Question
Question: If we define a relation R on the set \(N\times N\) as \(\left( a,b \right)R\left( c,d \right)\Leftri...
If we define a relation R on the set N×N as (a,b)R(c,d)⇔a+d=b+c for all (a,b),(c,d)∈N×N, then the relation is,
A. Symmetric only
B. Symmetric and transitive only
C. Equivalence relation
D. Reflexive only
Solution
For solving this type of questions you should know about the symmetric property, transitive property and reflexive property and the equivalence of the relation. So, according to the reflexive property a∼a, it is always valid for this. The symmetric property states that if a∼b, then b∼a. And the transitive property is defined as, if a∼b and b∼c, then a∼c. And all these are valid for the sets. If all these three are valid, then the relation will be an equivalence relation.
Complete step by step answer:
So, in the question it is given that, (a,b)R(c,d)⇔a+d=b+c.
So, if we take (a,a)R(a,a)⇔a+a=a+a. By the property of reflexive, we can say that here R is reflexive.
Now according to our question, if we apply the symmetric property, then let us consider,
(a,b)R(c,d)⇔a+d=b+c⇒c+b=d+a⇔(c,d)R(a,b)
So, here we can say that it is clear that R is symmetric. And this is the step to check the symmetricity of R.
Now if we check R by applying transitive property, then let us consider,
(a,b)R(c,d) and (c,d)R(e,f)
⇒a+d=b+c and c+f=d+e
If we merge both with one another with same sides, then we have,
a+d+c+f=b+c+d+e
So, here, a+f=b+e⇒(a,b)R(e,f)
By the property of transitive, it is clear that this is transitive also. So, we can say that all three properties transitive, reflexive and symmetric are completely satisfied by it. So, R is an equivalence relation.
So, the correct answer is “Option C”.
Note: When you solve this question then you should apply all the properties in its proper way because they are the only way to clear this type of questions and R will be equivalence when all three properties are completely satisfied and if any one or two or all properties do not satisfy, then it is not equivalence.