Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
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+…