Solution

A totally ordered set(TOS) is a partially ordered set in which every two elements are comparable.

If A is a set of positive integers less than 200, then Poset [A, ≤ ]

A has elements ranging from {1,2,3,4,.....200} each number is comparable with other number with the relation ‘≤’, thus its TOS

If A is a set of prime numbers less than 25, then poset [A,≤]

A ={2,3,5,7,11,13,17,19,23}.

Each number is comparable with other number, i.e we can compare with any two numbers with the relation ‘ ≤”

If A is a set of divisors of 20 then poset [A,/]

A={1,2,4,5,10,20}. Given relation is ‘/’. Here each number does not divide every number. I.e, they are not comparable. Eg: 5 cannot be divided by 4. Thus it is not a TOS

If A ={phi, {a},{a,b},{a,b,c} }

Each set can be compared with other set with the relation ⊆ . Thus its a TOS

Solution

The power set (or powerset) of any set S is the set of all subsets of S, including the empty set and S itself.

Consider a set A = {1,2,3,4}

For the set S = {1, 2, 3, 4} this means:

• Subsets with 0 elements: 0 (the empty set)
• Subsets with 1 element: {1}, {2}, {3}, {4}
• Subsets with 2 elements: {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}
• Subsets with 3 elements: {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}

Subsets with 4 elements: S

The number of subsets with 1 element are equal to the number of subsets with 3 elements . Thus it's wrong.

Solution

If |A| = n

AA contains, ‘n’ order pairs of type(a,a) i.e., diagonal elements

and (n^2-n) number of pairs of ordered pairs of type (x,y), (y,x)

eg: A = {1, 2}

A symmetric relation contains the pairs of ordered pairs, if we consider half of the non-diagonal ordered pair, thus total no. of symmetric relation

Solution

Reflexive :

A relation R on a set A is called reflexive if (a, a) ∈ R for every element a ∈ A.

We have all self loops like (a,a),(b,b),(c,c),(d,d) in R, so this is reflexive

Symmetric:

A relation R on a set A is called symmetric if (b, a) ∈ R whenever (a, b) ∈ R, for all a, b ∈ A

We have (a,b),(b,a), there are no other ordered pairs of this type. So, this is Symmetric

Antisymmetric:

A relation R on a set A such that for all a, b ∈ A, if (a, b) ∈ R and (b, a) ∈ R, then a = b is called antisymmetric.

Here, we have (a,b),(b,a) but a is not equal to b., So this is not antisymmetric

Transitive:

A relation R on a set A is called transitive if whenever (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R, for all a, b, c ∈ A. wehave (a,b),(b,a) as well as (a,a). So this is transitive

Solution

Solution

Complement of an element:

Let L be a bounded lattice, for any element a ∈ L, if there exists an element b ∈ L such that (a v b) = I and (aΛb) = O, then ‘b’ is called ‘complement of a’.

Here (5 v6) = lcm(5,6)=30 and (5 Λ 6) = gcd(5,6)=1.

Solution

An algebraic structure is called a Sumi Group if and only if a*(b*c) = (a*b)*c i.e. associative

I)

Set of integers contains Z = {... , -n ,.... ,-4 ,-3 ,-2 ,-1 ,0 ,1 ,2 ,3 ,4 ,... ,n, .......}

it must be a-(b-c) = (a-b)-c

Consider a=1, b=-2 c=3 then

1-(-2-3) =( 1-(-2)) -3

6 = 0 which is false. Thus its not semi group

II) Set of natural numbers N = {0, 1, 2, 3, ...}

a*(b*c) = (a*b)*c holds true as multiplication is associative Thus it is semi group

Solution

A group (G,*) is a cyclic group, if there exists an element a belongs to G such that every element of G can be written as a^n for some integer n.

Consider the element i,

(i)^2 = -1,

(i)^4 = 1

(i)^3 = -i

Also,

(-i)^2 = -1,

(-i)^3= -i

(-i)^4= 1

(-i)^5 = i

But, (1)^n is always 1, so we cannot derive i from it, same with -1 too.

So, generators ar i, -i. Thus number of generators are 2.

Solution

(P -> Q ) Λ (P -> R) ≡ P -> ( Q Λ R )

(P -> Q) Λ (Q -> R) ≡ (P v Q) -> R

(P -> Q) v (P -> R) ≡ P -> ( Q Λ R )

These are the standard equivalences of implication distribution.

Solution

Solution

a^2 yields same value for positive and negative values.

For eg, f(-1) = 1+1 =2 , f(1) = 1+1 = 2 so, its not one one.

Lets check onto,

1+a^2 an never be 0 ,

1+a^2=0,

a= sqrt(-1). Which is not a real value. Thus the value ‘0’ is not mapped to any value in the domain.

Thus its not onto.

Solution

The number of functions from A to B is |B|^|A|

Here, 5^3 = 125.

Solution

In a poset, an element not related to another element is called maximal element

In the given poset, 4, 6 does not related to any other elements, so both 4,6 are maximal elements

The sum of the maximal elements are asked in this question, thus 4+6 ie. 10 is the answer

Solution

A relation R is said to be Partial order if and only if the Relation R is Reflexive, Anti symmetric, Transitive . It is the property of the Partial order.

Solution

For any set A, ‘A”, “ ϕ” are called as the trivial subsets of A.

The III is called proper subset of A.