Question

Given S is the set of integers {2,3,4,5,6,7,8}. Let R be a relation defined on S by the following condition such that,

∀x,y∈S,xRy⟺xmod2=ymod2
i. Draw the digraph of R.
ii. Show that R is an equivalence relation.
iii. Find the equivalence classes for R item is R a partial order? Explain your answer

Answer 1

i. To draw the digraph, represent each element as a vertex and draw edges based on the relation. ii. R is an equivalence relation as it is reflexive, symmetric, and transitive. iii. The equivalence classes for R are [2,4,6,8] and [3,5,7]. R is not a partial order as it is not antisymmetric.

Step-by-step explanation:

i. To draw the digraph of relation R, we represent each element in set S as a vertex in the graph. Then, we draw directed edges between elements x and y if and only if xRy is true. In this case, xRy if x mod 2 = y mod 2.

ii. To show that R is an equivalence relation, we need to prove that it is reflexive, symmetric, and transitive. Reflexivity: For any x ∈ S, x mod 2 = x mod 2, so xRx. Symmetry: If x mod 2 = y mod 2, then y mod 2 = x mod 2, so yRx. Transitivity: If x mod 2 = y mod 2 and y mod 2 = z mod 2, then x mod 2 = z mod 2, so xRz.

iii. The equivalence classes for R are [2,4,6,8] and [3,5,7]. R is not a partial order because it is not antisymmetric. In a partial order, if xRy and yRx, then x = y. But in this case, for example, 2R3 and 3R2, so 2 ≠ 3.