A relation R is defined on the set A = \1, 2, 3, 4, 5\ by a R b a + b = 2n, n N. List all the equivalence classes of A under R.

You're on a roll — (ii) A relation R is defined on the set A = \1, 2, 3, 4, 5\ by a R b a + b = 2n, n N. List all the equivalence classes of A under R. Step 1: Interpret the relation condition. The condition a+b = 2n, n N means that a+b must be an even positive integer. Since a, b \1, 2, 3, 4, 5\, the smallest sum is 1+1=2 and the largest is 5+5=10. All these sums are positive. Therefore, the condition simplifies to a+b is an even number. This implies that a and b must have the same parity (both even or both odd). Step 2: Verify that R is an equivalence relation. • Reflexive: For any a A, a+a = 2a. Since a is an integer, 2a is always an even number. Thus, a R a for all a A. • Symmetric: If a R b, then a+b is an even number. Since addition is commutative, b+a is also an even number. Thus, b R a. • Transitive: If a R b and b R c, then a+b is even and b+c is even. If a+b is even, a and b have the same parity. If b+c is even, b and c have the same parity. Therefore, a, b, and c must all have the same parity. If a and c have the same parity, then a+c is an even number. Thus, a R c. Since R is reflexive, symmetric, and transitive, it is an equivalence relation. Step 3: Determine the equivalence classes. An equivalence class [x] consists of all elements y A such that y R x. This means y and x must have the same parity. • For x=1 (an odd number): [1] = \y A y is odd\ = \1, 3, 5\. • For x=2 (an even number): [2] = \y A y is even\ = \2, 4\. • For x=3 (an odd number): [3] = \y A y is odd\ = \1, 3, 5\. (Same as [1]) • For x=4 (an even number): [4] = \y A y is even\ = \2, 4\. (Same as [2]) • For x=5 (an odd number): [5] = \y A y is odd\ = \1, 3, 5\. (Same as [1]) The distinct equivalence classes are the sets of odd numbers and even numbers within A. The equivalence classes of A under R are \1, 3, 5\ and \2, 4\. What's next?