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Area is the quantity that expresses the extent of a two-dimensional figure or shape or planar lamina, in the plane. When its sides have lengths of irrational numbers, are their areas defined, for example, the area of a square with length $\pi$ is $\pi^2$ by definition?

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    Yes? Why wouldn't the area be defined? – PrincessEev Dec 01 '20 at 22:12
  • The formula for area of a square being $s^2~$ where $~s =~$ the length of the side of the square applies to all $s \in \mathbb{R^+}.$ – user2661923 Dec 01 '20 at 22:25
  • I was wondering how the formula was derived. When $s\in\mathbb{Q}$, it is understandable since the area means how many unit squares or parts of unit squares are needed to fill in the square of side length $s$ and the result is $s^2$. But when $s$ is irrational, do we have to define its area to fill in the gaps in $\mathbb{R}^2$? –  Dec 01 '20 at 22:35
  • You seek Integral calculus. –  Dec 01 '20 at 23:03
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    @SenZen I don’t see how integral calculus will help the op. In both the definition of the Riemann- or Lebesgue-Integral the volume of a cube/rectangle is assumed (or rather defined) to be the product of the length of edges, right? – Jonas Linssen Dec 01 '20 at 23:24

2 Answers2

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I'm not sure if I'm reading the question correctly, but I think my answer sufficiently general to cover the likely interpretations.

I'd say it's a corollary of the definition of the area. A definition wouldn't give a particular value since it's establishing a general concept. Given what an area is, there is the particular way the concept applies to a square, which includes a relationship between the side length of a square and its area. Then, given some side length, we can deduce the area. So we have the definition, then from there a general formula for a square, then finally the specific case of a square with a given side length.

So the area of a square having side length $\pi$ is not $\pi^2$ strictly by definition. This isn' t to say the area isn't $\pi^2$, but rather this isn't strictly a fundamental statement of the meaning of a base level term, i.e. a definition.

Whether we are permitted an irrational value for the area depends on the fundamental definition of the area and not specific values.

Since it is a given that we are allowed irrational lengths, specifically $\pi$ and we have some concept of area, at minimum, what do we need to have an area with an irrational value?

If we have a rectangle of side lengths 1m and $\pi^2$m, do we have an rea of $\pi^2m^2$?

We have no real lengths without a definition of the reals, often defined as the ordered field having rationals as a subset and which satisfies the least upper bound property. This in turn implies density of the rationals in the reals (by way of the Archimedean property). So if we have arbitrary rational lengths and areas, then we have arbitrary real areas by the least upper bound property.

So depending on our definitions, but not by definition, we can prove the area of a square of side length $\pi$ is $\pi^2$.

TurlocTheRed
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I think it is valid to question, whether a rectangle with (potentially) irrational sides $a$ and $b$ should have area $ab$.

The reason why this makes sense to me is that the product is a continuous function $\Bbb R \times \Bbb R \rightarrow \Bbb R$. So given any approximation $(a_n,b_n)_n$ of the rectangle by rectangles with rational sides, their areas will converge to the product $ab$. Hence it makes complete sense (or better is the only logical way) to define the area of a general rectangle to have area given by the product of its sides.

Jonas Linssen
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