Each element of set can be written as a power of g in multiplicative notation, or as a multiple of g in additive notation. This element g is called a generator of the group.
We can observe that, a is an identity element. ( a *x = x ). An identity element cannot be a generator, as it cannot produce any other element ( always a*a*... = a).
Also, b*b =a, so it also cannot produce all other elements ( always b*b*... =a , where a is identify element).
c,d are able to produce other elements like { c*c =b, c*(c*c) = c*b= d, c*(c*(c*c))) = c*(c*b)= c*d=a. }. Similar with d.

Given, (P∩Q∩R)∪(P^c∩Q∩R)∪Q^c∪R^c
It can be written as the p.q.r + p'.q.r +q'+r'
=> (p+p').q.r + q' +r'
=> q.r +(q'+r')
=> q.r + q'+r' = 1 i.e U

If a hasse diagram is a lattice then every pair of elements will have exactly one lub and gub.
(ii) and (iii), having more than one lub or glb for some pairs due to which they are not lattice.

A relation is said to be equivalent relation if it satisfies,
→ Reflexive
→ Symmetric
→ Transitive
Let a set S be,
S = {1, 2, 3}
Now, the smallest relation which is equivalence relation is,
S×S = {(1,1), (2,2), (3,3)}
= 3
= n (for set of n elements)
And, the largest relation which is equivalence relation is,
S×S = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}
= 9
= 3²
= n² (for set of n elements)

In this problem first of all we find the exponents of prime nos. that can be taken out from given number.
4 = 2²
So, prime no. is 2 and power of 2 is 2. So exponent value 2 is considered now.
Now the no. of ways we can divide 2 into sets will be the answer.
So division can be done as,
{1,1}, {0,2}
in two ways. Hence, answer is 2.

For ordered pair (a, b) to be in P, each digit in starting from unit place must not be larger than the corresponding digit in b.
The given condition can be satisfied by
(iii) (145, 265) → 5 ≤ 5, 4 < 6 and 1< 2
(iv) (0, 153) → 0 < 3

Let us consider
E = {(1,1), (1,2), (1,3), (2,2), (2,3), (3,3), (3,1)}
F = {(1,1), (2,2), (3,3)}
G = {(1,3), (2,1), (2,3), (3,1)}
X = (E∩F) - (F∩G)
= {(1,1), (2,2), (3,3) - ∅}
= {(1,1), (2,2), (3,3)} (✔️)
Y = (E - (E∩G) - (E - F))
= (E - {(1,3), (2,3), (3,1)} - {(1,2), (1,3), (2,3), (3,1)})
= {(1,1), (1,2), (1,3), (2,2), (2,3), (3,3), (3,1)} - {(1,3), (2,2), (2,1)} - (1,2), (1,3), (2,3), (3,1)}
= {(1,1), (1,2), (2,2), (3,3)} - {(1,2), (1,3), (2,3), (3,1)}
= {(1,1), (2,2), (3,3)} (✔️)
X = Y

X = (E∩F) - (F∩G) = {2,5} - {5} = {2}
Y = (E - (E∩G) - (E - F))
= {(1,2,4,5) - (4,5) - (1,4)}
= {(1,2) - (1,4)}
= {2}
X = Y

Proceed by taking f(x₁) = x,
LHS: f(x₁) = 0 where x₁ = 0
LHS: f(x₁) = 1 when x1 = 1
RHS: f(0) + f(1) = 0 + 1 = always 1

P = {1, 2, 4, 5}
Q = {2, 3, 5, 6}
R = {4, 5, 6, 7}
(1) P Δ (Q ∩ R) = (P Δ Q) ∩ (P Δ R)
P Δ ({2,3,5,6} ∩ {4,5,6,7}) = ({1,2,4,5} Δ {2,3,5,6} ∩ {1,2,4,5} Δ {4,5,6,7})
P Δ {5,6} = ({1,2,3,4,5,6} - {2,5}) ∩ ({1,2,4,5,6,7} - {4,5})
({1,2,4,5} Δ {5,6}) = {1,3,4,6} ∩ {1,2,6,7}
{1,2,4,5,6} - {5} = {1,6}
{1,2,4,6} ≠ {1,6}
Statement-1 is False.
(2) P ∩ (Q ∩ R) = (P ∩ Q) Δ (P Δ R)
LHS:
{1,2,4,5} ∩ {5,6} = {5}
RHS:
({1,2,4,5} ∩ {2,3,5,6}) Δ ({1,2,4,5} Δ {4,5,6,7})
{2,5} Δ ({1,2,4,5,6,7} - {4,5})
{2,5} Δ {1,2,6,7}
{1,2,5,6,7} - {2}
{1,5,6,7}
LHS ≠ RHS
Statement - 2 is also wrong.

It is lattice because both LUB and GLB exist for each pair of the vertex in above Hasse diagram.
But it is not distributed because for some element there are more than one complement. Moreover if it is not distributive then it can't be boolean.

abc != 0 & Identity matrix is identity then it is non-singular and inverse is also defined.
The algebraic structure is a group because the given matrix can have inverse and the inverse is non-singular.

a = cmod3
That means a=0,1,2 ⇒ |3|
b = dmod5
That means b=0,1,2,3,4 ⇒ |4|
→ Total no. of order pairs = 3 * 5 = 15
→ Ordered pair (c,d) has 1 combination.
Then total no. of combinations = 15+1 = 16

R∪S might not be transitive.
Let (a,b) present in R and (b,c) present in S and (a,c) is not present in either of them. Then R∪S will contain (a,b) and (b,c) but not (a,c) and hence not transitive.
And equivalence relation must satisfy 3 property:
(i) Reflexive
(ii) Symmetric
(iii) Transitive
But as we have seen that for R∪S, Transitivity is not satisfied.

Transitive means that x is related to greater value y and reflexive means x related to y.
Answer is option B.
{(x, y)|y ≥ x and x, y ∈ {0, 1, 2, ... }}

The last row is c e a b.

A lattice is complete if every subset of partialordered set is to be a supremum and infimum element.
For the set {1, 2, 3, 4, 5} there is no supremum element i.e., {1}.
Then clearly we need to add {1}, then it is to be a lattice.

In the concatenation '∊' is the identity element. And given one is not a group because no element has inverse element.
→ To perform concatenation with the given set can result a Monoid and it follows the property of closure, associativity and consists of identity element.

Total possible values = 3² = 9 i.e., (a,a), (b,b), (c,c), (a,b), (a,c), (b,a), (b,c), (c,a), (c,b)
In those (x, y) = (b,c) & (c,b) are the possible solution for the corresponding equations.
(x, y) = (b,c) ⇒ (a*b)+(a*c) ⇒ (b*b)+(c*c)
⇒ (b) + (c) ⇒ c + b
⇒ c (✔️) ⇒ c (✔️)
(x,y) = (c,b) ⇒ (a*c)+(a*b) ⇒ (b*c)+(c*b)
⇒ c+b ⇒ a+c
⇒ c (✔️) ⇒ c (✔️)

The answers can have only three possibilities of sequences i.e., "a", "a", "a", and there is no multiplies of 7 and 9. So eliminates option A & C.
And f(S) = 2³ = 8
So answer is 2⁸3⁸5⁸.

A group G should follow 4 properties:
a. Closure: result of a * b must be in group G.
b. There must be an identity element i.e. e * a = a * e = a
c. There must be an inverse element b for every element a such that a * b = b * a = e
d. Associativity i.e. (a * b) * c = a * (b * c)
Rational negative numbers don't form a group under multiplication, as multiplying two negative numbers results into a positive number, so closure property is not satisfied.
Set of non-singular matrices forms a group under multiplication.
The set of all matrices doesn't form a group under multiplication, since there may not be an inverse for a matrix (in particular, for singular matrices).

S = ɸ; A = {1,2,3}
A×S = {(ɸ,1), (ɸ,2), (ɸ,3), (1,ɸ), (2,ɸ), (3,ɸ)}
Not reflexive = (1,1), (2,2), (3,3) not there
Symmetric: if (a,b) is present then (b,a) is also present
Transitive: True; if (x,y), (y,z) then (z,x) is also present

(a) S = {(1,2), (2,1)}; A = {1,2,3}
The given relation S is irreflexive because no diagonal elements are present in the relation i.e., (x,x)∉R, ∀x∈A
If a relation is equivalence then it is
Reflexive
Symmetric
Transitive
Given relation S is not Reflexive & Transitive.
Reflexive closure on S = {(1,1), (2,2), (3,3), (1,2), (2,1)}
Transitive closure on S is does not change after performing reflexive closure on S.
S = {(1,1), (2,2), (3,3), (1,2), (2,1)}
(b) Given S = {a,b}
Powerset of S i.e., P(S) = {(ɸ), (a), (b), (a,b)}
Hasse diagram:

It is reflexive R(x+y) is even ⇒ R(x+x) is even.
It is symmetric R(x+y) is even ⇒ R(x+y) is even then R(y+x) is also v.
It is transitive R(x+y) is even ⇒ R((x+y)+z)) is even then R(x+(y+z)) is also even.
This given relation is equivalence.
Sum of two even numbers are even. Same sum of two odd numbers are even but one even, one odd is odd. It can have two equivalent classes.

Let us consider
S={1}
P(S)=[{ }, {1}]
P(P(S)) = [{ }, {1}, {{ }, 1}, {1, { }}]
Option A: P(P(S)) ≠ P(S) (wrong)
Option B: P(S) ∩ P(P(S)) ≠ ɸ (wrong)
Option C: P(S) ∩ S ≠ P(S) (wrong)
Option D: S ∈ P(S) (wrong)
None of the given is correct.

A1 is the identity element here. Inverse is does not exist for zero then it is not a group.

(a) No. of multiset of size 4 that can be constructed from n distinct elements so that atleast one element occurs exactly twice:
= multiset of size 4 with exactly one element occurs exactly twice + multiset of size 4 with exactly two element occurs exactly twice.

(b) Infinite, because size of multiset is not given.

In binary relation two elements are selected from a set. So total no. of pairs possible are
n×n = n²
Now, no. of binary relations possible is
2ⁿ²
Because each pair have two possibilities either chosen or not chosen.

Note: Options given are wrong.

In a largest equivalence relation the every element is related to other element
⇒ n×n
⇒ n²

If R₁ and R₂ be two equivalence relations on a set 'S' then the transitive property tells
(i) R₁ ∩ R₂ is also transitive.
(ii) R₁ ∪ R₂ is need not to be transitive.

→ Not reflexive because (4, 4) not present.
→ Not irreflexive becuaes (1, 1) is present.
→ Not symmetric because (2, 1) is present but not (1, 2).
→ Not antisymmetric - (2, 3) and (3, 2) are present.
It is transitive so correct option is (B).

Semigroup - Closed and associative
Monoid - Closed, Associative and has an identity
Group - Monoid with inverse
Abelian group - Group with commutative property
Go through with given:
Closure: Yes.
(m*n = max(m,n)) output is either m or n whichever is maximum since m,n belongs z. The result of the binary operation also belongs to z. So given is satisfying closure property.
Associative: Yes.
The output is max among the elements and it is associative.
Identity: No.
We don't have single unique element for all the elements which is less than all the elements.
Given one is semigroup only.

g, c, d are the complements of e.
No. of complements = 3
e∨g = a and e∧g = f
e∨c = a and e∧c = f
e∨d = a and e∧d = f

To make this partial order a total order, we need the relation to hold for every two element of the partial order. Currently between any a_i and a_j, there is no relation.
So for every a_i, a_j we have to add either (a_i, a_j) or (a_j, a_i in total order. So, this translates to giving an ordering for n elements between x and y, which can be done in n! ways. So answer is (A).

The no. of equivalence relation with 4 elements is 15.

(A – B) ∪ (B - A) ∪ (A∩B)

(A - B) = 1
(B - A) = 2
(A∩B) = 3
A∪B = (1∪2∪3)
(A – B) ∪ (B - A) ∪ (A∩B) = 1∪2∪3 = (A∪B)

No. of edges = 4

Rational number consists of number '0'. If 0 is present in a set inverse is not possible under multiplication.

Let A = {1, 2, 3} and B = {4, 5} and C = {1, 6, 7}
Now,
A ∩ B = ф
& B ∩ C = ф
But A ∩ B ≠ ф
So, R is not transitive.
A ∩ B = A, so R is not reflexive.
If A ∩ B = ф
then definitely B ∩ A = ф.
Hence, R is symmetric.
So, option (B) is true.

Option (C) is False because string concatatieon operation is monoid(doesn't have inverse to become a group).

R1:
a+a=2a
The set (a,a) is reflexive.
The set representation (a,a), (b,b) is also Symmetric.
The set representation is Transitive.
So, this is Equivalence.
R2:
a+a≠2a
The (a,a) is non-reflexive.
R3:
a⋅a>0 → Reflexive
a⋅b>0 and b⋅a>0 → Symmetric
a⋅b>0, b⋅c>0 then c⋅a>0 → Transitive
The relation R3 is equivalence relation.
R4:
|a - a| ≤ 2, which is not possible, not Reflexive.
R1 & R3 are equivalence, R2 & R4 are not.

S1: Consider A = {2,4,6,8,10,...} set of all even numbers
B = {2,3,5,7,11,13,...} set of prime numbers
C = {1,3,5,7,...} set of odd numbers
(B∪C) = {1,2,3,5,7,...}
A∩(B∪C) = {2}
Which is finite, S1 is true.
S2: Consider A=5+√3 Irrational
B=5-√3 Irrational
A+B=10 Rational
A+B, which is rational number, S2 is true.

S1: f(E∪F) = f(E)∪f(F)
The both LHS and RHS contains the same element in E and F.
So, S1 is true.
S2: f(E∩F) = f(E)∩f(F)
E and F are partitions of A.
f(E)∩f(F) = ϕ
f(ϕ) is not defined, but f(E)∩f(F)=1.
So, S2 is false.

Transitive closure of the relation {(1,2), (2,3), (3,4), (5,4)} ={(1,2), (2,3), (1,3), (3,4), (2,4), (1,4), (5,4)}
Transitive closure of R:
(a) It is transitive.
(b) It contains R.
(c) It is minimal, satisfies a & b.

Equivalence relation should satisfy Reflexive, Symmetric and Transitive property.
Reflexive ∪ Reflexive = Reflexive
Symmetric ∪ Symmetric = Symmetric
Transitive ∪ Transitive ≠ Transitive
Since Transitivity does not satisfy union, so union of two equivalence relation is not an equivalence relation.

True, every infinite cyclic group is isomorphic to the infinite cyclic group of integers under addition.