Equivalence Relations · step 7 of 13

Well-definedness of operations on classes

Defining an operation on classes by [a] + [b] = [a + b] uses a representative, and a different representative might give a different answer. Well-definedness is the proof that it does not: if [a] = [a′] and [b] = [b′], then [a + b] = [a′ + b′]. Skipping this check is how students "prove" false statements about quotients, and it is always an explicitly marked step.

Practise well-definedness of operations on classes free → All Equivalence Relations subtopics

✓ Unlimited questions · marked criterion by criterion · no card needed

Method: how to approach it

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

State what must be shownAssume a ∼ a′ and b ∼ b′; prove the results are equivalent. Writing this sentence earns the first mark.
Translate equivalence into equationsFor mod n, a − a′ = nk and b − b′ = nm. Now everything is algebra.
Compute the difference of the two resultsShow (a ⊕ b) − (a′ ⊕ b′) satisfies the equivalence condition, e.g. is a multiple of n.
Test suspect operations with a counterexampleIf the operation is not well-defined, two representatives of the same class give inequivalent results — exhibit them.

Worked example

Show that multiplication on ℤ/nℤ, defined by [a]·[b] = [ab], is well-defined.

Assume a ≡ a′ and b ≡ b′ (mod n), so a − a′ = nk and b − b′ = nm.
Then ab − a′b′ = ab − a′b + a′b − a′b′.
= b(a − a′) + a′(b − b′) = b·nk + a′·nm.
= n(bk + a′m), a multiple of n.

Answer. ab ≡ a′b′ (mod n), so [a]·[b] = [ab] does not depend on the chosen representatives.

Where marks get dropped

These are the specific errors that cost credit on well-definedness of operations on classes questions — QED's rubric penalises each of them separately.

Assuming well-definedness because the formula "looks fine". Exponentiation by class, [a]^[b] = [a^b], fails on ℤ/nℤ — the exponent must reduce mod φ(n), not mod n.
Proving only that one representative works. Both arguments must be allowed to vary simultaneously.
Confusing well-definedness with closure. Closure says the result is in the set; well-definedness says it does not depend on the representative.

Practise this until it is automatic

Unlimited fresh questions

QED generates new well-definedness of operations on classes problems on demand at warm-up, exam and challenge level, so you can drill this one skill until it stops costing you marks.

Marked like an examiner

Every answer is scored against a point-by-point rubric with partial credit, so you see exactly which step of the method broke down — not just a tick or a cross.

Answer in real notation

A one-tap symbol palette, a visual equation editor and a truth-table builder — or photograph your handwritten working and QED converts it to LaTeX.

Saved to your library

Every question you generate is kept and re-takeable as a timed exam, and your Equivalence Relations mastery is tracked so you know when this is exam-ready.

Well-definedness of operations on classes — frequently asked questions

Give an operation that is not well-defined.

On ℤ/6ℤ try [a] ↦ [a/2] for even representatives: [2] = [8], but 2/2 = 1 and 8/2 = 4, and [1] ≠ [4] mod 6. The rule depends on the representative, so it defines nothing.

Why does it always need proving?

Because the definition mentions an element while the object is a class. Nothing in the notation guarantees consistency — only the proof does.

Is there a way to avoid the check?

Yes, via the universal property: define the map on A first, show it is constant on classes, and it factors through the quotient. That is the same check in a different order.

The rest of Equivalence Relations

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

Ready to make well-definedness of operations on classes exam-proof?

Generate your first questions free — no card, no setup, no personal data stored. Practise until the method is second nature.

Start practising free →