CUSTOMER_CODE
SMUDE
DIVISION_CODE
SMUDE
EVENT_CODE
OCTOBER15
ASSESSMENT_CODE BC0039_OCTOBER15
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
105219
QUESTION_TEXT
Prove that if G is a finite group and H is a subgroup of G, then O(H) is
a divisor of O(G).
Proof: Let G be a finite group and H is a sub group of G with O(G) =n,
O(H)=m (Since G is finite, it is also finite). We know that any 2 right
cosets are either disjoint or identical.
Now suppose, Ha1, Ha2,…….. Hak are only distinct right coset of H in
G.
⟹G = Ha1 U Ha2U ………..UHak
⟹O (G) = O(Ha1) + O(Ha2) + ….. + O(Hak)
SCHEME OF
EVALUATION
= O(H) + O(H) + ……….. + O(H) (k times)
(Since every right coset has O(H) elements)
⇒O(G) = k. O(H)
⟹n = k. m ⟹ (n/m) = k
⟹Hence O(H) divides O(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,
SCHEME OF
EVALUATION
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,
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
105222
QUESTION_TEXT
Prove that the following properties of a lattice L are equivalent:
i.
a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c) for all a, b, c ∈ L ;
ii. (a ∧ b) ∨ c = (a ∨ c) ∧ (b ∨ c) for all a, b, c ∈ L ;
iii. (a ∧ b) ∨ (b ∧ c) ∨ (c ∧ a) = (a ∨ b) ∧ (b ∨ c) ∧ (c ∨ a) for all a, b, c ∈ L
(i)
(ii): suppose
for all a, b, c
L;
(by (i))
(by commutative and absorption laws)
(by (i))
(by associative law)
(by absorption law)
This proves (ii)
(ii)
(iii):
Suppose (ii)
.
(by (ii)
SCHEME OF
EVALUATION
(by commutative,
associative and absorption
(by (ii))
(by idempotent law)
(iii)
(i): Suppose that a ≤ c. Then
Also
Now
(by (A))
(by (iii))
(since a ≤ c)
(by absorption law).
Now we proved that
This shows that (i) is true. This completes the proof.
(Total 10 marks)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
105224
Prove that if V be a vector space over F and W is a nonempty subset of
V. Then the following conditions are equivalent.
QUESTION_TEXT
i.
W is a subspace of V
ii.
α, β ∈ F and W1, W2 ∈ W implies αW1 + βW2 ∈ W.
(i) ⇒ (ii): α ∈ F, w1 ∈ W ⇒ αw1 ∈ W; β ∈ F, w2 ∈ W ⇒ βw2 ∈ W.
Since αw1, βw2 ∈ W and W is subspace we have αw1 + βw2 ∈ W.
(ii) ⇒ (i): Let w1, w2 ∈ W. Since 1 ∈ F we have that w1 + w2 = 1.w1
+ 1. w2 = αw1 + βw2 ∈ W (here α = 1 and β = 1). Therefore + is a
closure operation on W.
Since W ⊆ V, the associative law holds.
SCHEME OF
EVALUATION
x ∈ W ⇒ x, x ∈ W ⇒ 0 = 1.x + (1) = αx + βx ∈ W (here α = 1 and β =
-1).
Therefore the additive identity 0 is in W.
Let x ∈ W. Then x, x ∈ W ⇒ -x = 0. x + (-1)x = αx + βx (here α = 1
and β = -1).
Therefore the additive inverse –x ∈ W for any element x ∈ W.
Hence (W, +) is a group. Since W ⊆ V and V is Abelian, we have that
(W, +) is also Abelian. Also it is easy to verify that the other conditions
of vector space.
Therefore W is a subspace of V.