Are These Equivalence Classes Correct for the Given Relation?

[Dustinsfl]

Define the relation ∼ on ℤ as follows: For a,b ∈ ℤ, a∼b iff. 2a + 3b ≡ 0 (mod 5). The relation ∼ is an equivalence relation on ℤ. Determine all the distinct equivalence classes for this equivalence relation.
Reflexive if a∼a.
2a + 3a ⇒ 5a ≡ 0 (mod 5); therefore, the relation is reflexive.
Symmetric if a∼b, then b∼a.
2a + 3b ≡ 4(2a + 3b) ≡ 8a + 12b ≡ 3a + 2b ≡ 0 (mod 5); therefore, the relation is symmetric.
Transitive if a∼b and b∼c, then a∼c.
a∼b ⇒ 2a + 3b ≡ 0 (mod 5)
b∼c ⇒ 2b + 3c ≡ 0 (mod 5) By adding the two, we obtain ⇒ 2a + 5b + 3c ≡ 2a + 3c ≡ 0 (mod 5); therefore, the relation is transitive.
2a + 3b ≡ 0 (mod 5) ⇒ 5 | (2a + 3b) ⇒ 5m = 2a + 3b
[0] = {a ∈ ℤ | a∼0} = {a ∈ ℤ | 5m = 2a} = {a ∈ ℤ | 2a = 5m} = {..., 5, 10, 15, ...}
[1] = {a ∈ ℤ | a∼1} = {a ∈ ℤ | 5n = 2a + 3} = {a ∈ ℤ | 2a = 5n - 3} = {..., 1, 6, 11, ...}
[2] = {a ∈ ℤ | a∼2} = {a ∈ ℤ | 5p = 2a + 6} = {a ∈ ℤ | 2a = 5p - 6} = {..., 2, 7, 12, ...}
[3] = {a ∈ ℤ | a∼3} = {a ∈ ℤ | 5r = 2a + 9} = {a ∈ ℤ | 2a = 5r - 9} = {..., -2, 3, 8, ...}
[4] = {a ∈ ℤ | a∼4} = {a ∈ ℤ | 5t = 2a + 12} = {a ∈ ℤ | 2a = 5t - 12} = {..., -1, 4, 9, ...}

Are these correct?

 

[Dick]

Yes, I think they are correct. Nice job.