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CUSTOMER_CODE
SMUDE
DIVISION_CODE
SMUDE
EVENT_CODE
JAN2016
ASSESSMENT_CODE BT0069_JAN2016
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
4742
QUESTION_TEXT
Define the following:
finite set, sub set, cardinality of set, union of set, intersection of set.
SCHEME OF
EVALUATION
(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.
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
4746
QUESTION_TEXT
Define group. Also prove the following identities:
i. If G is a group, then
a. The identity element of G is unique.
b. Every element in G has unique inverse in G.
ii. A non-empty subset H of a group G is a subgroup of G if and only
if:
a. a, b ∈ H ⟹ ab ∈ H and
b. a ∈ H ⟹ a –1 ∈ H
SCHEME OF
EVALUATION
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:
● Closure: a, b are elements of G, implies a*b is an element of G
● Associative: (a*b)*c=a*(b*c) for all elements a, b, c in G
● Identity: There exists an element ‘e’ in G such that a*e=e*a=a for all
a in G
● Inverse: For any element a in G there corresponds an element b in G
such that a*b=e=b*a
(3 marks)
Proof for the identities:
i. (4 marks)
a.Let e, f be two identity elements in G. since e is the identity, we have
e.f=f. since f is the identity, we have e.f=e. Therefore, e=e.f=f.
Hence the identity element is unique.
b.Let a be in G and a1, a2 are two inverses of a in G.
Now a1=a1.e (since e is the identity)
=a1.(a.a2) (since a2 is the inverse of a)
=(a1.a).a2 (by associativity)
=e.a2 (since a1 is the inverse of a)
=a2
Hence the inverse of an element in G is unique.
ii. (3 marks)
Suppose that H is a subgroup of G
Then H itself is a group under the product in G. Therefore (a) and (b)
holds
Converse:
Suppose H satisfies (a) and (b),
By (a), H satisfies the closure property.
For any a, b, c in H, we have that a, b, c belongs to H implies that
a(bc)=(ab)c
Therefore (H, .) is a subgroup of (G, .)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
74119
QUESTION_TEXT
Prove that a tree G with ‘n’ vertices has (n-1) edges.
SCHEME OF
EVALUATION
We prove this theorem by induction on the number vertices n.
If n = 1, then G contains only one vertex and no edge.
So the number of edges in G is n -1 = 1 - 1 = 0.
Suppose the induction hypothesis that the statement is true for all trees
with less than ‘n’ vertices. Now let us consider a tree with ‘n’ vertices.
Let ‘ek’ be any edge in T whose end vertices are vi and vj.
Since T is a tree, by Theorem 12.3.1, there is no other path between vi
and vj.
So by removing ek from T, we get a disconnected graph.
Furthermore, T- ek consists of exactly two
components(say T1 and T2).
Since T is a tree, there were no circuits in T and so there were no
circuits in T1 and T2.
Therefore, T1 and T2 are also trees.
It is clear that |V(T1)| + |V(T2)| = |V(T)| where V(T) denotes the set of
vertices in T.
Also |V(T1)| and |V(T2)| are less than n.
Therefore, by the induction hypothesis, we have –
|E(T1)| = |V(T1)| - 1 and |E(T2)| = |V(T2)| - 1.
Now |E(T)| - 1 = |E(T1)| + |E(T2)| = |V(T1)| - 1 + |V(T2)| - 1
 |E(T)| = |V(T1)| + |V(T2)| - 1
= |V(T)| - 1 = n-1.
This completes the proof.
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
112545
QUESTION_TEXT
If H is a non-empty finite subset of a group G and H is closed under
multiplication, then Prove that H is a subgroup of G
Suppose H is a non-empty finite sub set of a group G and H is closed
under multiplication. Now we have to show that H is a subgroup of G.
It is enough to show that aЄH →
ЄH
Since H is a non empty, there exists aЄH
Now a, aЄH →
.
ЄH
Similarly a3ЄH, …….., am ЄH ,…….
Therefore H
SCHEME OF
EVALUATION
{a,a2, -----}
Since H isfinite,we have that there must be repetitions in a, a2, …..
Therefore, there exists integers r,s with r > s > 0 such that ar= as
→ ar.as = a0
→ ar-s = e → e ЄH (since r-s > 0 and aЄ H→ ar-s Є H)
Since r-s-1 ≥0, we have ar-s-1 Є H and a. ar-s-1 =ar-s e Є H.
Therefore, ar-s-1 acts as the inverse of a Є H. Hence h is a subgroup. (10
marks)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
112548
QUESTION_TEXT
What do you mean by functions? Explain.
Definition
Let A and B be two non-empty sets. A function or a mapping f from A to B is a
rule, which associates every element of A with a unique element of B and is
denoted by
In other words, a function f from A to B is a relation satisfying the
following:
SCHEME OF
EVALUATION
1.
Every element of A is related to some element of B
2.
no element of A is related to two different elements of B.
3. If
is a function then A is called the domain and B is called the
co-domain of f. If
by the function f, then y is called the image of x
under f and is denoted by
Also x is called the pre-image of y under f.
The range of f is the set of those elements of B, which appears as the
image of at least one element of A and is denoted by f(A) is a subset of B.
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
112550
If
and
. Find
QUESTION_TEXT
i.
ii.
SCHEME OF
EVALUATION
=2+2x2+2x4+…