Download CUSTOMER_CODE SMUDE DIVISION_CODE SMUDE

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.