Sets · step 12 of 13

Symmetric difference A △ B

A △ B is the set of elements in exactly one of A and B: (A \ B) ∪ (B \ A), equivalently (A ∪ B) \ (A ∩ B). It is the set-theoretic exclusive or, and it has unusually good algebra — associative, commutative, with identity ∅ and every set its own inverse, which makes (𝒫(U), △) an abelian group where every element has order 2.

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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.

Use whichever definition is easierFor explicit sets, "in exactly one" is fastest. For proofs, (A ∪ B) \ (A ∩ B) often simplifies better.
Translate to indicators for identities1_{A △ B} = 1_A + 1_B mod 2. Symmetric-difference identities are then just arithmetic mod 2.
Exploit the group lawsA △ A = ∅ and A △ ∅ = A, so equations like A △ X = B are solved by X = A △ B — cancellation works exactly as in ordinary algebra.
Chain associativelyA △ B △ C is unambiguous and contains the elements lying in an odd number of the three sets.

Worked example

With A = {1,2,3,4} and B = {3,4,5}, compute A △ B, then solve A △ X = {1, 5} for X.

A \ B = {1, 2}; B \ A = {5}.
So A △ B = {1, 2, 5}.
For the equation, △ both sides by A: A △ A △ X = A △ {1,5}, and A △ A = ∅.
So X = A △ {1,5} = ({1,2,3,4} \ {1,5}) ∪ ({1,5} \ {1,2,3,4}) = {2,3,4} ∪ {5}.

Answer. A △ B = {1, 2, 5}, and X = {2, 3, 4, 5}.

Where marks get dropped

These are the specific errors that cost credit on symmetric difference a △ b questions — QED's rubric penalises each of them separately.

Computing A \ B and stopping. The symmetric difference needs both directions.
Assuming △ distributes over ∪. It does not; the correct distributive law is A ∩ (B △ C) = (A ∩ B) △ (A ∩ C).
Forgetting that △ is associative and trying to bracket it as though it were subtraction.

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Symmetric difference A △ B — frequently asked questions

Why is symmetric difference like XOR?

Because x ∈ A △ B exactly when the propositions x ∈ A and x ∈ B differ — the truth table of exclusive or. Under indicators it is literally addition mod 2.

Is (𝒫(U), △) really a group?

Yes: △ is associative and commutative, ∅ is the identity, and A △ A = ∅ makes every set its own inverse. It is an elementary abelian 2-group of order 2^|U|.

How is it used in practice?

Diffing and version control conceptually compute symmetric differences of change sets, and error-correcting codes use the same mod-2 structure over bit vectors.

The rest of Sets

Set-builder notation, operations, and set identities. Each subtopic below has its own method, worked example and mark-losing traps.

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