Students often read "antisymmetric" and assume it means "the opposite of symmetric," the way "irrational" means "not rational." It does not. Antisymmetric is a completely separate property with its own rule. "Not symmetric" is the negation of symmetry; antisymmetric is a stronger condition about what happens when pairs do reverse.

Both can fail. Both can hold. A relation can be neither symmetric nor antisymmetric. That alone proves the two ideas are different — if antisymmetric were just "not symmetric," every relation would satisfy one or the other, never both-fail.

The two definitions, side by side

Symmetric

R on A is symmetric if for every a, b \in A:

(a, b) \in R \implies (b, a) \in R

Every arrow has its reverse.

Antisymmetric

R on A is antisymmetric if for every a, b \in A:

(a, b) \in R \text{ and } (b, a) \in R \implies a = b

The only pairs allowed to reverse are the diagonal pairs (a, a).

Not symmetric is the logical negation of symmetric: there exists some pair (a, b) \in R with (b, a) \notin R. One witness is enough.

Antisymmetric is a for-every statement: for every pair of distinct elements a \neq b, you never find both (a, b) and (b, a) together in R.

Why: the two properties are in different logical moods. "Symmetric" is "for every pair, reverse is also in." "Antisymmetric" is "for every pair, if both directions are in, then the two elements were equal." "Not symmetric" just negates the first. Antisymmetric adds new structure — it restricts which pairs R is allowed to have — it does not simply drop the symmetric requirement.

The four-quadrant table

Every relation on A lands in exactly one of four boxes.

Four-quadrant table showing combinations of symmetric and antisymmetricA 2 by 2 grid. The rows are labelled symmetric and not symmetric. The columns are labelled antisymmetric and not antisymmetric. Each of the four cells contains an example of a relation: the top-left cell shows relations where the only pairs are on the diagonal, like equality. The top-right cell is labelled impossible. The bottom-left shows strict less-than. The bottom-right shows a sample relation with both a pair and its reverse where the two elements are unequal. antisymmetric not antisymmetric symmetric not symmetric both hold equality "=" only diagonal pairs in $R$ impossible see explanation below antisym, not sym strict $<$ on $\mathbb{R}$ $\le$, divides on $\mathbb{Z}^+$ neither holds $R = \{(1,2),(2,1),(2,3)\}$ on $\{1,2,3\}$
The four combinations of symmetric/not-symmetric and antisymmetric/not-antisymmetric. Three cells are non-empty; one (top-right) is impossible except in a trivial edge case. A relation can be neither symmetric nor antisymmetric, which immediately shows the two properties are not opposites.

Why is the top-right cell (symmetric and not antisymmetric) almost impossible? If R is symmetric and contains any pair (a, b) with a \neq b, then (b, a) \in R too — and that already violates antisymmetry. So the only way to be both symmetric and antisymmetric is to have no off-diagonal pairs at all: the equality-style relations where R \subseteq \{(a, a) \mid a \in A\}.

A relation that is neither

This is the clearest proof that "not symmetric" \neq "antisymmetric." Take A = \{1, 2, 3\} and

R = \{(1, 2),\; (2, 1),\; (2, 3)\}.

Both fail. Neither property holds. This is forbidden if antisymmetric were the negation of symmetric, so antisymmetric cannot be the negation of symmetric.

Interactive: flip through the four cases

Slider reading which symmetric or antisymmetric combination is activeA horizontal track with a draggable red point. The readout describes one of four quadrants: both, antisymmetric only, symmetric only (rare), or neither, depending on the slider position. both antisym sym neither drag through the four cells
Drag across the four cells. Each position names a relation example that lives there. The existence of a "neither" cell is the crucial fact — it shows antisymmetric and "not symmetric" are not the same property.

Quick recognition rules for exams

Symmetric and antisymmetric are two independent yes/no questions. "Not symmetric" answers only one of them and leaves the other open. To check antisymmetry, look at every off-diagonal pair in R and confirm its mirror is missing; do not assume it follows from symmetry failing somewhere.

Is "divides" antisymmetric on $\mathbb{Z}^+$?

Suppose a \mid b and b \mid a with a, b positive integers. Then b = ka and a = lb for positive integers k, l, giving a = l(ka) = (lk)a, so lk = 1. Since k, l are positive integers, k = l = 1, which forces a = b.

So yes — on \mathbb{Z}^+, the divisibility relation is antisymmetric. And it is definitely not symmetric (2 \mid 4 but 4 \nmid 2). So divisibility lives in the "antisymmetric, not symmetric" cell of the table.

Why: on \mathbb{Z} (allowing negatives), divisibility loses antisymmetry — for instance 3 \mid -3 and -3 \mid 3 but 3 \neq -3. Always check which set the relation lives on before concluding antisymmetry.

One-liner

"Not symmetric" means the mirror rule fails at least once. "Antisymmetric" means the mirror rule can only succeed on the diagonal. They live on different axes, and a single relation can independently satisfy neither, one, the other, or both.

Related: Relations · Symmetry Check Visualisation · A Relation Both Symmetric and Antisymmetric — Possible? · Equivalence Relations