CUSTOMER_CODE
SMUDE
DIVISION_CODE
SMUDE
EVENT_CODE
JULY15
ASSESSMENT_CODE MC0063_JULY15
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
33842
QUESTION_TEXT
Briefly explain with syntax:
i. Universal set
ii. Union of sets
iii. Intersection of sets
iv. Difference of sets
v. Complement of a set
SCHEME OF
EVALUATION
i. Universal set: If all the sets are subsets of a fixed set, then this set
is called the universal set and is denoted by U. (2 marks)
ii. Union of sets: The union of two sets A and B denoted by A∪B is
the set of elements which belong to A or B or both. (2 marks)
iii. Intersection of sets: The intersection of two sets A and B denoted
by A∩B is the set of elements which belong to both A and B. (2
marks)
iv. Difference of sets: The difference of two sets A and B denoted
by A–B is the set of elements of A which are not the elements of B. (2
marks)
v. Complement of a set: The complement of a set A with respect to
the universal set U is defined as U–A and is denoted by AI or AC (2
marks)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
33844
QUESTION_TEXT
Define:
i. Group
ii. Finite and infinite group
iii. Transposition
iv. Odd permutation
v. Even permutation
SCHEME OF
EVALUATION
i. Group: A non empty set G together with a binary operation * is
called a group if the algebraic system (G, *) satisfies the following four
axioms:
a. closure
b. Associative
c.
d.
Identity
Inverse (2 marks)
ii. Finite and infinite group: If G contains only a finite number of
elements then G is called a finite group. If G contains infinite number
of elements then G is called an infinite group. (2 marks)
iii. Transposition: The simplest permutation is a cycle of length 2.
Such cycles are called transpositions. (2 marks)
iv. Odd permutation: A permutation is said to be an odd permutation
if is the product of an odd number of transpositions. (2 marks)
v. Even permutation: A permutation is said to be even permutation if
is the product of an even number of transpositions. (2 marks)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
73746
QUESTION_TEXT
State and prove the fundamental theorem of homomorphism.
SCHEME OF
EVALUATION
Since F is an onto homomorphism from G to G1, we have F (G) = G1.
That is G1 is the homomorphic image of F. Define f: G/K – > G1 by
f(Ka) = F (a) for all Ka is in G/K.
F is well defined: Let a, b is in G and Ka = Kb
⟹ ab inverse ∈ K
⟹ F (ab inverse) = e1
⟹ F (a). [F (b)] inverse = e1
⟹ F (a) = F (b)
⟹ f (Ka) = f (Kb).
F is 1–1: Suppose f(Ka) = f(Kb)
⟹ F (a) = F (b)
⟹ F (a). [F (b)] inverse = e1
⟹ F (a). F (b inverse) = e1
⟹ F (ab inverse) = e1
⟹ ab inverse is in K
⟹ Ka = Kb
Therefore f is 1–1.
F is onto: Let y is in G1. Since F: G – > G1 is onto, we have that there
exists x is in G such that F (x) = y. Since x is in G, we have Kx is in G/K.
Now f(Kx) = F (x) = y.
Therefore f is onto.
F is homomorphism: Let Ka, Kb is in G/K.
F (Ka.Kb) = f (Kab) = F (ab) = F (a). F (b) (since F is homomorphism) =
f(Ka.f(Kb). Therefore f is a homomorphism.
Hence f: G/K – > G1 is an isomorphism.
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
73748
QUESTION_TEXT
What is tautology and contradiction and prove that [p ∧ (p ∨ q)] ∨ ~p
is a tautology and [p ∧ (p ∨ q)] ∧~p is a contradiction.
SCHEME OF
EVALUATION
A tautology is a propositional function (say p(x) with variable x)
whose truth value is true for all possible values of the propositional
variables.
A contradiction (or absurdity) is a propositional function whose truth
value is always false.
i. Show that [p ^ (p v q)] v ~p is a tautology.
Solution: Now we write down the truth table
p q p v q p ^ (p v q) ~p [p ^ (p v q)] v ~p
0 0 0
0
1 1
0 1 1
0
1 1
1 0 1
1
0 1
1 1 1
1
0 1
Observing the table we can conclude that [p ^ (p v q)] v ~p is always
true.
Hence [p ^ (p v q)] v ~p is a tautology
ii. Verify that [p ^ (p v q)] ^~p is a contradiction.
Solution: Now we write down the truth table
p q p v q p ^ (p v q) ~p [p ^ (p v q)] ^~p
0 0 0 0
1 0
0 1 1 0
1 0
1 0 1 1
0 0
1 1 1 1
0 0
Observing the table, we can conclude that [p ^ (p v q)] ^~p is always
false.
Hence [p ^ (p v q)] ^~p is contradiction.
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
105712
Explain the working of the RSA cryptosystem.
QUESTION_TEXT
SCHEME OF
EVALUATION
Suppose that we choose two random 150-digit prime numbers p and q.
Next, we compute the product n = pq and also compute
where is the Euler -function. Now we start
choosing random integers E until we find one that is relatively prime to
m; that is, we choose E such that gcd (E, m) = 1. Using the Euclidean
algorithm, we can find a number D such that DE = 1 (mod m). The
numbers n and E are now made public.
Suppose now that person B (Bob) wishes to send person A (Alice)
a message over a public line. Since E and n are known to everyone,
anyone can encode messages. Bob first digitizes the messages according
to some scheme, say A = 00, B = 02, …, Z = 25. If necessary, he will
break the message into pieces such that each piece is a positive integer
less than n. Suppose x is one of the pieces. Bob forms the number
mod n and sends y to Alice. For Alice to recover x, she need only
compute
mod n. Only Alice knows D.
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
105715
QUESTION_TEXT
Prove that the relation “a  b mod n” is an equivalence relation
on z.
i)reflexive
SCHEME OF
EVALUATION
ii) symmetric
iii) transitive should be shown