Discrete Mathematics Operators and Postulates - Discrete Mathematics

What are Operators & Postulates of Discrete Mathematics?

A structure of algebra named as group is defined by a branch of mathematics known as Group Theory. Group includes a set of elements and operation over that element.

Group is defined by the British Mathematician Arthur Cayley in 1854 as, A set of different symbols when put together, or the product of any two of them or the product of any one into itself also belongs to the set, then it is called as a group.

Operators and postulates are the basics of set theory, group theory and Boolean algebra. The elements of the mathematical system are defined using a set of operators and the postulates number.

A rule in which a unique element is assigned a pair of the element from the set of elements is defined as Binary operator. For instance, For the set A={1,2,3,4,5}, the binary operator is ⊗ for the operation c=a⊗b, for finding the value of c for the pair (a,b), by the condition that a,b,c∈A.

The basic assumptions from which the rules are deduced together form a mathematical system known as postulates.

What are the features of Postulates?

The features of postulates are:

Closure

If a unique element from the sent is found by the operator for every pair of elements in the set, the set is considered to be closed with respect to a binary operator.

Example

For instance, A={0,1,2,3,4,5,…}

This set is considered to be closed under the binary operator into (∗), for the operation, c=a∗b for any a,b∈A, the product c∈A.

The set is not considered to be closed under the binary operator divide (÷), for the operation c=a÷b, for any a,b∈A and the product c may not be in the set A.

If a=7,b=2a=7,b=2, then c=3.5c=3.5. Here a,b∈A but c∉A.

Associative Laws

When the following property is true for the binary operator ⊗ on set A, it is said to be associative.

(x⊗y)⊗z=x⊗(y⊗z), where x,y,z∈A

Example

Let A={1,2,3,4}

The operator plus (+) is associative because for any three elements, x,y,z∈A, the property (x+y)+z=x+(y+z) holds.

The operator minus (−) is not associative since

(x−y)−z≠x−(y−z)

Commutative Laws

When the following property is true for the binary operator ⊗ on set A, it is said to be commutative.

x⊗y=y⊗x, where x,y∈A

Example

Let A={1,2,3,4}

The operator plus (+) is commutative because for any two elements, x,y∈A, the property x+y=y+x holds.

The operator minus (−) is not associative since

x−y≠y−x

Distributive Laws

When the following property is true for two binary ⊗ and ⊛ on set A, the set is considered to be distributive over the operator ⊛.

x⊗(y⊛z)=(x⊗y)⊛(x⊗z), where x,y,z∈A

Example

Let A={1,2,3,4}

The operators into (∗) and plus (+) are distributive over operator + because for any three elements, x,y,z∈A, the property x∗(y+z)=(x∗y)+(x∗z) holds.

However, these operators are not distributive over ∗ since

x+(y∗z)≠(x+y)∗(x+z)

Identity Element

When the following property holds true, there exist an element e∈A, if the set A has an identity element with respect to the binary operator ⊗.

e⊗x=x⊗e, where x∈A

Example

Let Z={0,1,2,3,4,5,…}

With respect to the binary operation ∗, the element 1 is considered as an identity element, since for any element x∈Z,

1∗x=x∗1

For the operation minus (−), no identity element is found.

Inverse

A set is said to have an inverse, if it has an identity element e, with respect to binary operator ⊗. For every element x∈A, there is another element y∈A which satisfies the following property to be true -

x⊗y=e

Example

Let A={⋯−4,−3,−2,−1,0,1,2,3,4,5,…}

The inverse of any element x provided with the operation plus (+) and e=0 is (−x) since x+(x)=0.

De Morgan's Law

The transformations between the union and the intersections of two or more sets are given by De Morgan’s Laws. The laws are as follows:

(A∪B)′=A′∩B′(A∪B)′=A′∩B′

(A∩B)′=A′∪B′(A∩B)′=A′∪B′

Example

Let A={1,2,3,4},B={1,3,5,7}, and

Universal set U={1,2,3,…,9,10}

A′={5,6,7,8,9,10}

B′={2,4,6,8,9,10}

A∪B={1,2,3,4,5,7}

A∩B={1,3}

(A∪B)′={6,8,9,10}

A′∩B′={6,8,9,10}

Hence it is observed that (A∪B)′=A′∩B′

(A∩B)′={2,4,5,6,7,8,9,10}

A′∪B′={2,4,5,6,7,8,9,10}

It is observed that (A∩B)′=A′∪B.

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