Lesson 4 of 10 · 8 min
Equivalence classes
NCERT §1.2
On sports day the whole school sits by house: four blocks in the stands, nobody in two blocks, nobody left standing. Every equivalence relation seats its set exactly like that.
The lesson in notes
In short
For an equivalence relation R on X and a ∈ X, the equivalence class [a] is the set of all elements of X related to a.
Under a R b ⇔ 2 divides a − b on Z, every even integer is related to 0 and no odd one is, so [0] = E, the even integers, and [1] = O, the odd integers. E and O are disjoint and E ∪ O = Z; also [0] = [2r] and [1] = [2r + 1] for every r ∈ Z.
In general an equivalence relation cuts X into mutually disjoint subsets Aᵢ, the equivalence classes, such that any two members of the same Aᵢ are related, members of different parts Aᵢ and Aⱼ (i ≠ j) never are, and together the parts make up all of X.
The process runs backwards too: split Z into A₁ = {…, −3, 0, 3, 6, …}, A₂ = {…, −2, 1, 4, 7, …} and A₃ = {…, −1, 2, 5, 8, …}, the integers leaving remainder 0, 1 and 2 on division by 3. Calling two integers related when they sit in the same part gives a R b ⇔ 3 divides a − b, with A₁ = [0], A₂ = [1], A₃ = [2].
Any two classes are either identical or disjoint: [a] = [b] exactly when a R b. For divisibility of a − b by 3, [3r] = [0], [3r + 1] = [1] and [3r + 2] = [2].
On {1, 2, 3, 4, 5, 6, 7}, 'a and b are both odd or both even' has exactly two classes, {1, 3, 5, 7} and {2, 4, 6}; every element of one is related to every other element of it and to nothing in the other.
For points of a plane related when they are the same distance from the origin, the class of a point P other than the origin is the circle through P centred at the origin.