Relations

Step 1: Check reflexive property For any , which is irrational. Hence, for all , so the relation is reflexive.
Step 2: Check symmetric property If , then is irrational. Now, If is rational, then is irrational. If is irrational, then subtracting it from also gives an irrational number. Hence, so the relation is symmetric.
Step 3: Check transitive property Let and . Then But The above expression may be rational or irrational depending on . Thus, transitivity does not always hold.
Step 4: The relation is reflexive and symmetric but not transitive. Among the given options, the correct description is symmetric.

Step 1: Check symmetry A relation is symmetric if . If is brother of , then need not be brother of (since may be a sister).
Step 2: Check transitivity A relation is transitive if and . If is brother of and is brother of , it does not necessarily imply that is brother of in all cases (family relations may differ).
Step 3: Hence, the relation is neither symmetric nor transitive.

Step 1: Check reflexive property A relation on set is reflexive if for all . Given: Hence, is reflexive.
Step 2: Check symmetric property A relation is symmetric if . Here, Hence, is not symmetric.
Step 3: Check transitive property A relation is transitive if and . Consider: But, Hence, is not transitive.
Step 4: Since the relation is reflexive but neither symmetric nor transitive, it is reflexive only.

Step 1: Equivalence relation requires: 1. Reflexive 2. Symmetric 3. Transitive.
Step 2: Reflexive pairs needed:
Step 3: Symmetric closure of given:
Step 4: Transitivity from (1,2) and (2,3) gives
Step 5: But we are asked minimum addition. Best strategy: assume all elements equivalent → need only reflexive 3 if we add remaining via intention of single class. Hence → 3.