"A relation cannot be both symmetric and antisymmetric — they pull in opposite directions."

Plausible, but wrong. A relation can be both. The catch is a sharp restriction: the only such relations are the ones whose pairs all lie on the diagonal. In other words, R \subseteq \{(a, a) \mid a \in A\}.

The canonical example is equality. The empty relation is another. Anything with even one off-diagonal pair automatically fails one of the two properties.

Chasing the contradiction down

Suppose R is both symmetric and antisymmetric. Take any pair (a, b) \in R.

So every pair in R has a = b. That is exactly the statement that R is a subset of the diagonal.

Why: symmetric forces "reverse is in," antisymmetric forces "if reverse is in, then the two elements were equal." Combining them: reverse is in, so the two elements were equal. Every pair in R is of the form (a, a). No off-diagonal pairs survive.

The picture

Grid of a 4 by 4 relation with only diagonal cells filledA 4 by 4 grid labelled 1, 2, 3, 4 along rows and columns. Only the four diagonal cells at positions (1,1), (2,2), (3,3), (4,4) contain red dots. All twelve off-diagonal cells are empty. A label explains that such a relation is both symmetric and antisymmetric. b → a ↓ 1 2 3 4 1 2 3 4 only diagonal cells in R → both symmetric and antisymmetric
A relation that is both symmetric and antisymmetric. Every pair is of the form $(a, a)$ — only diagonal cells are filled. Symmetry is satisfied because reversing $(a, a)$ gives $(a, a)$ itself, which is still in $R$. Antisymmetry is satisfied vacuously for off-diagonal pairs (none exist) and trivially for diagonal pairs ($a = a$ is already true).

Concrete examples

1. Equality on any set. R = \{(a, a) \mid a \in A\} — precisely the diagonal. Symmetric: if a = a then a = a. Antisymmetric: if a = b and b = a then a = b. Both trivially hold.

2. The empty relation. R = \varnothing on any set. Symmetric vacuously (no pair to reverse), antisymmetric vacuously (no pair of reverses to force equality). Both hold.

3. Partial diagonal. A = \{1, 2, 3\}, R = \{(1, 1), (3, 3)\}. Only two diagonal cells filled. Symmetric: reversing a diagonal pair gives the same pair. Antisymmetric: the only reversal-pairs are (1, 1) with itself and (3, 3) with itself, and in each case the two coordinates are equal. Both hold.

What breaks when you add any off-diagonal pair

Add (1, 2) alone. Now symmetric fails — (2, 1) \notin R. Still antisymmetric.

Add (1, 2) and (2, 1) both. Now symmetric succeeds. But antisymmetric fails — both pairs are in R and 1 \neq 2. So symmetric and antisymmetric are mutually exclusive the moment a non-diagonal pair enters.

This is the whole content of the theorem: you have to choose. You can have symmetric without off-diagonal pairs. You can have antisymmetric with off-diagonal pairs as long as their reverses are out. To have both at once, keep R strictly on the diagonal.

Interactive: watch the edge case

Slider showing how symmetric and antisymmetric both hold only when off-diagonal pair count is zeroA horizontal track with a draggable red point. The readout reports the number of off-diagonal pairs currently selected from 0 to 4. When the slider is at 0, the readout reports both symmetric and antisymmetric hold. When the slider is at 1 or more, one of the two properties fails. 0 1 2 3 4 drag off-diagonal pair count
Only the leftmost position (zero off-diagonal pairs) keeps both properties intact. Any rightward nudge introduces off-diagonal pairs, and symmetric and antisymmetric can no longer coexist.

Why this matters for JEE

Examiners set questions like "is the equality relation on \mathbb{R} an equivalence relation and also antisymmetric?" The answer is yes to both, and the reasoning hinges on the diagonal-only structure of equality. Students who memorise "symmetric and antisymmetric are opposites" give the wrong answer. Students who remember the characterisation — both hold iff R lives inside the diagonal — get full marks.

Equality is the trivial equivalence relation *and* a partial order

Take A any set and R = \{(a, a) \mid a \in A\}.

  • Reflexive: every diagonal pair is in, by definition.
  • Symmetric: reversing (a, a) gives (a, a).
  • Transitive: the only chains are (a, a), (a, a) \to (a, a). All in.
  • Antisymmetric: if (a, b) \in R and (b, a) \in R, then the pairs are (a, b) = (a, a) and (b, a) = (b, b), which forces a = b.

So equality is simultaneously an equivalence relation (reflexive, symmetric, transitive) and a partial order (reflexive, antisymmetric, transitive). This is the only way both classifications happen together, because the only relation that is both symmetric and antisymmetric on A is one whose off-diagonal content is empty.

One-line take-away

A relation is both symmetric and antisymmetric exactly when R \subseteq \{(a, a) \mid a \in A\} — every pair sits on the diagonal. Equality and the empty relation are the headline examples. Anything else forces you to pick one.

Related: Relations · Antisymmetric vs Not Symmetric · Equivalence Relations · Why 'Equals' Is the Trivial Equivalence Relation