Question

Let R be a relation on $N\times N$ , defined by (a, b) R (c, d) $\Leftrightarrow$ a+d = b+c . Then R, is
(a) reflexive and symmetric but not transitive
(b) reflexive and transitive but not symmetric.
(c) symmetric and transitive but not reflexive.
(d) an equivalence relation.

Answer 1

Hint:Here, we will use the definitions of reflexive, symmetric and transitive relations to check whether the given relations are reflexive, symmetric or transitive.

Complete step-by-step answer:
A relation between two sets is a collection of ordered pairs containing one object from each set. If the object x is from the first set and the object y is from the second set, then the objects are said to be related if the ordered pair (x, y) is in the relation.
A relation R is reflexive if each element is related itself, i.e. (a, a) $\in$ R, where a is an element of the domain.
A relation R is symmetric in case if any one element is related to any other element, then the second element is related to the first, i.e. if (x, y) $\in$ R then (y, x) $\in$ R, where x and y are the elements of domain and range respectively.
A relation R is transitive in case if any one element is related to a second and that second element is related to a third, then the first element is related to the third, i.e. if (x, y) $\in$ R and (y, z) $\in$ R then (x, z) $\in$ R.
Here, the given relation is:
(a, b) R (c, d) $\Leftrightarrow$ a+d = b+c
Since, we know that (a +d)= (a + d).
Therefore, (a+d, a+d) $\in$ R.
So, R is reflexive.
Now, suppose that (a + d, b + c) $\in$ R.
Then we can write:
a+d=b+c
Or, b+c = a+d
So, (b+c, a+d) $\in$ R.
This means that R is symmetric.
Now, let us take an ordered pair (a + d, b + c) $\in$ R and (b + c, e+f) $\in$ R.
We can write:
$a+d=b+c..........\left( 1 \right)$
And, $b+c=e+f.............\left( 2 \right)$
From equations (1) and (2), we can say that a+d = e+f.
So, (a + d, e + f) $\in$ R.
This implies that R is transitive.
So, R is reflexive, symmetric as well as transitive.
We know that a relation which is symmetric, reflexive as well as transitive is an equivalence relation.
Hence, option (d) is the correct answer.

Note: Students should note here that for a relation to be a particular type of relation, i.e. reflexive, symmetric or transitive, all its elements must satisfy the required conditions. If we are able to find even a single counter example, i.e. if any of the elements doesn’t satisfy the criteria, then we can’t proceed further.