Relations  (Page 2/2)

$$R=\{\left(\mathrm{1,3}\right),\left(\mathrm{2,6}\right),\left(\mathrm{3,9}\right)\}$$

We can visualize the relation pictorially as shown in the figure.

Domain of the relation

The domain represents the valid values of the first element of the ordered pairs in the relation. Clearly, the elements of domain of a relation belong to “from” set “A”. But, elements in the domain are only those which are valid for the relation. It means that domain does not consist of all elements of “from” set “A”. Thus, domain set is a subset of “from” set “A”.

Domain

The set of first elements of all ordered pairs in the relation “R” from set “A” to “B” is called the domain of relation “R”.

We can write the domain set of relation “R” from set “A” to set “B” in set builder form as :

$$\mathrm{Domain}\left(R\right)=\{x:\left(x,y\right)\in R\}$$

Consider the example given earlier. The relation set is :

$$R=\{\left(\mathrm{1,3}\right),\left(\mathrm{2,6}\right),\left(\mathrm{3,9}\right)\}$$

The domain according to definition is :

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

$$\Rightarrow \mathrm{Domain}\left(R\right)\subset A$$

Co-domain

In a relation “R” from set “A” to “B”, the set “B” is called co-domain.

Range of the relation

The range represents the valid values of the second element of the ordered pairs in the relation.

Range

The set of second elements of all ordered pairs in the relation “R” from set “A” to “B” is called the range of relation “R”.

We can write the range set of relation “R” from set “A” to set “B” in set builder form as :

$$\mathrm{Range}\left(R\right)=\{y:\left(x,y\right)\in R\}$$

Consider the example given earlier. The relation set is :

$$R=\{\left(\mathrm{1,3}\right),\left(\mathrm{2,6}\right),\left(\mathrm{3,9}\right)\}$$

The range according to definition is :

$$\mathrm{Range}\left(R\right)=\left\{\mathrm{3,6,9}\right\}$$

Clearly, the elements of range of a relation belong to “to” set “B”. But, elements in the range are only those which are valid for the relation. It means that range does not consist of all elements of “to” set “B”. Thus, range set is a subset of “to” set “B”.

$$\mathrm{Range}\left(R\right)\subset B$$

Example

Problem 2 : Let $A=\left\{\mathrm{1,2,3,4,5}\right\}$ and $B=\left\{\mathrm{1,4,5}\right\}$ . Let a relation from “A” to “B” is :

$$R=\{\left(x,y\right):x<y,\left(x,y\right)\in AXB\}$$

Find R, Domain (R) and Range(R).

Solution : Let us find “y” for each value of “x”.

$$\mathrm{For}x=\mathrm{1,}y=\mathrm{4,5}$$

$$Forx=\mathrm{2,}y=\mathrm{4,5}$$

$$Forx=\mathrm{3,}y=\mathrm{4,5}$$

$$Forx=\mathrm{4,}y=5$$

$$Forx=\mathrm{5,}\text{There is no value of “y” in set “B”}$$

Hence,

$$R=\{\left(\mathrm{1,4}\right),\left(\mathrm{1,5}\right),\left(\mathrm{2,4}\right),\left(\mathrm{2,5}\right),\left(\mathrm{3,4}\right),\left(\mathrm{3,5}\right),\left(\mathrm{4,5}\right)\}$$

$$\mathrm{Domain}\left(R\right)=\left\{\mathrm{1,2,3,4}\right\}$$

$$\mathrm{Range}\left(R\right)=\left\{\mathrm{4,5}\right\}$$

Numbers of relations

Between two sets, the Cartesian product set consists of all possible instances of relation as ordered pair. Here, we need to find the total possible relations that can be generated from these ordered pairs. We have seen that total numbers of ordered pairs in the Cartesian product of sets “A” and “B” is “pq”, where “p” and “q” are the numbers of elements in two sets respectively.

Now, relation is nothing but a subset of the Cartesian product. It means that total numbers of relation is equal to total numbers of possible subsets of the Cartesian product. Recall that the set formed from all possible subsets is called power set. The numbers of subsets in the power set is given by

$$n=2^{pq}$$

Clearly, this number “n” denotes all possible relations (subsets) that can be generated from two finite sets. We should, however, be careful in interpreting this number as it also contains the mandatory empty set, which is not a meaningful set from the point of view of relation.

Inverse relation

Inverse relation of a given relation “R” from set “A” to set “B” is set of ordered pairs in which first and second elements exchanges their positions. The inverse set is defined in reference to a given relation. The inverse relation of a given relation “R” from “A” to “B” is denoted as “ $R^{-1}$ ”. Clearly,

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

where,

$$R=\{\left(x,y\right):\left(x,y\right)\in AXB\}$$

We should be careful to understand that “-1” is not a power, but a part of symbol to represent inverse relation with respect to a given relation. It is also clear that :

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

As the elements in the ordered pair of the relation exchanges positions, domain and range sets are exchanged across the sets.

$$\mathrm{Domain}\left(R^{-1}\right)=\mathrm{Range}\left(R\right)$$

$$\mathrm{Range}\left(R^{-1}\right)=\mathrm{Domain}\left(R\right)$$

Example

Problem 3 : Let $A=\{\mathrm{1,2,}\dots ..,10\}$ . A relation on “A” is defined as

$$R=\{\left(x,y\right):y=3x,wherex,y\in A\}$$

Find $R^{-1}$ , Domain ( $R^{-1}$ ) and Range ( $R^{-1}$ ).

Solution : In the earlier example, we determined the relation “R” as :

$$R=\{\left(\mathrm{1,3}\right),\left(\mathrm{2,6}\right),\left(\mathrm{3,9}\right)\}$$

According to the definition of inverse set, the elements of the ordered pair in the relation set are exchanged :

$$R^{-1}=\{\left(\mathrm{3,1}\right),\left(\mathrm{6,2}\right),\left(\mathrm{9,3}\right)\}$$

Clearly, inverse relation can be represented in set builder form as :

$$\Rightarrow R^{-1}=\{\left(y,x\right):y=x/\mathrm{3,}wherex,y\in A\}$$

Now, the domain and range of $R^{-1}$ are :

$$Domain\left(R^{-1}\right)=\left\{\mathrm{3,6,2}\right\}$$

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