2.1 Relation types  (Page 3/3)

$$Iff\left(x,y\right)\in R\Rightarrow \left(y,x\right)\in R\text{for all}x,y\in A$$

The symbol “Iff” means “If and only if”. Here one directional arrow means “implies”. Alternatively, the condition of symmetric relation can be stated as :

$$xRy\Rightarrow yRx\text{for all}x,y\in A$$

In words, we say that if (x,y) be an instance of relation, then (y,x) will also be the instance of a symmetric relation "R".

It is clear that identity relation is a symmetric relation. Also, universal set consists of the Cartesian product of a set with itself. It means that the relation consists of instances with mirror instances. Therefore, universal relation is also symmetric relation.

Symmetric and inverse relation

An inverse relation ( $R^{-1}$ ) consists of ordered pairs with exchange of positions of the elements in a given relation (R). Now let us consider a symmetric relation,

$$R=\{\left(\mathrm{1,2}\right),\left(\mathrm{1,3}\right),\left(\mathrm{2,1}\right),\left(\mathrm{3,1}\right),\left(\mathrm{3,3}\right)\}$$

By definition, its inverse relation is :

$$\Rightarrow R^{-1}=\{\left(\mathrm{2,1}\right),\left(\mathrm{3,1}\right),\left(\mathrm{1,2}\right),\left(\mathrm{1,3}\right),\left(\mathrm{3,3}\right)\}$$

Using the fact that order does not change a set, we conclude that :

$$\Rightarrow R=R^{-1}$$

We use this fact to identify symmetric relation. The given set is a symmetric relation, if it equals its inverse set.

Analytical proof

Let “R” be a symmetric relation on set “A”. In order to prove that $R=R^{-1}$ , we consider an arbitrary instance of relation “R” :

$$\left(x,y\right)\in R$$

According to definition of symmetric relation,

$$\left(y,x\right)\in R$$

According to definition of inverse relation,

$$\left(x,y\right)\in R^{-1}$$

But, we had started with “R” and used definitions to show that “(x,y)” belongs to another set “ $R^{-1}$ ”. It means that the “ $R^{-1}$ ”set consists of the elements of set “R” – at the least. Thus,

$$R\subset R^{-1}$$

Similarly, we can start with “ $R^{-1}$ ”set and reach the conclusion that :

$R^{-1}\subset R$

If sets are subsets of each other, then they are equal. Hence,

$$\Rightarrow R=R^{-1}$$

Asymmetric relation

A relation “R” on a set “A” is asymmetric for the following condition :

$$Iff\left(x,y\right)\in R\mathrm{and}\left(y,x\right)\Rightarrow a=b\text{for all}x,y\in A$$

It means that possibility of symmetry in asymmetric relation exists only if elements are equal.

Transitive relation

If “R” be the relation on set A, then we state the condition of transitive relation as :

$$Iff\left(x,y\right)\in R\mathrm{and}\left(y,z\right)\in R\Rightarrow \left(x,z\right)\in R\text{for all}a,b,c\in A$$

Alternatively,

$$xRyandyRz\Rightarrow xRz\text{for all}x,y,z\in A$$

In words, we say that if (x,y) and (y,z) be the instances of a relation R such that (a,z) is also the instance of the relation, then that relation is transitive.

The identity and universal relations are transitive. Some other important transitive relations are :

• “is equal to”
• “is greater than”
• “is at least as great as”
• “is a subset of”
• "divides"

Example

Problem 3 : Determine whether “divides” is a transitive relation for natural number?

Solution : Let us consider three elements “x”,”y” and “z” of set “N” of natural numbers such that a relation “R” on “N” is :

$$\left(x,y\right)\in R,\left(y,z\right)\in R,“\mathrm{divides}”,x,y,z\in N$$

This means that :

$$“\text{x divides y}”\mathrm{and}“\text{y divides z}”$$

Let us now consider two natural numbers “a” and “b” such that :

$$y=ax\mathrm{and}z=by$$

$$z=abx$$

This means that “x divides z”. Hence, we conclude that the relation "divides" is transitive relation.

Equivalence relation

A relation is equivalence relation if it is reflexive, symmetric and transitive at the same time. In order to check whether a relation is equivalent or not, we need to check all three characterizations.