Example 2: Congruence of triangles is an equivalence relation

Basics you need

Solution

1. Turn "congruent" into equal side lengths

   For a triangle $T$⁠, list its three side lengths in increasing order and call the list $s(T) = (p, q, r)$ with $p \le q \le r$⁠. By the SSS rule (Basic 1),

   $$(T_1, T_2) \in R \iff T_1 \cong T_2 \iff s(T_1) = s(T_2).$$

   For example a triangle with sides 3, 5, 4 cm has $s = (3, 4, 5)$⁠, and so does any triangle with sides 4, 3, 5 cm, so they are congruent.

   The word "congruent" describes a geometric act (moving one triangle onto another), which is awkward to reason with. The SSS rule converts it into equality of three numbers, which we already know how to handle. Sorting the sides into increasing order is what takes care of "in some order": two triangles whose sides are equal in some order have exactly the same sorted list.
   SSS: if the three sides of one triangle are equal to the three sides of another, the triangles are congruent; conversely, congruent triangles have equal corresponding sides.

2. Reflexive: every triangle is congruent to itself

   Take any triangle $T_1 \in \text{T}$⁠. Its list of side lengths equals itself:

   $$s(T_1) = s(T_1).$$

   So $T_1 \cong T_1$⁠, that is $(T_1, T_1) \in R$⁠. Geometrically, the "do nothing" motion places $T_1$ exactly on itself. Since $T_1$ was arbitrary, $R$ is reflexive.

   Reflexivity asks about every triangle, so we take an arbitrary one. The only fact needed is that a list of numbers equals itself, which does not depend on which triangle we chose.
   Equality is reflexive: $x = x$ (Basic 3). The identity motion (moving nothing) is a rigid motion.

3. Symmetric: if $T_1 \cong T_2$ then $T_2 \cong T_1$

   Suppose $(T_1, T_2)\in R$⁠. Then $s(T_1) = s(T_2)$⁠. Reading the same equation from right to left,

   $$s(T_2) = s(T_1),$$

   so $T_2 \cong T_1$⁠, that is $(T_2, T_1)\in R$⁠. Geometrically: if a motion carries $T_1$ onto $T_2$⁠, the reverse motion carries $T_2$ back onto $T_1$⁠. Hence $R$ is symmetric.

   $$(T_1,T_2)\in R \;\Rightarrow\; T_1 \cong T_2 \;\Rightarrow\; T_2 \cong T_1 \;\Rightarrow\; (T_2,T_1)\in R.$$
   Symmetry is an "if … then" statement, so we start from the "if" part, $(T_1,T_2)\in R$⁠, and derive the "then" part. The key point is that congruence has no direction: the condition "same side lengths" treats the two triangles exactly alike.
   Equality is symmetric: $x = y \Rightarrow y = x$ (Basic 3). Every rigid motion can be undone by another rigid motion (undo a turn by turning back, undo a flip by flipping again).

4. Transitive: $T_1 \cong T_2$ and $T_2 \cong T_3$ give $T_1 \cong T_3$

   Suppose $(T_1,T_2)\in R$ and $(T_2,T_3)\in R$⁠. Then

   $$s(T_1) = s(T_2) \qquad\text{and}\qquad s(T_2) = s(T_3).$$

   Both lists equal the same list $s(T_2)$⁠, so $s(T_1) = s(T_3)$⁠. Hence $T_1 \cong T_3$⁠, that is $(T_1, T_3)\in R$⁠, and $R$ is transitive.

   $$(T_1,T_2),\,(T_2,T_3)\in R \;\Rightarrow\; T_1\cong T_2 \text{ and } T_2 \cong T_3 \;\Rightarrow\; T_1 \cong T_3 \;\Rightarrow\; (T_1,T_3)\in R.$$

   Geometrically: do the motion that carries $T_1$ onto $T_2$⁠, then the one that carries $T_2$ onto $T_3$⁠. The two motions done one after the other form a single rigid motion carrying $T_1$ onto $T_3$⁠.

   The middle triangle $T_2$ is the link. The goal $(T_1,T_3)\in R$ does not mention $T_2$⁠, so we need a fact that eliminates it, and "two things equal to the same thing are equal to each other" does exactly that.
   Equality is transitive: $x = y$ and $y = z \Rightarrow x = z$ (Basic 3). Doing one rigid motion after another is again a rigid motion, because each one keeps all distances unchanged.

5. Conclude, and see the classes

   $R$ is reflexive (Step 2), symmetric (Step 3) and transitive (Step 4). By Definition 4, $R$ is an equivalence relation.

   Its equivalence classes are the groups of triangles with one fixed list of side lengths. For instance, the class of a 3-4-5 triangle consists of every triangle in the plane with sides 3, 4 and 5, however it is positioned, turned or flipped.

   An equivalence relation always splits its set into disjoint classes (Basic 5). Seeing what the classes are is the best way to understand what the relation "means": here, a class is "one size and shape of triangle".
   Every equivalence relation partitions its set into equivalence classes (Section 1.2, after Example 5). Basics: equivalence classes

Every fact used, and where it comes from