I'm not a math expert but I know a little bit of calculus and theorems. I've heard things like "this result is "dual"", or this "theorem is "dual"". Often people say "dual comes for free". Like you swap variables or something and you get another result.

I've never understood this. Can anyone explain what "dual" means precisely? Examples would be nice as well.


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    $\begingroup$ +1 I love this question. The entry on duality at Wikipedia is pretty comprehensive although it requires a fair amount of prior knowledge to read it: en.wikipedia.org/wiki/Duality_%28mathematics%29 $\endgroup$ Nov 8, 2015 at 7:19
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    $\begingroup$ It has a lot of different meanings. $\endgroup$
    – copper.hat
    Nov 8, 2015 at 7:24
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    $\begingroup$ "Can anyone explain what "dual" means precisely?" No, because it is not a single precise term. It has many different meanings in different contexts (some precise, some not), and the only thing they all have in common is that there is some sort of symmetry involving two things. $\endgroup$ Nov 8, 2015 at 7:27
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    $\begingroup$ Platonic solids have duals too. For example, the cube and octahedron are dual and dodecahedron and icosahedrons are dual if you map vertices to faces. Under this same mapping the tetrahedron is self dual. $\endgroup$ Nov 8, 2015 at 7:35
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    $\begingroup$ Also, planar graphs have "duals". Related: What's the difference between "duality" and "symmetry" in mathematics? $\endgroup$
    – hardmath
    Nov 8, 2015 at 14:26

2 Answers 2


A duality is a pair of related concepts that display a one-to-one translation symmetry, usually (not always) as the result of some form of involution operator.

In classical logic, the operators, $\vee$ and $\wedge$ form a dual, and negation is their involution operator.   This is expressed through deMorgan's Laws:$$\neg(A\vee B) = \neg A\wedge \neg B\\\neg(A\wedge B)=\neg A\vee \neg B$$

Similarly in set algebra, the $\cup$ and $\cap$ operators form a dual, with complementation being the involution. $$(A\cup B)^\complement = A^\complement\cap B^\complement\\ (A\cap B)^\complement=A^\complement\cup B^\complement$$

In predicate logic, the quantifiers, $\forall$ and $\exists$ are a dual, again with negation being the transformation. $$\neg \forall x\;P(x) \iff \exists x\;\neg P(x)\\\neg \exists x\;P(x) \iff \forall x\;\neg P(x)$$

And such.


Broadly, two mathematical set-ups are dual when they can be transformed into each other by a simple exchange of symbols and terminology. For example, a general topological space may be defined in terms of its open sets, or dually in terms of its closed sets: corresponding to infinite unions and finite intersections of open sets, in the first formulation, are respectively finite unions and infinite intersections in the second one. Another example is plane projective geometry: a theorem about lines through points, and points where lines intersect, can be transformed into a distinct dual theorem, respectively about points where lines intersect and lines through points. The dual of a finite-dimensional vector space is the space of linear functionals on it; and the dual of the latter space is isomorphic to the original space via a canonical mapping.

Unfortunately, the word dual is not confined to such simple cases where, if B is the dual of A, then A is the dual of B. For example, in the case of infinite-dimensional vector spaces, the dual of a space (defined as above) can be much "bigger" than the original, and its dual in turn is bigger still. However, when the word is used, there is usually a clear link to the simpler idea of duality.

  • $\begingroup$ I don't think this theory covers dual graphs or dual polyhedra or duality of vector spaces. $\endgroup$
    – MJD
    Nov 8, 2015 at 14:42
  • $\begingroup$ @MJD: Agreed. In the case of graphs and infinite-dimensional vector spaces, duality goes in only one direction. So what I described initially might better be called reciprocal duality, although I don't know whether this term is used. This sort of duality does hold at least for the regular polyhedra. $\endgroup$ Nov 8, 2015 at 15:33

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