Assessment Name: BT0069   Assessment ID: 32994   Question ID: 536808   Question Label: Q11   Question Text: Define the following: finite set, sub set, cardinality of set, union of set, intersection of set.

   

Answer Text:

(2 marks each) 1. Finite set: If the number of elements in a set is finite, then it is said to be a finite set. 2. Sub set: if every element of a set A is also an element of a set B then A is said to be a subset of B. 3. Cardinality of set: If A is a finite set, then the cardinality of A is the total number of elements that comprise the set. 4. Union of set: The union of two sets A and B denoted by AUB is the set of elements which belong to A or B or both. 5. Intersection of set: The intersection of two sets A and B denoted by A∩B is the set of elements which belong to both A and B.

   Score: 10.00

Assessment Name: BT0069   Assessment ID: 32994   Question ID: 536809   Question Label: Q12   

Question Text:

Let (L, ≤) be a lattice. Then for a,b,c,d ∈ L ,

i. a ≤ b → a v c ≤ b v c

ii. a ≤ b and c ≤ d → a v c ≤ b v d

Hint:

≤ – <=

→ = ––>

↔ = <––>

   

Answer Text:

i. We know that a <= b <––> a v b = b. Now( a v c) v ( b v c) = (a v c) v ( c v b) (by commutative ) = a v ( c v c ) v b ( by associative ) = a v ( c v b) = ( a v b ) v c = b v c Therefore we have a v c <= b v c. (5 marks) ii. We know that a <= b <––> a v b = b and c ≤ d ––> c v d = d. Now (a v c) v ( b v d )= a v (c v b ) v d (by associative ) = a v ( b v c ) v d (commutative ) = (a v b ) v (c v d ) (associative ) =b v d ( since a <= b and c <= d) Therefore a v c <= b v d (5 marks)

   Score: 10.00

Assessment Name: BT0069   Assessment ID: 32994   Question ID: 536811   Question Label: Q14   

Question Text:

Suppose there are 6 boys and 5 girls. In how many ways can they sit

a. in a row?

b. in a row if the boys and girls are each to sit together?

c. in a row if the girls are to sit together and the boys do not sit together?

d. where no two girls are sitting together?

   

Answer Text:

Solution: a. There are 6+5 = 11 persons and they can sit in P (11, 11) = 11! ways. b. The boys among themselves can sit in 6! ways and the girls among themselves can sit in 5! ways. They can be considered as 2-units and can be permuted in 2! Ways. Thus the required seating arrangements can be in 2! 6! 5! Ways. c. The boys can sit in 6! Ways and girls in 5! Ways. Since girls have to sit together they are considered as one unit. Among the 6 boys either 0 or 1 or 2 or 3 or 4 or 5 or 6 have to sit to the left of the girls unit. Of these seven ways, 0 and 6 cases have to be omitted as the boys do not sit together. Thus the required number of arrangements = 5 x 6! x 5!. d. The boys can sit in 6! Ways. There are seven places where the girls can be placed. Thus total arrangements are P(7, 5) x 6!.

   Score: 10.00

Assessment Name: BT0069   Assessment ID: 32994   Question ID: 536802   Question Label: Q5   

Question Text:

Find the solution of the recurrence relation

a. an =an-1 + 2an-2 with a0=2 & a1=7

b. Find the solution of the recurrence relation

an = 6an-1– 9an-2 with initial conditions a0=1 & a1=6

(Note: write r2 as r2, a1 as a1, a2 as a2, an as an, 2n as 2n, 3n as 3n, α1 as q1, α2 as q2, ∴ as therefore)

   

Answer Text:

a. The characteristics eqn of the recurrence relation is r2-r-2=0 Its roots are r=2 & r=-1 Therefore the sequence {an} is a solution to the recurrence if and only if an= q12n+ q2(-1)n for some constants q1 & q2. Now a0=2=q1+q2, a1=7=q1=3 and q2=-1. (5 mark) b. The characteristics equation r2-6r+9=0 The only root is r =3 Therefore the solution to the recurrence relation an=q13n+q2n3n, for some constants q1 & q2. using the initial condition we get a0=1=q1, a1=6=q1.3+q2.3 Solving these simultaneous equations we get q1=1 and q2= 1 Therefore the solution to the recurrence relation is an=3n+n3n (5 marks)

   Score: 10.00

Assessment Name: BT0069   Assessment ID: 32994   Question ID: 536803   Question Label: Q6   

Question Text:

In a finite set S, show that there is always at least one maximal element and one minimal element.

(Note: Write x1 as x1, x2 as x2, x3 as x3, xi as xi, xi+1 as xi+1, ∈ as belongs to)

   

Answer Text:

For maximal element, suppose S contains no maximal element, Let x1 belongs to S. Since x1 is not maximal there exists x2 belongs to S such that x2>x1 (3 mark) Since x2 is not maximal, there exists x3 in S such that x3>x2 If we continue the process we get an infinite sequence of distinct element x1, x2, x3……….such that xi+1 > xi for each i (4 mars) This is a contradiction to the fact that S contains only a finite number of elements.(Since S is a finite Poset). Hence we conclude that S contains a maximal set.(3 mars)

   Score: 10.00

Assessment Name: BT0069   Assessment ID: 32994   Question ID: 536804   Question Label: Q7   

Question Text:

a. Prove that the set N, of natural numbers is a semigroup, under the operation *, Where x*y = max(x, y)

b. Test whether the set Z(the set of integers) with binary operation * such that x*y=xy is a semigroup.

(Note: write 22 as 2 to the power of 2)

   

Answer Text:

a. (x*y)*z = max {max(x, y) z} (1 mark)=max {x,y,z} (1 mark)=max{x, max y,z} (1 mark)=x*(y*z) (1 mark)therefore * is associative .Thus (N, *) is a semigroup. (1 mark)

b. Consider (2+2)*3= 2 to the power of 2 *3=4*3 (1 mark)=4 to the power of 3 =64 (1 mark)2*(2*3)=2*2 to the power 3 (1 mark)=2*8 =2 to the power of 8= 256 (1 mark)therefore (Z,*) is NOT a semigroup. (1 mark)

   Score: 10.00