# Thread: Binary Relation and Range

1. ## Binary Relation and Range

I have a question with solution and want to understand how we get the range given in the solution ?

Question
If L={a,b,c,d} and M={3,4} then find two binary relations of L*M

Solution
L*M={(a,3),(a,4),(b,3),(b,4),(c,3),(c,4),(d,3),(d, 4)}
Two binary relations of L*M are as follows:

R1={(a,3),(b,4),(c,3)}

R2={(a,4),(b,3),(c,4)}

2. ## Re: Binary Relation and Range

Originally Posted by haftakhan
Question
If L={a,b,c,d} and M={3,4} then find two binary relations of L*M

Solution L*M={(a,3),(a,4),(b,3),(b,4),(c,3),(c,4),(d,3),(d, 4)}
Two binary relations of L*M are as follows:
R1={(a,3),(b,4),(c,3)}
R2={(a,4),(b,3),(c,4)}
The range is simply the set of second terms: $\text{Rng}(R_1)=\{3,4\}~.$

3. ## Re: Binary Relation and Range

Originally Posted by Plato
The range is simply the set of second terms: $\text{Rng}(R_1)=\{3,4\}~.$
Actually my question should have been that how we get
R1={(a,3),(b,4),(c,3)}
R2={(a,4),(b,3),(c,4)} ?

4. ## Re: Binary Relation and Range

Originally Posted by haftakhan
Actually my question should have been that how we get
R1={(a,3),(b,4),(c,3)}
R2={(a,4),(b,3),(c,4)} ?
There are $12$ ordered pairs in $L\times M$.
A relation is any subset of those pairs. So there are $2^{12}$ possible relations.

Those are just two of them. There is no rule.

5. ## Re: Binary Relation and Range

Originally Posted by Plato
There are $12$ ordered pairs in $L\times M$.
A relation is any subset of those pairs. So there are $2^{12}$ possible relations.

Those are just two of them. There is no rule.
It means that i can write any ordered pair in R1={(a,3),(b,4),(c,3)} instead of the three given?

6. ## Re: Binary Relation and Range

Yes, a "binary relationship" between set A and set B is any subset of $A\times B$. That is, any set of ordered pairs where the first member of each pair is from A and the second member if from B.