.

Direct Proof

This form of proof is based entirely on modus ponens. Let us formally spell out the strategy.

Definition: A direct proof of p ⇒ q is a logically valid argument that begins with the

assumptions that p is true and, in one or more applications of the law of detachment, concludes

that q must be true.

So, to construct a direct proof of p ⇒ q, we start by assuming that p is true. Then, in one or more

steps of the form p ⇒ q1, q1⇒ q2, ……., qn ⇒ q, we conclude that q is true. Consider the following

examples

Example : Give a direct proof of the statement ‘The product of two odd integers is odd’.

Solution: Let us clearly analyse what our hypotheses are, and what we have to prove.

We start by considering any two odd integers x and y. So our hypothesis is p: x and y are odd.

The conclusion we want to reach is

q: xy is odd.

Let us first prove that p ⇒ q.

Since x is odd, x = 2m + 1 for some integer m.

Similarly, y = 2n + 1 for some integer n.

Then xy = (2m + 1) (2n + 1) = 2(2mn + m + n) +1

Therefore, xy is odd.

So we have shown that p ⇒ q.

Now we can apply modus ponens to p ∧ ( p ⇒ q) to get the required conclusion.

b) Show whether 17 is rational or irrational.

Ans:

Let us try and prove the given statement by contradiction. For this, we begin by assuming that

17 is rational. This means that there exist positive integers a and b such that 17 = a/b, where

a and b have no common factors.

This implies a = 17 b

⇒ a

2

= 17b

2

⇒ 17 | a

2

⇒ 17 | a.

Therefore, by definition, a = 17c for some c ∈ Z.

Therefore, a

2

= 289c

2

.

But a

2

= 17b

2

also.

So 289c

2

= 17b

2

⇒ 17c

2

= b

2

⇒ 17 | b

2

⇒ 17 | b

But now we find that 17 divides both a and b, which contradicts our earlier assumption that a and

b have no common factor.

Therefore, we conclude that our assumption that 17 is rational is false, i.e, 17 is irrational.

Question 3:

Marks (5 + 5)

a) What is Boolean algebra? Explain how Boolean algebra methods are used in logic circuit design.

Ans :

Definition: A Boolean algebra B is an algebraic structure which consists of a set X (≠ Ø) having

two binary operations (denoted by ∨ and ∧), one unary operation (denoted by ' ) and two specially

defined elements O and I (say), which satisfy the following five laws for all x, y, z ∈ X.

B1. Associative Laws: x ∨ (y ∨ z) = (x ∨ y) ∨ z,

x ∧ (y ∧ z) = (x ∧ y) ∧ z

B2. Commutative Laws: x ∨ y = y ∨ x,

x ∧ y = y ∧ x

B3. Distributive Laws: x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z),

x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z)

B4. Identity Laws: x ∨ O = x,

x ∧ I = x

B5. Complementation Laws: x ∧ x' = O,

x ∨ x'= I.

We write this algebraic structure as B = (X, ∨, ∧, ' , O, I), or simply B, if the context makes the

meaning of the other terms clear. The two operations ∨ and ∧ are called the join operation and

meet operation, respectively. The unary operation ' is called the complementation.

Boolean algebra methods to circuit design.

While expressing circuits mathematically, we identify each circuit in terms of some Boolean variables.

Each of these variables represents either a simple switch or an input to some electronic switch.

Definition: Let B = (S, ∨, ∧, ', O, I) be a Boolean algebra. A Boolean expression in variables

x1, x2 , . . . , xk (say), each taking their values in the set S is defined recursively as follows:

i) Each of the variables x1 , x2 , . . . , xk , as well as the elements O and I of the Boolean algebra B are

Boolean expressions.

ii) If X1 and X2 are previously defined Boolean expressions, then X1 ∧ X2 , X1 ∨ X 2 and X'1 are also

Boolean expressions.

For instance, x1 ∧ x'3 is a Boolean expression because so are x1 and x'3 , Similarly,

because x1 ∧ x2 is a Boolean expression, so is (x1 ∧ x2 ) ∧ (x1 ∧ x'3 ).

If X is a Boolean expression in n variables x1 , x2 , . . . , xn (say), we write this as

X = X(x1 , . . . , xn ) .

In the context of simplifying circuits, we need to reduce Boolean expressions to simpler ones.

`Simple' means that the expression has fewer connectives, and all the literals involved are distinct.

We illustrate this technique now.

Example : Reduce the following Boolean expressions to a simpler form.

X(x1, x2 ) = (x1 ∧ x2 ) ∧ (x1 ∧ x'2 );

Solution: (a) Here we can write

(x1 ∧ x2 ) ∧ (x1 ∧ x'2 ) = ((x1 ∧ x2 ) ∧ x1 ) ∧ x'2 (Associative law)

= (x1 ∧ x2) ∧ x'2 (Absorption law)

= x1 ∧ (x2 ∧ x'2 ) (Associative law)

= x1 ∧ O (Complementation law)

= O. (Identity law)

Thus, in its simplified form, the expression given in (a) above is O, i.e., a null expression.

b) If p and q are statements, show whether the statement [(~p→q)  (q)] → (p  ~q)

is a tautology or not.

Ans : Tautology :

A compound proposition that is true for all possible truth values of the simple propositions

involved in it is called a tautology.

p q ~p ~q ~p ->q [(~p→q)  (q)] (p  ~q) [(~p→q)  (q)] → (p  ~q)

0 0 1 1 0 0 1 1

0 1 1 0 1 1 0 0

1 0 0 1 1 0 1 1

1 1 0 0 1 1 1 1

Question 4:

Marks (4 + 4 +2)

a) Make logic circuit for the following Boolean expressions:

i) (x′.y + z) + (x+y+z)′ +(x+y+z)

X Y Z

O

ii) ( x'+y).(y′+ z).(y+z′+x′)

X Y Z

O

b) What is dual of a Boolean expression? Find dual of boolean expression of the

output of the following logic circuit:

Ans: If p is a proposition involving ~, ∧ and ∨, the dual of p, denoted by pd, is the proposition

obtained by replacing each occurrence of ∧ (and/or ∨) in p by ∨ (and/or ∧, respectively) in pd .

For example, x ∨ (x ∧ y) = x is the dual of x ∧(x ∨ y) = x.

The principle of duality: If s is a theorem about a Boolean algebra, then so is its dual sd .

Corresponding dual expressions of given figures:

Fig(a) : Given expression = [{(A’ + B) + B’}’ . C]’

Dual of expression = [{(A’. B) . B’}’ + C]’

Fig(b) : Given expression = [(A.B)+{(B’+C)}’]’

Dual of expression = [(A+B).{(B’.C)}’]’

c) Set A,B and C are:

A = {1, 2, 3, 4, 5,6,9,19,15}, B = { 1,2,5,22,33,99 } and C { 2, 5,11,19,15},

Find A  B C and A  B  C

Ans : A  B C => {1,2,5}U{2,5,11,19,15}

 {1,2,5,11,19,15}

A  B  C => {1,2,3,4,5,6,9,19,15,22,33,99} U {2,5,11,15,19}

 {1,2,3,4,5,6,9,11,15,19,22,33,99}

Question 5: Marks (3+4 +4)

a) Draw a Venn diagram to represent followings:

i) (A  B)  (C~B)

A B

C

U

ii) (A  B)  (B  C)

A B C

b) Give geometric representation for following

i) R x { 3}

321

-3-2-10123

-1

-2

-3

Y

X

ii) {-1, -2) x (-3, -3)

321 -

3

-

2

-

1

0

1

2

3

-1

-2

-3

Y

X

(-1,-3)

(-2,-3)

c) What is counterexample? Explain the use of counterexample with the help of an example.

Ans:

In logic, and especially in its applications to mathematics and philosophy, a counterexample is an exception to a

proposed general rule. For example, consider the proposition "all students are lazy". Because this statement makes

the claim that a certain property (laziness) holds for all students, even a single example of a diligent student will

prove it false. Thus, any hard-working student is a counterexample to "all students are lazy". More precisely, a

counterexample is a specific instance of the falsity of a universal quantification (a "for all" statement).

In mathematics, this term is (by a slight abuse) also sometimes used for examples illustrating the necessity of the full

hypothesis of a theorem, by considering a case where a part of the hypothesis is not verified, and where one can

show that the conclusion does not hold.[citation needed]A counterexample may be local or global in an argument.

eg

Suppose that a mathematician is studying geometry and shapes, and she wishes to prove certain theorems about

them. She conjectures that "All rectangles are squares". She can either attempt to prove the truth of this statement

using deductive reasoning, or if she suspects that her conjecture is false, she might attempt to find a counterexample.

In this case, a counterexample would be a rectangle that is not a square, like a rectangle with two sides of length 5

and two sides of length 7. However, despite having found rectangles that were not squares, all the rectangles she did

find had four sides. She then makes the new conjecture "All Rectangles have four sides". This is weaker than her

original conjecture, since every square has four sides, even though not every four-sided shape is a square.

The previous paragraph explained how a mathematician might weaken her conjecture in the face of

counterexamples, but counterexamples can also be used to show that the assumptions and hypothesis are needed.

Suppose that after a while the mathematician in question settled on the new conjecture "All shapes that are

rectangles and have four sides of equal length are squares". This conjecture has two parts to the hypothesis: the

shape must be 'a rectangle' and 'have four sides of equal length' and the mathematician would like to know if she can

remove either assumption and still maintain the truth of her conjecture. So she needs to check the truth of the

statements: (1) "All shapes that are rectangles are squares" and (2) "All shapes that have four sides of equal length

are squares". A counterexample to (1) was already given, and a counterexample to (2) is a parallelogram or a

diamond. Thus the mathematician sees that both assumptions were necessary.

Question 6:

Marks (5+4)

a) What is inclusion-exclusion principle? Also explain one application of inclusion-exclusion principle.

b) Find inverse of the following functions

i) f(x) =

3

3 5





x

x

x  3

ii) f(x) =

4

7

2

3





x

x

x  2

Question 7:

Marks ( 4 + 3 + 3)

a) Find how many 4 digit numbers are even? How many 4 digit numbers are composed of odd

digits.

Ans

4 digit numbers number means we have to put 0-9 digit at 4 places....ABCD

at the position of A (first place) we can put any digit 1-9 not 0, otherwise it will not a 4 digit number, so by

9 ways

at the position of B (second place) we can put 0-9 i.e by 10 ways...

at the position of C (third place) we can put 0-9 i.e by 10 ways...

at the position of D (fourth place) we can put 0,2,4,6 or 8 i.e by 5 ways...

so, total no. of digit = 9*10*10*5 = 4500

Total no. of four digit even numbers =4500

Ii)

Since we need only those four digit numbers ,whose digits are odd,

This can be in following ways

There can only be 1,3,5,7,9 numbers at each place in following ways

First place= 5ways

Second place = 5 ways

Third place=5 ways

Fourth place=5ways

So total numbers = 5x5x5x5=625

b) How many different 15 persons committees can be formed each containing at least

2 Accountants and at least 3 Managers from a set of 10 Accountants and 12 Managers.

Ans:

3 Accountants and 12 Managers in C(10,3)*C(12,12)=120 WAYS

4 Accountants and 11 Managers in C(10,4)*C(12,11)=2520 WAYS

5 Accountants and 10 Managers in C(10,5)*C(12,10) =16632 ways

6 " and 9 " in C(10,6)*C(12,9)=46200 ways

7 " and 8 " in C(10,7)*C(12,8)=59400 ways

8 " and 7 " in C(10,8)*C(12,7)=35640 ways

9 " and 6 " in C(10,9)*C(12,6)=9240 ways

10 " and 5 " in C (10,10)*C(12,5)=792 ways

therefore different committees can be formed in

(120+2520+16632+46200+59400+35640+9240+792)= 170544 ways

b) What is a function? Explain one to one mapping with an example.

Question 8:

Marks ( 4 +4 +2)

a) What is Demorgan’s Law? Also explain the use of Demorgen’s law with example?

b) How many ways are there to distribute 15 district object into 5 distinct boxes with

i) At least two empty box.

Ans: 15P3= 15!/(15-3)!

15!/12! = 2730

ii) No empty box.

Ans c(15,5)=15!/5!(15-5)!

= 21 x 13x 11

= 3003

c) In a fifteen question true false examination a student must achieve five correct

answers to pass. If student answer randomly what is the probability that student will fail.

Ans:

5C1 + 5C2 + 5C3 +5C4+ 5C5

so, n(s) = 15C5

so, P(E) = (5C1 + 5C2 + 5C3 +5C4+ 5C5)/(15C5)

= (5+10+10+5+1)/2983

= 0.010