As usual for us, we will start refreshing the relevant definitions to solve this exercise.
Definition: “A relation on a set A is called an equivalence relation if it is reflexive, symmetric, and transitive.”
Definition: “A relation R on a set A is called reflexive if (a,a) ∈ R for every element a ∈ A.”
Definition: “A relation R on a set A is called symmetric if (b,a) ∈ R whenever (a,b) ∈ R, for all a,b ∈ A.”
Definition: “A relation R on a set A is called transitive if whenever (a,b)∈R and (b,c)∈R, then (a,c) ∈ R, for all a,b,c ∈ A.”
The definitions above are from the textbook “Discrete Mathematics and its applications” By Rosen.
Now, let’s solve the exercise.
Reflexive:
((a,b),(a,b))∈R because a+b=b+a. Therefore, the relation is reflexive.
Symmetric:
((a,b),(c,d))∈R, implies that a+d=b+c (1). From (1), we get that c+b=d+a, so ((c, d), (a, b)) ∈ R. Therefore, the relation is symmetric.
Transitive:
if ((a, b), (c, d)) ∈ R and((c,d),(e,f))∈R, then a+d=b+c (1) and c+e=d+f.
Adding c+e and d+f to both sides of the first equation we get a+d+c+e = b+c+d+f. Subtracting d and c in both sides we get:
a+d+c+e = b+c+d+f
a+d+c+e-d-c = b+c+d+f-d-c
a+e = b+f, it follows that ((a, b), (e, f )) ∈ R.
Therefore, the relation is transitive.
R is an equivalence relation because is reflexive, symmetric, and transitive.
Related exercises:
- Which of these relations on the set of all people are equivalence relations? Determine the properties of an equivalence relation that the others lack
- Suppose that A is a nonempty set, and f is a function that has A as its domain. Let R be the relation on A consisting of all ordered pairs (x, y) such that f (x) = f (y)
- Show that the relation of logical equivalence on the set of all compound propositions is an equivalence relation. What are the equivalence classes of F and of T?
- Which of these relations on {0, 1, 2, 3} are equivalence relations? Determine the properties of an equivalence relation that the others lack.
