Equivalence Relations

Probably one of the lessons in Mathematics possessing obscure importance is equivalence relations. Fundamentally, an equivalence relation must satisfy three properties: reflexivity, symmetry, and transitivity. Practice makes perfect, so they say, and I guess this is true also in proving that a certain relation satisfies the RST properties. Let’s take on an easy one first: congruent segments. By definition, two segments are congruent if they have the same length.

Claim: Congruence between segments is an equivalence relation.

Reflexivity: any segment is congruent to itself.
This is trivially true, since a segment can only have one measure.

Symmetry: if segment AB is congruent to segment MN, then segment MN is congruent to segment AB.
Again true, since, if AB=MN, then MN=AB by properties of real numbers, and we convert back again to the congruence statement.

Transitivity: if segment AB is congruent to segment MN, and segment MN is congruent to segment XY, then segment AB is congruent to segment XY.
From the two given statements, AB=MN, and MN=XY, so that we can add the left-hand side and right-hand sides of the equations, then cancel MN, then convert back to the congruence statement.

That was rather easy; in fact, proving the same for congruence between angles, congruence between triangles, similarity between triangles, and proportionality of sequences would not take much time, as long as the definitions are clear. Actually, it is more exciting to prove that something is not an equivalence relation. This exercise also invites the person to summon Always-Sometimes-Never powers.

Show: Perpendicularity of lines is not an equivalence relation.

Reflexivity: a line is perpendicular to itself.
Never true. This statement doesn’t make any sense, since a right angle needs to be formed. We can stop right now in our exercise, since an equivalence relation must satisfy all three properties. But let’s still see inspect the other two properties.

Symmetry: if line A is perpendicular to line B, then line B is perpendicular to line A.
Always true.

Transitivity: if line A is perpendicular to line B, and line B is perpendicular to line C, then line A is perpendicular to line C.
This statement is tricky. A and C will never be perpendicular – if there is an additional given that all three lines are coplanar. However, the three lines can be arranged in space so as to satisfy the transitivity statement: they may represent the x, y, and z axes of three-dimensional analytic geometry (or much simpler, the lines containing three edges of a cube sharing a common vertex). Thus, the statement is sometimes true; our two possibilities are enough basis.
[Warning: the transitivity statement above can be restated so as the answer is “never”]

We now move on to equivalence relations applied in “real-life” situations (I put the words “real life” in quotation marks since I find it hard to stomach that not everyone considers Math as real life. Oops haha). Let’s try if we can have similar assessments to other non-Math relations.

Prove/disprove: Being siblings is an equivalence relation.

Reflexivity: any person is a sibling of himself/herself.
Never true, by definition.

Symmetry: if A is a sibling of B, then B is a sibling of A.
Always true.

Transitivity: if A is a sibling of B, and B is a sibling of C, then A is a sibling of C.
Always true. However, if the definition of a sibling is tweaked so as to allow half-brothers and half-sisters, we will have a new answer.

Prove/disprove: Being cousins is an equivalence relation.

Reflexivity: any person is a cousin of himself/herself.
Never true.

Symmetry: if A is a cousin of B, then B is a cousin of A.
Always true.

Transitivity: if A is a cousin of B, and B is a cousin of C, then A is a cousin of C.
This allows for the possibility that A and C are siblings, making them non-cousins. A and C may also be totally unrelated, if they are cousins of B on different parental sides (thus the term “pinsan ng pinsan”). But A, B, and C may all come from the second generation of actual siblings, so the statement is sometimes true.

Prove/disprove: Being a “better half” is an equivalence relation. (for purposes of clarity, we limit our domain to married people)

Reflexivity: any person is the better half of himself/herself.
Never true.

Symmetry: if A is the better half of B, then B is the better half of A.
Always true. Probably A and B may argue that, if both of them are better halves, then what’s the use of using “better”?

Transitivity: if A is the better half of B, and B is the better half of C, then A is the better half of C.
Never true. It would take a totally strange society that will allow the statement to be possible.

Prove/disprove: Being a best friend is an equivalence relation.

Reflexivity: any person is the best friend of himself/herself.
In my opinion, intrapersonal skills are very important, if not mandatory, but I guess we can have a definition to make the statement never true. Think Johari’s window.

(Let’s skip symmetry first)
Transitivity: if A is the best friend of B, and B is the best friend of C, then A is the best friend of C.
Strictly speaking, the statement should never be true if “best” implies only one. This usually happens in an intimate circle of friends (barkada). But probably to make it a “sometimes” statement, all the “the’s” should be replaced with “a’s”. It might be pointless to use the word “best” if it will not refer to an upper bound, and maybe “better fiends” will be a more appropriate terms. But we are allowed to make our own definitions by now, and you can have a best friend from childhood, high school, college, choir, and so on.

Symmetry: if A is the best friend of B, then B is the best friend of A.
Surprisingly, this is not always true (think “A Separate Peace”). Although most best friends-ship are mutual, there might be some that can be hard to fathom. Though settling the question is not as important as the friendship.

If you have found that one best friend, congratulations.