Agni College of Technology

1

DEPARTMENT OF CSE

Subject Name: DISCRETE MATHEMATICS

Subject code :MA 6566 Important Question and Answers PART-A

1. Define a rule of Universal specification. Ans: If a statement of the form (𝑥)𝐴(𝑥) is assumed to be True then the universal quantifier may be dropped from the statement to obtain A(x) which is true for each object in the universe, or to obtain P(x) which is true for a specific object of the universe. 2. What are the contrapositive, the converse, and inverse of the conditional statement.” If you work hard then you will be rewarded.” ¬𝑝: You will not work hard

Ans: P: You work hard

¬𝑞:You will not be rewarded

Q: You will be rewarded Converse:𝑞 → 𝑝 You will be rewarded only if you work hard. Contrapositive: ¬𝑞 → ¬𝑝 If you will not be rewarded then you will not work hard Inverse: ¬𝑝 → ¬𝑞

3. Using truth table, show that the proposition 𝑝 ∨ ¬(𝑝 ∧ 𝑞) is a tautology. Solution: ¬(𝑝 ∧ 𝑞)

𝑝∧𝑞

𝑝 ∨ ¬(𝑝 ∧ 𝑞)

P

q

T

T

T

F

T

T

F

F

T

T

F

T

F

T

T

F

F

F

T

T

∴ 𝑝 ∨ ¬(𝑝 ∧ 𝑞)is a tautology 4. Write the negation of the statement ∃𝑥 ∀𝑦 𝑝 𝑥, 𝑦 . Solution: Given ∃𝑥 ∀𝑦 𝑝 𝑥, 𝑦 . Negation: ∀𝑥 ∃𝑦 𝑝 𝑥, 𝑦 .

5. In how many ways can all the letters in MATHEMATICAL be arranged. Ans: In the word MATHEMATICAL has 12 letters. M,T appears 2 times &A appears 3 times ∴ The required number of permutations =

12! 2!3!2!

= 19958400

6. Twelve students want to place order of different ice-creams in a ice-cream parlour,which has six type of icecreams. Find the number of orders that the twelve students can place. Ans: Here n=12,r=6 The number of possible selection is,

17! 12 + 6 − 1 = 6!11! 6

7. State Pigeonhole principle. If (n+1) pigeon occupies ‘n’ holes than atleast one hole has more than 1 pigeon. 8. Solve: 𝑎𝑘 = 3𝑎𝑘−1 , for k≥ 1, 𝑤𝑖𝑡𝑕 𝑎𝑜 = 2. Ans: The characterisitics equation is 𝑟 − 3 = 0 𝑟 = 3 ⇒ 𝑎𝑘 = A3𝑘

Agni College of Technology DEPARTMENT OF CSE

Subject Name: DISCRETE MATHEMATICS

Subject code :MA 6566 Important Question and Answers PART-A

Given 𝒂𝒐 = 𝟐 ⟹ 𝑎0 = A30 ⟹2=A 1 ⟹2=A ∴ 𝑎𝑛 = 2 3𝑘 9. Define a regular graph.Can a complete graph can be a regular graph? If every vertex of a simple graph has the same degree, then the graph is called a regular graph.If every vertex in a regular graph has degree k, then the graph is called K-regular. Every complete graph is regular. 10. State the handshaking theorem. Ans: Let 𝐺 = (𝑉, 𝐸) be an undirected graph with e edges then 𝑣∈𝑉 deg 𝑣 = 2𝑒.The sum of degrees of all the vertices of an undirected graph is twice the number of degrees of the graph and hence even. 11. Obtain the adjancey of the following graph.

Solution: 𝑣1 𝑣2 𝑣3 𝑣4 𝑣5 𝑣1 𝑣2 𝐴 𝐺 = 𝑣3 𝑣4 𝑣5

0 1 0 0 1

1 0 1 0 1

0 1 0 1 1

0 0 0 0 1

1 1 1 1 0

12. Give an example of a non-Eulerian graph which is Hamiltonian

Solution: 𝑣1 𝑣1

𝑣4

𝑣2

𝑣3

deg 𝑣1 = 3; deg 𝑣2 = 3; deg 𝑣3 = 3; deg 𝑣4 = 3 Here the vertices are not even degree. ∴ The given graph is non-Eulerian graph

Deg 𝑣1 + deg 𝑣2 ≥ 𝑛 − 1 Deg 𝑣2 + deg 𝑣3 ≥ 𝑛 − 1 Deg 𝑣3 + deg 𝑣4 ≥ 𝑛 − 1 Deg 𝑣4 + deg 𝑣1 ≥ 𝑛 − 1

2

Agni College of Technology

3

DEPARTMENT OF CSE

Subject Name: DISCRETE MATHEMATICS

Subject code :MA 6566 Important Question and Answers PART-A

∴ The given graph is Hamiltonian. The Hamiltonian circuit is 𝑣1 𝑣2 𝑣3 𝑣4 𝑣1 . 13. Define Isomorphism of two graphs.

Solution: Two graphs G and 𝐺 ′ are isomorphic. If there is a function 𝑓: 𝑉(𝐺) → 𝑉(𝐺 ′ ) from the vertices of G (i) f is one-one (ii) f is onto and (iii) f preserves adjacency (iv) 14. Prove or disprove,”Every subgroup of an abelian group is normal” Solution: If G is abelian,then every subgroup of G is normal in G,(as 𝐻𝑎 = 𝑕𝑎/𝑕 ∈ 𝐻 = {𝑎𝑕/𝑕 ∈ 𝐻} since 𝑕𝑎 = 𝑎𝑕 = 𝑎𝐻, 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑎 ∈ 𝐺) 15. Define a Semi group. Solution: If S is nonempty set s together with the binary operation * satisfying the following properties(𝑖)𝑎 ∗ 𝑏 = 𝑏 ∗ 𝑎 (𝑖𝑖) (𝑎 ∗ 𝑏) ∗ 𝑐 = 𝑎 ∗ (𝑏 ∗ 𝑐) 𝑎, 𝑏, 𝑐 ∊ 𝑆 is called semigroup it is denoted by (S,*).

16. Give an example of a ring which is not a field. Solution:

The ring Z of all integers in an integral domain but not a field.

17. In Lattices (𝑳, ≤) ,prove that 𝒂 ∧ (𝒂 ∨ 𝒃) = 𝒂𝒇𝒐𝒓𝒂𝒍𝒍𝒂, 𝒃 ∊ 𝑳. Solution: since 𝑎 ∧ 𝑏 is the GLB of {𝑎, 𝑏} 𝑎∧𝑏 ≤𝑎 ……………………. 1 obviously𝑎 ≤ 𝑎 ……………………. 2 from (1) and (2) we have 𝑎 ∨ (𝑎 ∧ 𝑏) ≤ 𝑎 by the definition of LUB we have 𝑎 ≤ (𝑎 ∧ 𝑏) = 𝑎 Similarly we can prove that 𝑎 ∧ (𝑎 ∨ 𝑏) = 𝑎.

18. When is a lattice said to be bounded? Ans: Let (𝐿,∧,∨) be a given lattice. If it has both 0 element and 1 element then it is said to be bounded lattice. It is denoted by (𝐿,∧,∨ ,0,1) 19. When is a lattice said to be a Boolean Algebra? Ans: A lattice which is complemented and distributive is called a Boolean algebra. 20. Give an example of a distributive lattice but not complemented. Solution:

No complement exist for 0,b,c,d,1 The element ‘a’ is a complement of d and vice versa. ∴ The above graph is not complemented. 21. Define Semigroup and monoid. Give an example of a semigroup which is not a monoid. Sol. Semigroup: A non empty set S together with a binary operation is called a semigroup if the following conditions are satisfied. a. a b  S a,bS ( Closure) b. a (b c) = (a b) c a,b,cS ( Associative )

Agni College of Technology DEPARTMENT OF CSE

Subject Name: DISCRETE MATHEMATICS

Subject code :MA 6566 Important Question and Answers PART-A

Monoid : A non empty set M together with a binary operation is called a monoid if the following conditions are satisfied. a b  M a,bM ( Closure) a (b c) = (a b) c a,b,cM ( Associative ) There exists an element eM such that a  e = e  a = a aM ( Identity ) Example : (N, +) is a semigroup but not monoid For, we know that N is the set of all positive integers

i) ii) iii)

(i.e.) N = {1,2,3,…….}

4

Agni College of Technology DEPARTMENT OF CSE

Subject Name: DISCRETE MATHEMATICS Subject code :MA 6566 Important Question and Answers PART-A

The additive identity is zero because a +0 = 0 +a = a aN. But 0N (i.e.) 0 is not an element of N

(N , +) is not a monoid. 22. Show that the intersection of two subgroups is a subgroup. Sol. Let H and K are subgroups of G. Then atleast the identity element eH and eK.

eH K Thus H K is a non empty subset of G. Let a,bH K a,bH and a,bK ab-1H and ab-1K ab-1H K

H K

is a subgroup of G.

23. Define distributive lattice. Sol.

A lattice (L, ,) is said to be distributive lattice if for any a,b,cL a  (b  c)  (a  b)  (a  c) a  (b  c)  (a  b)  (a  c)

24 . If B is a Boolean algebra, then for aB, a + 1 = 1, a.0 = 0 Sol.

a + 1 = (a + 1).1

a.0 = a.0 + 0

= (a + 1).(a + a’)

= a.0 + (a.a’)

= a + (1.a’)

= a. (0 + a’)

= a + (a’.1)

= a. (a’ + 0)

= a + a’

= a. a’

=1

= 0

Agni College of Technology DEPARTMENT OF CSE

Subject Name: DISCRETE MATHEMATICS Subject code :MA 6566 Important Question and Answers PART-A

PART-B 1.(i) Verify the validity of the following argument. Every living thing is a plant or an animal. John’s gold fish is alive and it is not a plant. All animals have hearts. Therefore John’s gold fish has a heart. Sol. Let P(x) : x is a plant A(x) : x is an animal H(x) : x has a heart, g : John’s gold fish We need to show ( x)[P(x) A(x)] , 7P(g) ,( x)[A(x) H(x)] H(g) Argument 1. ( x)[P(x) A(x)] Rule P 2. P(g) A(g) Rule US 3. 7P(g) Rule P 4. A(g) Rule T [ From 2,3 i.e. 7P, P QQ] 5. ( x)[A(x) H(x)] Rule P 6. A(g) H(g) Rule US 7. H(g) Rule T The argument is valid. ii) Prove that ( x)[P(x) Q(x)], ( 𝑦)P(y) ( x)Q(x) . Sol. 1. 7( x)Q(x) Rule P (assumed premise) 2. ( x)7Q(x) Rule T 3. 7Q(a) Rule US 4. ( 𝑦)P(y) Rule P 5. P(a) Rule ES 6. ( x)[P(x) Q(x)] Rule P 7. P(a) Q(a) Rule US 8. Q(a) Rule T [ From 5,7] 9. Q(a) 7Q(a) Rule T [ From 8,3] which is a contradiction 2.(i) Prove that 2 is irrarational by giving a proof by contradiction. Solution: Assume the contrary that 2 is rational number. 𝑝

∴ 2 = 𝑞 for some integer p&q such that p &q have no common factors. ∴

𝑝2 = 2 ⟹ 𝑝2 = 2𝑞 2 𝑞2

Since 𝑝2 is an even integer, p is an even integer.

Agni College of Technology DEPARTMENT OF CSE

Subject Name: DISCRETE MATHEMATICS Subject code :MA 6566 Important Question and Answers PART-A ∴ 𝑝 = 2𝑚for some integer m ∴ (2𝑚)2 = 2𝑞 2 ⟹ 2𝑚2 Since 𝑞 2 is an even integer, q is an even integer. ∴ 𝑞 = 2𝑘for some integer k Thus p and q are even. Hence they have a common factor 2. This contradicts the assumption p and q have no common factors. Thus our assumption 2 Is rational is wrong. Hence 2 is irrational. (ii) Prove by mathematical induction that 6𝑛+2 + 72𝑛 +1 is divisible by 43 for each positive integer n. Solution: 𝑃 𝑛 : 6𝑛+2 + 72𝑛 +1 is divisible by 43 Step 1: To prove 𝑃 1 is true 𝑃 1 : 61+2 + 72+1 is divisible by 43 𝑃 1 : 63 + 73 is divisible by 43 ∴ 𝑃 1 is true Sep 2: Assume that 𝑃(𝑘) is true Ie., 𝑃 𝑘 : 6𝑘+2 + 72𝑘+1 is divisible by 43 6𝑘+2 + 72𝑘+1 = 𝑐 43 Step 3: prove that 𝑃(𝑘 + 1) is true Ie., 𝑃 𝑘 + 1 : 6𝑘+3 + 72𝑘+3 is divisible by 43 6𝑘+3 + 72𝑘+3 = 6𝑘+2 6 + 72𝑘+1 72 = 6 𝑐 43 − 72𝑘+1 + 72𝑘+1 72 258𝑐 − (6)72𝑘+1 + 72𝑘+1 72

Agni College of Technology DEPARTMENT OF CSE

Subject Name: DISCRETE MATHEMATICS Subject code :MA 6566 Important Question and Answers PART-A = 258𝑐 + 72𝑘+1 49 − 6 258𝑐 + 72𝑘+1 . 43 = 43(6𝑐 + 72𝑘+1 ) which is divisible by 43 Hence the proof. 3.(i) Determine the number of positive integers n, 1 ≤ 𝑛 ≤ 2000 that are not divisible by 2,3 or 5 but are divisible by 7. Solution:

Let A denote the number of integers not divisible by 2

Let B denote the number of integers not divisible by 3 Let C denote the number of integers not divisible by 5 Let D denote the number of integers divisible by 7 Therefore, 𝐴 =

2000 2

= 1000 , 𝐵 =

2000 3

= 666

2000 2000 = 400, 𝐷 = = 285. 5 7 𝐴∩𝐵∩𝐶∩D = 𝐴∪𝐵∪𝐶∩D |𝐶| =

= |𝐷 ∩ 𝐴 ∪ 𝐵 ∪ 𝐶 | = 𝐷 − 𝐷∩ 𝐴∪𝐵∪𝐶 = 𝐷 −| 𝐷∩𝐴 ∪ 𝐷∩𝐵 ∪ 𝐷∩𝐶 | = 𝐷 −[ 𝐷∩𝐴 + 𝐷∩𝐵 + 𝐷∩𝐶 − 𝐴∩𝐵∩𝐷 − 𝐴∩𝐶∩𝐷 − 𝐵∩𝐶∩𝐷 + 𝐴 ∩ 𝐵 ∩ 𝐶 ∩ 𝐷 ]……….(1) 2000 𝐻𝑒𝑛𝑐𝑒, | 𝐷 ∩ 𝐴 | = = 142 2×7 2000 |𝐷∩𝐵| = = 95 3×7 2000 |𝐷∩𝐶|= = 57 5×7 2000 |𝐴 ∩ 𝐵 ∩ 𝐷| = = 47 2×3×7 2000 |𝐴 ∩ 𝐶 ∩ 𝐷| = = 28 2×5×7 2000 |𝐵 ∩ 𝐶 ∩ 𝐷| = = 19 3×5×7

Agni College of Technology DEPARTMENT OF CSE

Subject Name: DISCRETE MATHEMATICS Subject code :MA 6566 Important Question and Answers PART-A 2000 =9 2×3×5×7

|𝐴 ∩ 𝐵 ∩ 𝐶 ∩ 𝐷| =

(1)⇒ 𝐴 ∩ 𝐵 ∩ 𝐶 ∩ D = 285 − 142 + 95 + 57 − 47 − 28 − 19 + 9 = 76 (ii) Solve the recurrence relation relation 𝑆(𝑛) = 𝑆(𝑛 − 1) + 2(𝑛 − 1)with 𝑆 0 = 3, 𝑆 1 = 1, 𝑏𝑦 𝑡𝑕𝑒 𝑚𝑒𝑡𝑕𝑜𝑑 𝑜𝑓 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠. Solution: 𝑛 Let 𝐺 𝑥 = ∞ 𝑛=0 𝑎𝑛 𝑥 ………….(1) Where 𝐺 𝑥 is the G.F for the sequence 𝑎𝑛 . Given 𝑎𝑛 = 𝑎𝑛−1 + 2𝑛 − 2, 𝑛 ≥ 1…………(2) ∞

∞

∞

𝑛

𝑛

𝑎𝑛 𝑥 = 𝑛=1 𝑛 =1 ∞ 𝑛 𝑛=1 𝑛𝑥 − 2

𝐺 𝑥 − 𝑎0 = 𝑥𝐺 𝑥 + 2 Since 𝑥𝐺 𝑥 =

∞ 𝑛 +1 𝑛=0 𝑎𝑛 𝑥

(2𝑛 − 2)𝑥 𝑛

𝑥 𝑎𝑛−1 +

𝑛 =1 ∞ 𝑛 𝑛=1 𝑥 [by (1)]

∞ 𝑛 𝑛 =1 𝑎𝑛−1 𝑥

=

2 +2 1−𝑥 2 𝐺 𝑥 (1 − 𝑥) = 3 + 2𝑥 1 − 𝑥 2 − +2 1−𝑥 2𝑥 2𝑥 =5+ − 2 (1 − 𝑥) 1−𝑥

𝐺 𝑥 − 𝑥𝐺 𝑥 = 3 + 2𝑥 1 − 𝑥

⇒𝐺 𝑥 = ∞

∞ 𝑛

∴ 𝑛 =0

∞

∞

−2 𝑛

𝑥 −2 𝑛=0

−

5 2 2𝑥 − − (1 − 𝑥) (1 − 𝑥)2 (1 − 𝑥)3

𝑛

𝑎𝑛 𝑥 = 5

2

(1 − 𝑥)−3 𝑥 𝑛

(1 − 𝑥) 𝑥 − 2𝑧 𝑛=0

𝑛 =0

Hence𝑎𝑛 = 5 1𝑛 − 2 𝑛 + 1 + 2𝑛(𝑛 + 1) which is the required solution. = 2𝑛2 + 3 4.(i) Prove that a given connected graph G is an Euler graph if and only if all the vertices of G are of even degree. Proof: Let G be an Eulerian graph. We have to prove all vertices are of even degree. Since G is Eulerian, G contains an Euler circuit, say,𝑣0 , 𝑒1 , 𝑣1 , 𝑒2 … , 𝑣𝑛 −1 , 𝑒𝑛 , 𝑣0 . Both the edges 𝑒1 and 𝑒𝑛 contributes one to the degree of 𝑣0 and so deg𝑣0 is at least two. In tracing this circuit we find an edge enters a vertex and another edge leaves the vertex contributing 2 to the degree of the vertex.

Agni College of Technology DEPARTMENT OF CSE

Subject Name: DISCRETE MATHEMATICS Subject code :MA 6566 Important Question and Answers PART-A This is true for all vertices and so each vertex is of degree 2, an even integer. Conversely, Let the graph G be such that all its vertices are of even degrees. We have to prove: G is an Eulerian graph We shall construct an Euler circuit and prove. Let 𝑣 be an arbitrary vertex in G. Beginning with 𝑣 form a circuit 𝐶: 𝑣, 𝑣1 , 𝑣2 , 𝑣3 … . 𝑣𝑛 −1 , 𝑣 This is possible because every vertex of even degree. We can leave a vertex (≠ 𝑣) along an edge not used to enter it.This tracing clearly stops only at the vertex 𝑣 because 𝑣 is also of even degree and we started from 𝑣.Thus we get a circuit or cycle 𝐶. If 𝐶 includes all the edges of G, then 𝐶 is an Euler circuit and so G is Eulerian. If 𝐶 does not contain all the edges of G,consider the subgraph H of G obtain by deleting all the edges of 𝐶 from 𝐺 and vertices not incident with the remaining edges. Note that all vertices of H have even degree. Since G is connected, H and C must have a common vertex 𝑢. Beginning with 𝑢 construct a circuit 𝐶1 for H. Now combine 𝐶and 𝐶1 to form a larger circuit 𝐶2 .If it is Eulerian. ie., if it contains all the edges of 𝐺,then 𝐺 is Eulerian. Otherwise continue this process until we get an Eulerian circuit. Since 𝐺 is finite this procedure must come to an end with a Eulerian circuit. Hence 𝐺 is Eulerian. (ii)

Let 𝑓: 𝐺 → 𝐺 ′ be a homomorphism of groups with kernel K. Then prove that K is a

normal subgroup of G and 𝐺/𝑘 is isomorphic to the image of f. Proof: Given 𝑓: 𝐺 → 𝐺 ′ be a homomorphism from the group 𝐺,∗ to the group 𝐺 ′ , ∆ . Then 𝑘 = ker 𝑓 = 𝑥 ∈ 𝐺/𝑓(𝑥) = 𝑒 ′ is a normal sub-group of 𝐺,∗ Also we know that the quotient set (𝐺/𝑘, ⨂) is a group 𝐺

Define 𝜙: 𝐺/𝑘 → 𝐺 ′ by 𝜙 𝑘𝑎 = 𝑓 𝑎 ∀ 𝑘𝑎 ∈ 𝑘 , 𝑎 ∈ 𝐺 𝝓is well-defined: Let 𝑘𝑎 = 𝑘𝑏 ∀𝑎, 𝑏 ∈ 𝐺

Agni College of Technology DEPARTMENT OF CSE

Subject Name: DISCRETE MATHEMATICS Subject code :MA 6566 Important Question and Answers PART-A 𝑘𝑎 = 𝑘𝑏 ⟹ 𝑏 ∈ 𝑘𝑎, 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑘 ∈ 𝐾[∵ 𝐻 𝑖𝑠 𝑎 𝑠𝑢𝑏𝑔𝑟𝑜𝑢𝑝 𝑜𝑓 𝑎 𝑔𝑟𝑜𝑢𝑝 𝐺 &𝑎, 𝑏 ∈ 𝐺 𝑡𝑕𝑒𝑛 𝑎 ∈ 𝐻 ⟺ 𝐻𝑎 = 𝐻] ⟹ 𝑓 𝑏 = 𝑓 𝑘𝑎 = 𝑓 𝑘 .𝑓 𝑎 = 𝑒 ′ . 𝑓(𝑎)[∵ 𝑘 = 𝑘 𝑘 ∈ 𝐺&𝑓 𝑘 = 𝑒 ′ ] ∴𝑓 𝑏 =𝑓 𝑎 Thus 𝑘𝑎 = 𝑘𝑏 ⟹ 𝜙 𝑘𝑏 = 𝜙 𝑘𝑎 ∴ 𝜙is well defined. 𝝓 𝒊𝒔 𝟏 − 𝟏: Let 𝜙 𝑘𝑎 = 𝜙 𝑘𝑏 ⇒𝑓 𝑎 =𝑓 𝑏 ⇒ 𝑓 𝑎 . [𝑓 𝑏 ]−1 = 𝑓 𝑏 . [𝑓 𝑏 ]−1 ⇒ 𝑓 𝑎 . 𝑓(𝑏 −1 ) = 𝑒 ′ ⇒ 𝑓(𝑎𝑏 −1 ) = 𝑒 ′ ⇒ 𝑎𝑏 −1 ∈ ker 𝑓 = 𝑘 ⇒ 𝑎𝑏 −1 ∈ 𝑘 ⟹ 𝑎 ∈ 𝑘𝑏 ⟹ 𝑘𝑎 = 𝑘𝑏 Thus 𝜙 𝑘𝑎 = 𝜙 𝑘𝑏 ⟹ 𝑘𝑎 = 𝑘𝑏 ∴ 𝜙 is 1-1 𝝓 is onto: Let 𝑎′ ∈ 𝐺 ′ & 𝑓: 𝐺 → 𝐺 ′ is an epimorphism ⟹ ∃ an element 𝑎 ∈ 𝐺 ∋: 𝑓 𝑎 = 𝑎′ Hence 𝜙 𝑘𝑎 = 𝑓 𝑎 = 𝑎′ 𝑎′ ∈ 𝐺 ′ ⟹ ∃an element 𝑘𝑎 ∈ 𝐺/𝑘 ∋: 𝜙 𝑘𝑎 = 𝑎′

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Subject Name: DISCRETE MATHEMATICS Subject code :MA 6566 Important Question and Answers PART-A ∴ 𝜙 is onto. 𝝓 is homomorphism: Let 𝑘𝑎, 𝑘𝑏 ∈ 𝐺

𝑘, 𝑎, 𝑏 ∈ 𝐺 𝜙 𝑘𝑎𝑘𝑏 = 𝜙 𝑘𝑎𝑏 = 𝑓 𝑎𝑏 =𝑓 𝑎 𝑓 𝑏 = 𝜙 𝑘𝑎 𝜙 𝑘𝑏

∴ 𝜙 is homomorphism ∴ 𝜙 is well defined,1-1,onto&homomorphism. ∴ 𝜙 is an isomorphism Thus 𝐺/𝑘 ≅ 𝐺′. 5.(i) State and prove the Lagrange’s theorem . Statement : Lagrange’s theorem The order of a subgroup H of a finite group G divides the order of the group. ie) 𝑂(𝐻)/ 𝑂(𝐺). Proof:Given G is a finite group of order n . H is a subgroup of G, Let 𝑂 𝐺 = 𝑛 & 𝑂 𝐻 = 𝑚 ie |𝐺| = 𝑛 & |𝐻| = 𝑚 To prove: 𝑂(𝐻)/𝑂(𝐺). ie ) to prove :𝑚/𝑛 G is finite ⇒ There are only a finite number of distinct left cosets of H in G, let it be 𝑎1 𝐻, 𝑎2 𝐻,………, 𝑎𝑘 𝐻. These cosets are pairwise disjoint and 𝐺 = 𝑎1 𝐻 ∪ 𝑎2 𝐻 ∪ … … ∪ 𝑎𝑘 𝐻 Then 𝑛 = 𝑂 𝐺 = 𝑂 𝑎1 𝐻 + 𝑂 𝑎2 𝐻 + ⋯ + 𝑂 𝑎𝑘 𝐻 = 𝑂 𝐻 + 𝑂 𝐻 + ⋯ + 𝑂 𝐻 𝑘 𝑡𝑖𝑚𝑒𝑠 = 𝑘𝑂 𝐻 =𝑘∙𝑚 ⟹ 𝑚/𝑛ie) 𝑂(𝐻)/𝑂(𝐺). (ii) Prove that every chain is distributive lattice.

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Subject Name: DISCRETE MATHEMATICS Subject code :MA 6566 Important Question and Answers PART-A Solution: Let 𝐿, ≤ be a chain and 𝑎, 𝑏, 𝑐 ∈ 𝐿 It enough to verify one of the conditions say𝑎 ∗ 𝑏⨁𝑐 = 𝑎 ∗ 𝑏 ⨁ 𝑎 ∗ 𝑐 … … … … . (1) This is symmetric in b and c Since L is a chain either 𝑏 ≤ 𝑐 and 𝑐 ≤ 𝑏 Because of the symmetric of the roles of b and c We shall consider only of them say 𝑏 ≤ 𝑐 Then 𝑏⨁𝑐 = 𝑐 … … … … (2) We have the following cases . Case i: 𝑎 ≤ 𝑏 ≤ 𝑐 Then 𝑎 ∗ 𝑏 = 𝑎, 𝑎 ∗ 𝑐 = 𝑎 𝑎 ∗ 𝑏⨁𝑐 = 𝑎 ∗ 𝑐 = 𝑎 [𝑏𝑦 2 ] 𝑎 ∗ 𝑏 ⨁ 𝑎 ∗ 𝑐 = 𝑎⨁𝑎 = 𝑎 𝑎∗ 𝑏∗𝑐 = 𝑎∗𝑏 ⨁ 𝑎∗𝑐 Case ii: 𝑏 ≤ 𝑐 ≤ 𝑎 Then 𝑎 ∗ 𝑏 = 𝑏, 𝑎 ∗ 𝑐 = 𝑐 𝑎 ∗ 𝑏⨁𝑐 = 𝑎 ∗ 𝑐 = 𝑐 𝑎 ∗ 𝑏 ⨁ 𝑎 ∗ 𝑐 = 𝑏⨁𝑐 = 𝑐 𝑎 ∗ 𝑏⨁𝑐 = 𝑎 ∗ 𝑏 ⨁ 𝑎 ∗ 𝑐 Case iii: 𝑏 ≤ 𝑎 ≤ 𝑐 Then 𝑎 ∗ 𝑏 = 𝑏, 𝑎 ∗ 𝑐 = 𝑎 𝑎 ∗ 𝑏⨁𝑐 = 𝑎 ∗ 𝑐 = 𝑎 𝑎 ∗ 𝑏 ⨁ 𝑎 ∗ 𝑐 = 𝑏⨁𝑎 = 𝑎[∵ 𝑏 ≤ 𝑎] 𝑎 ∗ 𝑏⨁𝑐 = 𝑎 ∗ 𝑏 ⨁ 𝑎 ∗ 𝑐 So, the distributive law hold in all the cases. Hence the result.

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Subject Name: DISCRETE MATHEMATICS Subject code :MA 6566 Important Question and Answers PART-A 6.(i) In any Boolean algebra, show that 𝑎 + 𝑏 ′ 𝑏 + 𝑐 ′ 𝑐 + 𝑎′ = (𝑎′ + 𝑏)(𝑏 ′ + 𝑐)(𝑐 ′ + 𝑎) Solution: LHS= 𝑎 + 𝑏 ′ + 0 𝑏 + 𝑐 ′ + 0 𝑐 + 𝑎′ + 0 = 𝑎 + 𝑏 ′ + 𝑐. 𝑐′ 𝑏 + 𝑐 ′ + 𝑎. 𝑎′ 𝑐 + 𝑎′ + 𝑏. 𝑏′ ′ = 𝑎 + 𝑏 + 𝑐 . 𝑎 + 𝑏 ′ + 𝑐′ . 𝑏 + 𝑐 ′ + 𝑎 . 𝑏 + 𝑐 ′ + 𝑎′ . 𝑐 + 𝑎′ + 𝑏 . 𝑐 + 𝑎′ + 𝑏′ = 𝑎′ + 𝑏 + 𝑐 . 𝑎′ + 𝑏 + 𝑐 ′ . 𝑏′ + 𝑐 + 𝑎 . 𝑏′ + 𝑐 + 𝑎′ . { 𝑐′ + 𝑎 + 𝑏 . 𝑐′ + 𝑎 + 𝑏 ′ } = 𝑎′ + 𝑏 + 𝑐𝑐′ . 𝑏′ + 𝑐 + 𝑎𝑎′ . 𝑐′ + 𝑎 + 𝑏𝑏′ = 𝑎′ + 𝑏 + 0 . 𝑏′ + 𝑐 + 0 . 𝑐′ + 𝑎 + 0 = (𝑎′ + 𝑏)(𝑏 ′ + 𝑐)(𝑐 ′ + 𝑎) = 𝑅. 𝐻. 𝑆 (ii)

Show that in a distributive and complemented lattice satisfied DeMorgan’s laws.

Proof: Let (L, ) be a Boolean lattice . Ie., L is complemented and distributive lattice. The De-Morgan’s laws are (i) 𝑎⨁𝑏 = 𝑎 ∗ 𝑏 ;(ii)𝑎 ∗ 𝑏 = 𝑎⨁𝑏 ∀𝑎, 𝑎, 𝑏 ∈ 𝐿 Assume that 𝑎, 𝑏 ∈ 𝐿. There exists elements 𝑎, 𝑏 ∈ 𝐿 such that 𝑎⨁𝑎 = 1; 𝑎 ∗ 𝑎 = 0; 𝑏⨁𝑏 = 1; 𝑏 ∗ 𝑏 = 0. Claim:𝑎⨁𝑏 = 𝑎 ∗ 𝑏 Now 𝑎 ⊕ 𝑏 ⊕ 𝑎 ∗ 𝑏 = 𝑎 ⊕ 𝑏 ⊕ 𝑎 ∗ 𝑎 ⊕ 𝑏 ⊕ 𝑏 = 𝑎⨁𝑎 ⊕ 𝑏 ∗ [𝑎 ⊕ 𝑏 ⊕ 𝑏 ] = 1⊕𝑏 ∗ 𝑎⊕1 =1∗1=1 𝑎⊕𝑏 ∗ 𝑎∗𝑏 = 𝑎⊕𝑏 ∗𝑎 ∗ 𝑎⊕𝑏 ∗𝑏 = (𝑎 ∗ 𝑎) ⊕ (𝑏 ∗ 𝑎) ∗ [(𝑎 ∗ 𝑏 ) ⊕ (𝑏 ∗ 𝑏) ] = 0 ⊕ (𝑏 ∗ 𝑎) ∗ [(𝑎 ∗ 𝑏 ) ⊕ 0 ] = 𝑏 ∗ 𝑎 ∗ (𝑎 ∗ 𝑏 ) = 𝑏∗ 𝑎∗𝑎 ∗𝑏 = 𝑏∗ 0 ∗𝑏 =0 Hence (i) is proved (ii)Claim:𝑎 ∗ 𝑏 = 𝑎⨁𝑏 Now 𝑎 ∗ 𝑏 ⊕ 𝑎⨁𝑏 = 𝑎 ∗ 𝑏 ⊕ 𝑎 ∗ 𝑎 ∗ 𝑏 ⊕ 𝑏 = 𝑎⨁𝑎 ∗ (𝑏 ⊕ 𝑎) ∗ [(𝑎 ⊕ 𝑏) ⊕ (𝑏 ⊕ 𝑏) ] = 1 ∗ (𝑏 ⊕ 𝑎) ∗ 𝑎 ⊕ 𝑏 ∗ 1 = (𝑏 ⊕ 𝑎) ⊕ (𝑎 ⊕ 𝑏) =𝑏⊕ 𝑎⊕𝑎 ⊕𝑏 = 𝑏⊕1⊕𝑏 = 1 𝑎 ∗ 𝑏 ∗ 𝑎⨁𝑏 = 𝑎 ∗ 𝑏 ∗ 𝑎 ∗ 𝑎 ∗ 𝑏 ∗ 𝑏 = 𝑎 ∗ 𝑎 ∗ 𝑏 ⊕ [𝑎 ∗ 𝑏 ∗ 𝑏) ] = 0 ∗ 𝑏 ⊕ [𝑎 ∗ 0) ] = 0+0 =0

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Subject Name: DISCRETE MATHEMATICS Subject code :MA 6566 Important Question and Answers PART-A

Hence (ii) proved. 7.(i) Show that every chain is a lattice. Solution:Let (𝑃, ≤) be a chain.so any two elements of p are comparable. So far any two elements 𝑎, 𝑏 ∈ 𝑃, 𝑒𝑖𝑡𝑕𝑒𝑟 𝑎 ≤ 𝑏 𝑜𝑟 𝑏 ≤ 𝑎. Case i: Let 𝑎 ≤ 𝑏.Then b is an upper bound of{a,b} If c is any other upperbound of {a,b} then 𝑎 ≤ 𝑐 and 𝑏 ≤ 𝑐 Thus 𝑏 ≤ 𝑐 for every upper bound c of a and b. Hence b is the LUB {a,b} ⟹𝑎⊕𝑏 =𝑏 Since 𝑎 ≤ 𝑏, a is a lower bound of {a,b}. If d is any other lower bound of {a,b} then 𝑎 ≤ 𝑎 and 𝑑 ≤ 𝑏 Thus 𝑑 ≤ 𝑎 for every lower bound d. Hence a is the GLB {a,b} ⟹𝑎∗𝑏 =𝑎 Thus we have proved for any two elements a,b if 𝑎 ≤ 𝑏, then GLB and LUB exist for {a,b}. Caseii: Let 𝑏 ≤ 𝑎.proceeding as above we can prove GLB and LUB exist for {a,b}. Thus for any 𝑎, 𝑏 ∈ 𝑃 by cases (i) and (ii) 𝑎 ∗ 𝑏 𝑎𝑛𝑑 𝑎 ⊕ 𝑏 exist. Hence the chain (𝑃, ≤) is a lattice.

 7(7Q  7P)] Sol. P  [(P  Q)  7(7Q  7P)]  7P  [(P  Q)  7(7Q 7P)]  7P [(7P Q)  (Q  P)] ii) Obtain PCNF and PDNF of P  [(P  Q)

 [7P (7P Q)]  [7P (Q  P)]  [(7P 7P)  Q]  [(7P Q)  (7P P)]  (7P Q)  [(7P Q)  T]

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Subject Name: DISCRETE MATHEMATICS Subject code :MA 6566 Important Question and Answers PART-A

 (7P Q)  (7P Q)  (7P Q), which is PCNF. To find PDNF

Now,7A  (P  Q)  (P  7Q)  (7P  7Q) 7(7A)  7 [(P  Q)  (P  7Q)  (7P  7Q)] A  7(P Q)  7(P 7Q)  7(7P 7Q) A  (7P  7Q)  (7P  Q)  (P  Q) which is PDNF.