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Polygons are, in this question, defined as non-unique if they similar to another (by rotation, reflection, translation, or scaling).

Would this answer be any different if similar but non-identical polygons were allowed? And if only if rotated/translated by rational coefficients?

Would this answer be any different if we constrained the length and internal angles of all polygons to rational numbers?

Assume the number of sides is finite but unbounded, and greater than two.

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    Very interesting question!2016-11-16

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There are uncountably many, because for example one can have rectangles with arbitrary side ratios. For your second question, if everything is constrained to be rational, there will be countably many, because a polygon is uniquely determined by its ordered collection of sides and angles.

This is part of a general fact: if $A$ is a countable set, then the collection of ordered $n$-tuples of elements of $A$ for all $n$ is still countable.

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    This popped up as I was thinking of writing it. +1 :)2010-07-23
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    Assuming finitely-many sides; otherwise, the collection of side-lengths becomes the (ordered) power-set of the rationals.2010-07-23
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    Does this general fact still apply if your elements are ordered *infinite*-tuple elements? Because the number of sides of polygons are unbounded. Would it still be countable?2010-07-23
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    Justin: No, because the set of all infinite sequences over $Q$ is of the same cardinality as $R$. In fact, the set of all sequences over $\{0,1\}$ is of the same cardinality as $R$---this is by the binary encoding. If you are allowing polygons with infinitely many sides (finite but unbounded is OK), then I'll edit my answer - could you please clarify?2010-07-23
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    I mean finite, but unbounded.2010-07-23
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    Then it's countable, because for each $n$ the set of all such polygons is countable, and taking the union over all $n$ gives a countable union of countable sets, which is (by a diagonal argument, countable - see http://en.wikipedia.org/wiki/Countable_set.2010-07-23