Question

If r is a relation from R (set of real numbers) to R defined by{$$r=(a,b):a,b\in R,a-b+\sqrt{3}$$ is an irrational number} . Then the relation r is
1. An equivalence relation
2. Only reflexive
3. Only symmetric
4. Only transitive

Answer 1

Hint: We have to check whether the relation is reflexive, symmetric, transitive or equivalence relation. In order to check this, we must know that a relation R on a set A is said to be an equivalence relation if and only if the relation R is reflexive, symmetric and transitive.
 Reflexive: A relation is said to be reflexive, if $$(a,a)\in R$$ for every $$a\in R$$ .
Symmetric: A relation is said to be symmetric, if $$(a,b)\in R$$ then $$(b,a)\in R$$ .
Transitive: A relation is said to be transitive if $$(a,b)\in R$$ and $$(b,c)\in R$$ then $$(a,c)\in R$$ .

Complete step-by-step solution:
We are given that r is a relation defined by{$$r=(a,b):a,b\in R,a-b+\sqrt{3}$$ is an irrational number}.
Reflexivity:
$$a\in R$$
$$a-a+\sqrt{3}=\sqrt{3}$$ is an irrational number
Therefore $$\left(a,a\right)\in r$$
Hence r is Reflexive.
Symmetry:
Let $$\left(a,b\right)\in r$$
$$a-b+\sqrt{3}$$ is an irrational number.
Therefore $$b-a+\sqrt{3}$$ is an irrational number.
Therefore $$\left(b,a\right)\in r$$
Hence r is Symmetric.
Transitivity:
Let $$\left(a,b\right)\in r$$ and $$\left(b,c\right)\in r$$
Therefore
$$a-b+\sqrt{3}$$ is an irrational number.
$$b-c+\sqrt{3}$$ is an irrational number.
Now adding both the equations we get ,
$$a-c+2\sqrt{3}$$ is an irrational number.
Therefore $$\left(a,c\right)\in r$$
Hence r is Transitive.
Thus r being Reflexive, symmetric, and transitive is an equivalence relation.
Therefore option (1) is the correct answer.

Note: Reflexivity, symmetricity, and transitivity are the three parameters that are to be checked before a relation can be called an equivalence relation. Each equivalence relation provides a partition of the underlying set into disjoint equivalence classes. Two elements of A are equivalent if and only if their equivalence classes are equal. Any two equivalence classes are either equal or they are disjoint.
For example: Let $$R=\{(0,0),(0,4),(1,1),(1,3),(2,2),(3,1),(3,3),(4,0),(4,4)\}$$
Then, we can notice that the relation is reflexive, symmetric, and transitive. Hence, it is an equivalence relation.
Now, we can find the equivalence classes of this relation.
So, zero is related to four and itself. So, $\left[0\right]=\{0,4\}$
Similarly, $\left[4\right]=\{0,4\}$
Also, equivalence class of 3 is $\left[3\right]=\{1,3\}$
Similalrly, equivalence class of 1 is $\left[1\right]=\{1,3\}$
Also, equivalence class of 2 is $\left[2\right]=\{2\}$
Hence, the equivalence classes of the relation are: $\{0,4\}$, $\{1,3\}$ and $\{2\}$.