To show that R is an equivalence relation, we must show that it is symmetric, reflexive, and transitive.

Let A = P({a,b,c}) and let xRy iff |x| = |y| for all x, y in A.

Symmetric: Let x, y be let elements of A, and let xRy. Therefore |x| =...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

To show that R is an equivalence relation, we must show that it is symmetric, reflexive, and transitive.

Let A = P({a,b,c}) and let xRy iff |x| = |y| for all x, y in A.

Symmetric: Let x, y be let elements of A, and let xRy. Therefore |x| = |y|, or |y| = |x|. Therefore yRx. Since xRy => yRx, R is symmetric.

Reflexive: Let x be an element of x. |x| = |x|, and therefore xRx. Since xRx, R is reflexive.

Transitive: Let x, y, z be elements of A. Suppose xRy. It follows then that |x| = |y| = n, for some integer n. Suppose yRz. Since |y| = n and yRz, it follows that |z| = n. Therefore xRy and yRz implies xRz, and thus R is transitive.

A graph G of R on A has elements of A as nodes, and edges between nodes u,v iff uRv. We can sketch G as follows:

{}

{a}<---->{b}<---->{c}

^--------------------^

{a,b}<---->{a,c}<---->{b,c}

^----------------------------^

{a,b,c}