QED
Equivalence Relations · step 1 of 13

Verifying an equivalence relation

A relation is an equivalence relation exactly when it is reflexive, symmetric and transitive. Proving it is a three-part obligation, and exam schemes award a mark for each part — so a proof that establishes two of the three scores two thirds, no matter how elegant. The standard sources of equivalence relations are "same something": same remainder, same length, same image under a function.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

  1. Restate the rule symbolicallyTurn "a ∼ b iff a and b leave the same remainder mod 5" into an equation you can manipulate: 5 ∣ (a − b).
  2. Reflexivity — plug in a for bothShow a ∼ a follows immediately. This is usually one line, but it must appear.
  3. Symmetry — assume a ∼ b, derive b ∼ aTypically you negate or reverse an equation. If a − b = 5k then b − a = 5(−k).
  4. Transitivity — chain two hypothesesAssume a ∼ b and b ∼ c, add or substitute the two equations, and conclude a ∼ c. The middle term must cancel.

Worked example

Prove that a ∼ b iff 5 ∣ (a − b) is an equivalence relation on ℤ.

  1. Reflexive: a − a = 0 = 5·0, so 5 ∣ (a − a) and a ∼ a.
  2. Symmetric: if a − b = 5k then b − a = 5(−k) with −k ∈ ℤ, so b ∼ a.
  3. Transitive: if a − b = 5k and b − c = 5m, then a − c = (a − b) + (b − c) = 5(k + m).
  4. Since k + m ∈ ℤ, 5 ∣ (a − c) and a ∼ c.

Answer. All three properties hold, so ∼ is an equivalence relation — congruence modulo 5.

Where marks get dropped

These are the specific errors that cost credit on verifying an equivalence relation questions — QED's rubric penalises each of them separately.

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Verifying an equivalence relation — frequently asked questions

Do I really have to prove all three?

Yes. Each is independently marked, and each can fail on its own: divisibility is reflexive and transitive but not symmetric, while "differs by at most 1" is reflexive and symmetric but not transitive.

Is reflexivity implied by symmetry and transitivity?

No — a common false shortcut. If a ∼ b then b ∼ a and hence a ∼ a, but only for elements that relate to something. The empty relation is symmetric and transitive without being reflexive.

What is the quickest source of examples?

Any function f gives one: a ∼ b iff f(a) = f(b). All three properties follow from properties of equality, which is why this "kernel" construction shows up everywhere.

The rest of Equivalence Relations

Equivalence classes, partitions and quotient sets. Each subtopic below has its own method, worked example and mark-losing traps.

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