Why an equivalence relation cuts its set into disjoint classes, and why the cut can be run backwards

What to assume they know

• Reflexive, symmetric and transitive as three demands that can fail independently — reflexive, symmetric and transitive as three separate demands, and how each is tested and refuted
• A relation as a subset of a product set, and the two extreme cases — a relation as a subset of A × A
• Parity, divisibility, and negative integers on a number line
• The union and intersection of sets, and what disjoint means
• Similar and congruent triangles, and the distance of a point from the origin
• The equation of a straight line in the plane, and slope

What they should be able to do

• State Definition 4 and verify all three demands for a relation defined by a divisibility condition
• Write the class of a chosen element as a set, and list the classes of a relation on a small finite set
• Prove that any two classes of an equivalence relation are either identical or share nothing, and say where each of the three demands is spent
• Explain why every element lies in exactly one class, and why the classes therefore cover the set without overlap
• Run the construction backwards: from a stated division into boxes, define the relation it induces and verify that it is an equivalence relation
• Show that two differently described relations on the same set are the same relation, by proving containment each way
• Identify the classes of a geometric equivalence relation, and say why one element of the set has to be excluded from the description
• Count the equivalence relations on a small set by counting the divisions instead

Where it usually goes wrong

• "The classes all have the same size." Example 6 gives four against three; Exercise 1.1 Q8 gives three against two; Exercise 1.1 Q9(i) gives one box of four and three boxes of three. Equal size is a feature of the tidy integer examples, not part of the definition.
• "Two classes can overlap a bit." They cannot, and the reason is transitivity, not tidiness. Section 6's proof shows a single shared element drags the two classes into being the same set. Overlapping-but-different is the one configuration that is ruled out, and it is ruled out by a two-line argument.
• "[0] and [2] are different because 0 and 2 are different." A class has as many names as it has members. In Example 5's relation the chapter itself writes the even class as the class of any 2r. Under division by three, [0], [3] and [–6] all name the same box.
• "An equivalence relation partitions the set, so a partition is a different thing that happens as a result." They are the same information twice. The chapter makes the point by starting from three boxes on Part I p. 4 and getting the relation back, and Miscellaneous Example 20 does it again on nine elements.
• "Reflexivity is the easy condition, so it does not really do anything." It is what guarantees the classes cover the set. Drop it and an element can belong to no class at all, and the boxes no longer add up to X.
• "Similar and congruent give the same classes." Exercise 1.1 Q12's first and third triangles are similar without being congruent — one has sides double the other. And Q13's classes are coarser still: every triangle in one box, regardless of shape.
• "Every point of the plane is on a circle about the origin." The origin is not, which is why Exercise 1.1 Q11 excludes it by name. Its class is a single point.
• "The union of two equivalence relations must be one too, since the intersection is." It need not be; the counterexample above is three elements wide. Intersection preserves all three demands because each is a condition of the form "these pairs must be present", and being present in both is enough. Union has no such argument available.

Questions to check understanding

• Show that a stated relation is an equivalence relation, verifying each demand separately — the commonest single question form in this chapter's exercise
• Produce everything a named element is related to, and list all the classes
• Show that two differently described relations on one set are equal, by proving containment in each direction — the form of Miscellaneous Example 20
• Decide whether a stated relation is an equivalence relation and name the demand it fails — the form of Miscellaneous Exercise Q3
• Count the equivalence relations on a small set subject to a stated constraint
• Identify the geometric description of a class, and say which element of the set the description does not cover

Examples worth working on the board

Values marked verified are worked out here on the chapter's own data; no answer key was consulted, and the chapter prints no answers to its exercises.

• Definition 4 (Part I p. 3). Passing all three demands earns a relation the name; the definition adds nothing else. Every consequence in this topic comes out of those three and nothing more.
• Example 5 (Part I p. 3). In Z, the pair (a, b) is admitted when 2 divides a – b. The chapter's verification: 2 divides a – a because that difference is zero; if 2 divides a – b then it divides b – a; and if it divides a – b and b – c, then it divides their sum, which is a – c. The book flags that last step with a parenthetical question.
• The forward reading (Part I p. 4). E is the set of even integers, O the set of odd. The chapter records three facts: everything inside each of E and O is related to everything else inside it; nothing in one is related to anything in the other; and E and O are disjoint with their union the whole of Z. It then names E the class containing zero, written [0], and O the class containing one, written [1], and states that [0] and [1] are different while [0] equals the class of any even integer 2r and [1] the class of any odd integer 2r + 1.
• The general statement (Part I p. 4). For an arbitrary equivalence relation on an arbitrary set X, the chapter asserts that X splits into mutually disjoint subsets — it uses both partitions and subdivisions for these — satisfying the same three conditions, and calls them the equivalence classes. No proof is printed. Sections 5 and 6 of the explanation supply it.
• The proof to supply (not in the book; the chapter asserts the result without argument). Write [a] for the set of all x related to a. Reflexivity gives a in [a], so no element is left out and the classes cover X. Now suppose [a] and [b] share an element c. Then c is related to a and c is related to b; symmetry turns the first into a related to c, and transitivity chains it with the second to give a related to b. Take any x in [a]: x is related to a, and a is related to b, so transitivity puts x in [b]. That gives [a] inside [b], and the same argument with a and b exchanged gives [b] inside [a]. So the two classes are equal. Each demand is spent on a named step — reflexivity on the covering, symmetry on reversing a single arrow, transitivity on the two chainings — and deleting any one leaves a step with nothing behind it, which is the cleanest possible demonstration that Definition 3 has no redundant clause.
• Running it backwards (Part I p. 4). The chapter starts from three sets whose union is Z and which share nothing: A1 collects the multiples of 3 and is listed as ..., –6, –3, 0, 3, 6, ...; A2 collects the x for which x – 1 is a multiple of 3 and is listed as ..., –5, –2, 1, 4, 7, ...; A3 collects the x for which x – 2 is a multiple of 3 and is listed as ..., –4, –1, 2, 5, 8, .... It then defines the relation admitting (a, b) when 3 divides a – b, notes that the verification runs as in Example 5, and identifies A1 with [0], A2 with [1] and A3 with [2] — equally with the class of 3r, of 3r + 1 and of 3r + 2 for any integer r. Verified: the three listed sets do partition Z, since every integer leaves exactly one of the remainders 0, 1, 2 on division by 3; and the general backwards construction works for any division into boxes — declare a and b related when they sit in the same box, and reflexivity holds because the boxes cover, symmetry because "same box" reads the same in either order, and transitivity because the boxes do not overlap. That general statement is added here, not the chapter's.
• Example 6 (Part I pp. 4–5). On A = {1, 2, 3, 4, 5, 6, 7}, the pair (a, b) is admitted when a and b are both odd or both even. It is an equivalence relation; the subsets {1, 3, 5, 7} and {2, 4, 6} each have all their elements related to one another, and nothing in one is related to anything in the other. Verified: the two classes have four and three members. That inequality is worth pointing at, because the classes of a partition need not be the same size.
• Exercise 1.1 Q8 (Part I p. 6). On A = {1, 2, 3, 4, 5}, the pair is admitted when the absolute value of a – b is even. Verified: the difference of two integers is even exactly when they share a parity, so the classes are {1, 3, 5} and {2, 4} — again three against two.
• Exercise 1.1 Q9 (Part I p. 6). A is the set of integers x with 0 at most x and x at most 12. Two relations: (i) admits (a, b) when the absolute value of a – b is divisible by 4; (ii) admits (a, b) when a = b. In each case the exercise asks for everything that 1 is related to. Verified: A holds thirteen elements, 0 through 12. Under (i) the classes are {0, 4, 8, 12}, {1, 5, 9}, {2, 6, 10} and {3, 7, 11} — four boxes of sizes 4, 3, 3, 3, summing to 13 — and the answer sought is {1, 5, 9}. Under (ii) every class is a single element and the answer sought is {1}. Item (ii) is the best degenerate case in the chapter: thirteen classes, each of size one, and it is still a genuine partition.
• Exercise 1.1 Q11 (Part I p. 6). A is the set of points of a plane; the pair (P, Q) is admitted when P and Q are the same distance from the origin. The exercise then asks the student to show that, for any P away from the origin, the class of P is the circle drawn through it about the origin as centre. Verified: the classes are the circles about the origin, one for each positive radius, together with the origin's own class, which is the single point {(0, 0)} and is not a circle. That is exactly why the exercise excludes the origin. Treat the exclusion as content, not fine print; it is the one place in this chapter where a printed restriction can be motivated in ten seconds.
• Exercise 1.1 Q12 and Q13 (Part I p. 6). Q12: similarity of triangles, with three right-angled triangles supplied — sides 3, 4, 5; sides 5, 12, 13; sides 6, 8, 10 — and the question of which are related. Q13: polygons, related when they have the same number of sides, with the class of the triangle of sides 3, 4, 5 asked for. Verified: all three of Q12's triangles are right-angled, since 9 + 16 = 25, 25 + 144 = 169 and 36 + 64 = 100. The third has sides exactly double the first, so those two are similar; the second has side ratios 5 : 3, 3 : 1 and 13 : 5 against the first and is not similar to either. Under Q13 the class is every three-sided polygon in A, similar or not — so the two questions put side by side show two different equivalence relations cutting the same objects differently.
• Exercise 1.1 Q14 (Part I p. 6). L is the set of lines of the plane; the pair is admitted when the first is parallel to the second. The class of the line y = 2x + 4 is asked for. Verified: parallel lines are those of equal slope, so the class is every line y = 2x + c as c runs over the reals — a family of parallels, the given line included.
• Exercise 1.1 Q7 (Part I p. 6). Books in a college library, related when they have the same page count. An equivalence relation whose classes are indexed by a number, which is a good bridge to section 11.
• Miscellaneous Example 18 (Part I p. 13). The intersection of two equivalence relations on one set is an equivalence relation; the chapter checks all three demands. Verified: the union is not, in general. On {1, 2, 3} take the relation whose non-diagonal pairs are (1, 2) and (2, 1) and the one whose non-diagonal pairs are (2, 3) and (3, 2), each with the three diagonal pairs. Both are equivalence relations. Their union holds (1, 2) and (2, 3) but not (1, 3), so it is not transitive. The chapter does not ask this; it is added here, and it is the right way to show that Example 18 is a real result rather than a formality.
• Miscellaneous Example 20 (Part I pp. 13–14). X = {1, 2, ..., 9}. One relation admits (x, y) when x – y is divisible by 3; the other admits (x, y) when x and y both lie in {1, 4, 7}, or both in {2, 5, 8}, or both in {3, 6, 9}. The chapter proves the two relations are the same set by showing each is contained in the other. This is the whole topic in one example — a relation described by arithmetic and the same relation described by its boxes.
• Miscellaneous Example 21 (Part I p. 14). For a function f from X to Y, the relation admitting (a, b) when f sends a and b to the same value is an equivalence relation. Verified: the classes are the sets of inputs sharing an output. This is the general machine behind Q7, Q11, Q12, Q13 and Q14 — each of those relations is "same page count", "same distance", "same shape", "same side count", "same slope" — and it is the bridge from this module to the next.
• Miscellaneous Example 24 and Miscellaneous Exercise Q7 (Part I pp. 14, 16). Example 24 shows there are two equivalence relations on {1, 2, 3} containing (1, 2) and (2, 1); Q7 asks, with four options, how many contain (1, 2). Verified: since symmetry forces (2, 1) as soon as (1, 2) is in, the two questions have the same answer. Counting divisions instead of relations settles it fast: {1, 2, 3} divides into boxes in five ways — three singletons; one pair and a singleton, in three ways; and one box of all three. Requiring 1 and 2 in the same box leaves two of those five. So the count is two.
• Miscellaneous Exercise Q3 (Part I p. 15). X is a non-empty set and P(X) the set of its subsets; the relation admits (A, B) when A is contained in B, and the question is whether it is an equivalence relation. Verified: containment passes the first and third demands and fails the second — take any element of X, and the empty set is contained in the one-element set holding it while the reverse containment fails. So the answer is no, and it fails on exactly one of the three demands. Note that this book's containment sign allows equality: it writes the empty relation as contained in A × A on p. 2, and on p. 14 it concludes two relations are equal from containment in each direction. Reflexivity therefore holds.

Figures to have open

• A strip of integers from about –6 to 8 that can be recoloured into two boxes and then into three. This carries sections 2, 4 and 7 and should be the same strip throughout so the recolouring is visibly the same set being cut twice. Standard schematic; the chapter prints the element lists but draws nothing.
• A two-panel diagram for the equal-or-disjoint proof: two overlapping blobs with a shared element, then the same two blobs collapsed onto one. An added device, and the most important picture in the topic.
• Concentric circles about the origin with the origin marked as an isolated point, for section 10. Standard schematic; Exercise 1.1 Q11 describes this and the chapter prints no figure for it.
• The five divisions of a three-element set drawn as boxes, for section 12. Not in the book.

Where this sits in the book

• NCERT Class 12 Mathematics, Part I, Chapter 1 "Relations and Functions", §1.2 Types of Relations, Definition 4 and Example 5, p. 3
• The equivalence-class and partition passage, and Example 6, pp. 4–5
• Exercise 1.1, questions 7, 8, 9, 11, 12, 13 and 14, p. 6
• Miscellaneous Examples 18, 20, 21 and 24, pp. 13–14
• Miscellaneous Exercise, questions 3 and 7, pp. 15–16