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CUSTOMER_CODE SMUDE DIVISION_CODE SMUDE EVENT_CODE APR2016 ASSESSMENT_CODE BC0039_APR2016 QUESTION_T DESCRIPTIVE_QUESTION YPE QUESTION_ID 73264 QUESTION_T EXT Define each of the following and give the truth table of each. a. Conjunction b. Negation c. Disjunction d. Conditional (implication) e. bi-conditional A) the negation of statement is formed by means of the world “not” b)let P&Q be any two statements then “P” and “Q” is conjunction of P and Q c) junction of two statements P and Q is “P or Q” SCHEME OF EVALUATION D) let P & Q any two statements then “if P then Q” is called conditional or implication. Let P and Q be any two statements then P if Q is called conditional statement. QUESTION_TYPE DESCRIPTIVE_QUESTION QUESTION_ID 105218 QUESTION_TEXT Prove that if H is a non empty finite subset of a group G and H is closed under multiplication, then H is a subgroup of G. Proof: Suppose H is a non empty finite subset 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 a ∈H ⟹ a–1 ∈H. Since, H is a non empty, there exists a ∈ H. Now, a∈ H ⟹ a2 ∈H. Similarly a3 ∈H, …… am ∈H, …….. SCHEME OF EVALUATION Therefore H≥ {a, a2, …….}. Since H is finite, we name that there must be repetitions in a, a2….. Therefore there exists integers r, s with r>s > 0 Such that ar = as ⟹ ar. a–s = a0 ⟹ ar–s = e ⟹ e ∈H (since r–s >0, a∈ H ⟹ar–s ∈H) Since r–s–1 ≥0, we have ar–s–1 ∈H and a. ar–s–1 = ar–s = e ∈H. Therefore ar–s–1 acts as the inverse of a in H. Hence H is a subgroup of G QUESTION_TYPE DESCRIPTIVE_QUESTION QUESTION_ID 105220 QUESTION_TEXT Prove that a finite integral domain is a field. Proof: We know that in an integral domain we have, ab = 0 ⇒ a=0 or b =0 Now it suffices to show that every non–zero element has multiplicative inverse. Let D be an integral domain. Now we show, SCHEME OF EVALUATION i. There exist 1 ∈D such that a.1 = a for all a ∈D, ii. 0≠ a ∈D ⟹ there exists b ∈D such that ab = 1 Let D = {x1, x2,……xn} and 0 ≠a ∈D. Now, x1a, x2a, ….. xna are all distinct ∴ D = {x1 a, x2 a, ……, xn a}. Since a ∈D, a = xka, for some 1 ≤k ≤n. Again Since D is commutative, we have, xka = a= axk We show xk is the identity element. For this, let y∈D, then y =xja for some i. Now consider, y.xk = (xja) xk = xj (axk) = xja = y Thus yxk = y for all y∈D. Therefore xk is the identity element. For xk ∈D = {x1 a, x2a, ………. xna} ⟹ xk = xja for some 1≤ j ≤n Therefore, xi is the multiplicative inverse of a. Hence D is a field ∴ A finite integral domain is a field QUESTION_TYPE DESCRIPTIVE_QUESTION QUESTION_ID 105221 QUESTION_TEXT Using the principle of mathematical induction show that 102n–1+1 is divisible by 11 for all n ∈N. Let Now which is divisible by 11. Therefore for some integer k. Consider Therefore SCHEME OF EVALUATION which is clearly divisible by 11. Therefore Therefore P(m+1) is divisible by 11. Hence by the principle of mathematical induction P(n) is divisible by 11 for all n. (Total 10 marks) QUESTION_TYPE DESCRIPTIVE_QUESTION QUESTION_ID 105225 QUESTION_TEXT Define the following: Relation, Domain and range of a relation, inverse of a relation, reflexive relation, Symmetric relation. Ans: Relation: Let A and B be 2 non–empty sets, then a relation R from A to B is a Subset of A x B containing the ordered pairs (a, b) ∈ A x B such that some relation exists between a and b. Domain and Range of a relation: Let R be a relation from A to B. The domain of R is the Set of all first coordinates of the ordered pairs of R and the range of R is the set of second coordinates of the pairs of R. SCHEME OF EVALUATION Inverse of a relation: Let R be a relation from a set A to set B. The inverse relation R denoted by R–1 is the relation from B to A defined by R–1 = {(b, a) / (a, b) ∈ R} Reflexive relation: A relation R in a Set A is said to be reflexive if for every a ∈ A, (a, a) ∈ A Symmetric relation: A relation R in a Set A is said to be symmetric if (a, b) ∈ R implies (b, a) ∈ R. That is R is said to be symmetric if aRb implies bRa. QUESTION_TYPE DESCRIPTIVE_QUESTION QUESTION_ID 105226 QUESTION_TEXT Define the following with one example. Set, Subset, cardinality of a set, Universal set Set: A well defined collection of objects is called a set. Ex: B= {1, 3, 5, 7, 9} SCHEME OF EVALUATION Subset: If every element of a set A is also an element of a Set B, then A is said to be a sub set of B. Ex: N is the Set of Natural no’s: Z is the Set of integers. Then NCZ (N is a Subset of Z) Cardinality of a set: If A is a finite set, then the Cardinality of A is the total number of elements that comprise the set and denoted by N(A). Ex: A = {1, 2, 3, 4, 5} Then n(A) = 5 Universal Set: If all the Sets are subsets of a fixed Set, then this Set is called the universal set & denoted by U Ex: In the Study of theory of numbers the set Z of integers is considered as the universal set.