Two operations that trip up almost every student in a sets chapter: set difference A - B and symmetric difference A \triangle B. The names rhyme, the symbols are one character apart, and both "remove the overlap" in some sense. A side-by-side picture settles the difference in a glance — one shades a single crescent, the other shades two.

The two definitions

A - B = \{ x \mid x \in A \text{ and } x \notin B \}
A \triangle B = (A - B) \cup (B - A) = \{ x \mid x \in \text{exactly one of } A, B \}

In words: A - B keeps what is in A but not B; A \triangle B keeps what is in exactly one of the two sets, regardless of which one. The second operation is completely symmetric in A and B; the first prefers A.

Why "symmetric": swap A and B in A \triangle B and nothing changes, because "exactly one of the two" does not care which is first. Swap them in A - B and you get a different set, B - A.

The side-by-side picture

Slide the overlap control below. The two Venn diagrams update together — left shades A - B, right shades A \triangle B. The "Animate" button sweeps the overlap from disjoint to heavy overlap so you can watch the shaded regions grow, shrink, and eventually differ by two crescents instead of one.

Left: $A - B$ shades only the left crescent — "$A$ with $B$-overlap removed." Right: $A \triangle B$ shades both crescents — "everything in exactly one set." Drag the slider or press Animate to sweep separation. When circles are fully disjoint both panels shade the same region (the whole of $A$, since $B-A$ is empty); at partial overlap the symmetric difference has up to twice the area of set difference.

The picture shows the structural fact cleanly: A \triangle B is the union of two set differences.

A \triangle B = (A - B) \cup (B - A)

If you shade only the first term, you get the left diagram. Add the second term and you get the right.

A numerical example

Take A = \{1, 2, 3, 4\} and B = \{3, 4, 5, 6\}.

Operation Result
A - B \{1, 2\}
B - A \{5, 6\}
A \triangle B \{1, 2, 5, 6\}

The symmetric difference has twice the elements of A - B, because it collects both the "A-only" and "B-only" pieces. If you ran the same problem with A = B, then A - B = \varnothing and A \triangle B = \varnothing would agree — but that is a coincidence of the equal-sets case, not a general rule.

Cardinality formulas

Both operations have clean counting rules.

|A - B| = |A| - |A \cap B|
|A \triangle B| = |A| + |B| - 2|A \cap B|

Notice the factor of 2 in the second formula. That is the signature of symmetric difference — you subtract the overlap twice because it appears in both |A| and |B| and must be removed from both crescents.

Why twice: each overlap element gets counted once in |A| and once in |B|. Neither appearance belongs to A \triangle B, so you remove both. Inclusion-exclusion only removes the overlap once because the union includes it; symmetric difference excludes it.

Commutativity — the other big divider

Set difference is not commutative: A - B \neq B - A in general. Drag the slider far enough and you could swap the circle labels, and the left diagram would shade the right crescent instead — a completely different set.

Symmetric difference is commutative: swapping A and B in A \triangle B leaves the shaded region unchanged (both crescents, regardless of which one you call A - B). This symmetry is not a detail — it is the whole reason the operation deserves its own name.

When are they equal?

Only in two special cases.

  1. Disjoint sets. If A \cap B = \varnothing, then B - A is empty, so A \triangle B = A - B \cup \varnothing = A - B = A. The slider shows this: push the circles fully apart and both diagrams shade the same region (the whole of A).
  2. B \subseteq A. If every element of B is already in A, then again B - A = \varnothing and the two operations agree, both equal to A - B.

In all other cases the two operations give different sets, and the size gap is |B - A|.

A worked JEE-style problem

$A = \{x \in \mathbb{N} : x \le 8\}$ and $B = \{x \in \mathbb{N} : 4 \le x \le 10\}$. Find $A - B$, $A \triangle B$, and their cardinalities.

Step 1. Roster the two sets.

A = \{1, 2, 3, 4, 5, 6, 7, 8\}, \quad B = \{4, 5, 6, 7, 8, 9, 10\}

Step 2. Find A - B: keep elements of A not in B.

A - B = \{1, 2, 3\}, \quad |A - B| = 3

Why: walk through A and drop any element that also appears in B. Elements 4 through 8 appear in B, leaving \{1, 2, 3\}.

Step 3. Find B - A.

B - A = \{9, 10\}, \quad |B - A| = 2

Step 4. Union the two to get A \triangle B.

A \triangle B = \{1, 2, 3\} \cup \{9, 10\} = \{1, 2, 3, 9, 10\}, \quad |A \triangle B| = 5

Cardinality check. |A| + |B| - 2|A \cap B| = 8 + 7 - 2 \cdot 5 = 5. Match.

Result. A - B = \{1, 2, 3\} (three elements) and A \triangle B = \{1, 2, 3, 9, 10\} (five elements). The difference is the two extra elements from B - A that the symmetric difference captures but set difference does not.

The two answers differ by exactly |B - A| = 2 elements — which is what the side-by-side Venn picture predicts.

Why the confusion persists

The English vocabulary betrays both operations. "Set difference" suggests subtraction — but subtraction of numbers is not commutative, so students correctly infer that A - B and B - A differ. "Symmetric difference" sounds like "the thing that makes it symmetric" — which it does, but at the cost of adding back the other crescent, not by merely reversing directions.

The picture dissolves both confusions. Whenever a question asks for "everything in A but not B," you want one crescent — A - B. Whenever it asks for "everything that is in exactly one set" or "the disagreement between the two," you want both crescents — A \triangle B.

Related: Set Operations · Symmetric Difference · A − B vs A ∩ B′ · Animated Venn Diagrams