Set, Relations and FunctionsPYQ June 24Question 1948 of 217
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Consider the following relations on A={1,2,3}\displaystyle A = \{1,2,3\}. R={(1,1),(1,2),(2,1),(2,2),(3,3)}\displaystyle R = \{(1,1), (1,2), (2,1), (2,2), (3,3)\}, S={(1,1),(1,2),(2,2),(2,3)}\displaystyle S = \{(1,1), (1,2), (2,2), (2,3)\} and Φ=empty set\displaystyle \Phi = \text{empty set}. Which one of these forms an equivalence relation?

Options

AR\displaystyle R
BS\displaystyle S
CT\displaystyle T
DΦ\displaystyle \Phi
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Correct Answer

✅ Option a — R\displaystyle R

All Options:

  • AR\displaystyle R
  • BS\displaystyle S
  • CT\displaystyle T
  • DΦ\displaystyle \Phi

Detailed Solution & Explanation

Let us test the relation R={(1,1),(1,2),(2,1),(2,2),(3,3)}\displaystyle R = \{(1,1), (1,2), (2,1), (2,2), (3,3)\} on A={1,2,3}\displaystyle A = \{1,2,3\} for the three properties of an equivalence relation:
1. **Reflexive**: For R\displaystyle R to be reflexive on A\displaystyle A, we must have (1,1),(2,2),(3,3)∈R\displaystyle (1,1), (2,2), (3,3) \in R. These are all present. So R\displaystyle R is reflexive.
2. **Symmetric**: We must check if (x,y)∈R  ⟹  (y,x)∈R\displaystyle (x, y) \in R \implies (y, x) \in R.
Here, (1,2)∈R  ⟹  (2,1)∈R\displaystyle (1,2) \in R \implies (2,1) \in R. Yes, it is symmetric.
3. **Transitive**: We check if (x,y)∈R and (y,z)∈R  ⟹  (x,z)∈R\displaystyle (x, y) \in R \text{ and } (y, z) \in R \implies (x, z) \in R.
- (1,2)∈R\displaystyle (1,2) \in R and (2,1)∈R  ⟹  (1,1)∈R\displaystyle (2,1) \in R \implies (1,1) \in R (True)
- (2,1)∈R\displaystyle (2,1) \in R and (1,2)∈R  ⟹  (2,2)∈R\displaystyle (1,2) \in R \implies (2,2) \in R (True)
Yes, R\displaystyle R is transitive.
Therefore, R\displaystyle R is an equivalence relation.

Now let's check the relation S={(1,1),(1,2),(2,2),(2,3)}\displaystyle S = \{(1,1), (1,2), (2,2), (2,3)\}:
- It is not reflexive because (3,3)∉S\displaystyle (3,3) \notin S.
- It is not symmetric because (1,2)∈S\displaystyle (1,2) \in S but (2,1)∉S\displaystyle (2,1) \notin S.
- It is not transitive because (1,2)∈S\displaystyle (1,2) \in S and (2,3)∈S\displaystyle (2,3) \in S but (1,3)∉S\displaystyle (1,3) \notin S.
So S\displaystyle S is definitely not an equivalence relation.
*Note: The relation R\displaystyle R is an equivalence relation, corresponding to Option A. The textbook key incorrectly lists Option B (S\displaystyle S) as correct, which is a typographical error.*
Hence, **Option A** is the correct answer.

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