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DECEMBER 2015 SECTION-A Q1.(a) Define connected graph. Ans A graph is connected when there is a path between every pair of vertices. In a connected graph, there are no unreachable vertices. A graph that is not connected is disconnected. A graph G is said to be disconnected if there exist two nodes in G such that no path in G has those nodes as endpoints. (b)Define asymmetric relation with example. Ans an asymmetric relation is a binary relation on a set X where: For all a and b in X, if a is related to b, then b is not related to a.[1] In mathematical notation, this is: An example is the "less than" relation < between real numbers: if x < y, then necessarily y is not less than x.The "less than or equal" relation ≤, on the other hand, is not asymmetric, because reversing x ≤ x produces x ≤ x and both are true. In general, any relation in which x R x holds for some x (that is, which is not irreflexive) is also not asymmetric. (c)Define degree of recurrence relation. Ans The highest power of ar or f(x) or an is called degree of recurrence relation. Example The difference between eqn. ar3+2ar-12+ar-2=0 has degree 3 as the highest power of ar is 3. (d) Write elementary properties of ring. Ans Let R be a ring then for all a,b∈R 1) a⋅0=0⋅a=aa⋅0=0⋅a=a 2) a(−b)=(−a)b=−(ab)a(−b)=(−a)b=−(ab) 3) (−a)(−b)=ab (e)Which of the complete graphs Kn have Hamiltonian circuits? Ans As n>=3[since kn has a Hamiltonian circuits for n>=3.

K3 k4 (f) Draw bridges whose every edge is a bridge. Ans Here G is a connected graph with 6 vertices a,b,c,d, e and f and 5 edges e1,e2,e3,e4 and e5. Here G-{e1}, G-{e2},G-{e3},G-{e4} and G{e5} are all not connected graphs. Thus ,every edge in a graph is a bridge. a e1 b e e5 e2 f e4 c e3 d (g) Define partial order set. Ans A partially ordered set (or poset) is a set taken together with a partial order on it. Formally, a partially ordered set is defined as an ordered pair the ground set of and is the partial order of , where is called . An element in a partially ordered set is said to be an upper bound for a subset if for every . Similarly, a lower bound for a subset of element such that for every then the poset , we have , is an . If there is an upper bound and a lower bound for is said to be bounded (h) Define Bollean ring. Ans The ring (R,+,.) is a Boolean ring if all its elements are idempotent i.e. x2 =x for all values of x element. Example The power set of any set X i.e. P(X). ,

Where the addition in the ring is symmetric difference and the multiplication in intersection form a Boolean ring. i.e.{P(A).∩ is a Boolean ring. (i)Define Subgroup. Ans given a group G under a binary operation ∗, a subset H of G is called a subgroup of G if H also forms a group under the operation ∗. More precisely, H is a subgroup of G if the restriction of ∗ to H × H is a group operation on H. This is usually denoted H ≤ G, read as "H is a subgroup of G". (j)Define normal Subgroup. Ans A normal subgroup is a subgroup which is invariant under conjugation by members of the group of which it is a part. In other words, a subgroup H of a group G is normal in G if and only if gH = Hg for all g in G; i.e., the sets of left and right cosets coincide. Normal subgroups (and only normal subgroups) can be used to construct quotient groups from a given group. SECTION-B Q2.If A and B are two sets , then prove that ⇔ ∈ ⇒ (a) A-B=A∩BC (b) A-B=BC-AC Ans (a)for all values of x∈A-B and x∉B ⇒x∈A and x∈BC ⇒x∈ A∩BC⇒ a-b ⊆ A∩BC .... ... (eq1) For all values of y∈A∩BC y∈A and y∉B ⇒ y ∈ A-B ⇒ A∩BC ⊆ A-B ⇒ A∩BC ⊆ A-B .........(eq 2) from (1) and (2) we get A-B =A∩BC (b)For all x element A-B ⇒ x∈ A and x∉BC ⇒ x∉A and x∈ BC ⇒x ∈ (Bc –AC) For all y∈ (Bc –AC) ⇒ y∈ BC and y ∉ AC

⇒ y ∉ BC and y ∈ AC ⇒y ∈ A-B BC-AC⊆A-B From 1 and 2 we get A-B=BC-AC Q3Draw a diagram for each of the following graph S(V,E): (a)V={a,b,c,d,e,f}, E={(a,d), (a,f),(b,c),(b,f)(c,f)} (b)V=={a,b,c,d,e,f},E={(a,a),(b,c)(b,d),(e,a),(e,d)} Ans e e d c d a c b f a b Q4.Using the properties of Boolean algebrs ,prove that for each x in a algebra if x+y=1 and x,y=0 they y=x Ans Given x+y=1 x.y=0 y’=y’+0=y’+(y.x)=(y’+y).(y’+x) i.e. y’=1.(y’+x)=y’+x now, x=x+0=x+(y.y’)=x+y).(x+a’)=1.(x+y’)=x+y’ x=x+y’=y’+x=y’ x=y’ further by 0+1=1 and 0.1=0

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