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Can some provide a proof for the formula $A = s^2$ for the area of a square?

Jamāl
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Greg Price
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    This is a definition of area, so there's no way to “prove” it. Lots of other area formulas can be proved from it, though. – Matthew Leingang Mar 17 '20 at 17:33
  • I can’t edit it but it seems a square with sides of 2 should be 2 * 4 – Greg Price Mar 17 '20 at 17:34
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    Humph.. usually these things have a proof somewhere that’s shocking to hear – Greg Price Mar 17 '20 at 17:36
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    See the question "Area in axiomatic geometry" and my answer. The thing is, from an axiomatic standpoint, there's no inherent area calculation; we choose one with desired properties ("Invariance under congruence" and "Additivity"). The nuances of the underlying geometry can lead to dramatically different "formulas". In Euclidean geometry, "base-times-height" (a product of lengths) works nicely; in spherical and hyperbolic geometry, a polygon's area is instead computed from its angle sum(!). – Blue Mar 17 '20 at 18:08
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    To ask this question, I suppose one should know what area is. – Lubin Mar 17 '20 at 20:04

2 Answers2

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$$A = \int\limits_{x=0}^s \int\limits_{y=0}^s 1\ dx\ dy = s^2$$

David G. Stork
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    To prove the area formula for a rectangle by using an integral seems...circular. – Matthew Leingang Mar 17 '20 at 17:37
  • "Circular"—great pun! But no it isn't "circular." And it most certainly isn't "a definition." Is the area of a rectangle a similar "definition"?! Or a parallelogram? Or every geometric figure!?! If I create a new class of polygon and derive their area by calculus, is that somehow another "definition"? Gee... then what isn't a definition in your world? – David G. Stork Mar 17 '20 at 17:39
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    In the most rigorous treatments of area that I've seen (measure theory, though it's been a long time), area is defined by a set of axioms. One axiom is that you can cut a set up into finitely many pieces and add up the areas of all the pieces. Another is that the area of a rectangle is the product of its length and width. You can use these two to derive the area of a parallelogram (cut off a corner and make a rectangle), or a triangle (two congruent triangles make a parallelogram), or every geometric figure (because it can be triangulated). But you have to start somewhere. – Matthew Leingang Mar 17 '20 at 17:46
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    I agree with David here. Calculus works pretty well for structural/civil engineering. When creating something so crucial I would much rather have a proof than just a definition someone said was true. Peace to all good points all around – Greg Price Mar 17 '20 at 17:50
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This depends a lot on your assumptions about area, and there are probably a handful of very different approaches.

Here is one: suppose we want to define "area" for at least some subsets of the plane $\mathbb R^2$, like the ones we would call "squares". There are some properties we might find intuitive, like:

  1. The area of a set (when it exists) is always a nonnegative real number or $\infty$. (Maybe we want to say things like "the area of the plane is $\infty$.")
  2. The area of the empty set is $0$.
  3. The complement of something with area has area.
  4. The intersection of two things with area has area.
  5. The area of (your favorite version of) the unit square is $1$. (If we don't pin down the area of anything with nonzero area in particular, then we could accidentally define something like "twice area" by mistake.)
  6. If you translate/shift a set around in the plane, its area shouldn't change.
  7. If you break a set $S$ into (countably-many) disjoint subsets $S_1,S_2,\ldots$, each of which has an area (possibly $0$ or $\infty$), then the area of $S$ should be the sum of all of the areas. We use sometimes ideas like that when working with integrals or thinking about complicated areas like that of a circle.
  8. For convenience, any subset of a set with area $0$ should definitely have an area; and by the above property, that area is forced to be $0$. (Intuitively, one of those subsets should have area $0$ since it's just a piece of something with area $0$.)

With fixed assumptions about set theory, the above assumptions pin down the notion of area in the plane uniquely. There are no more choices to be made. The name for this concept in arbitrary numbers of dimensions (length in the real line, area in the plane, volume in 3D, etc.) is the Lebesgue measure, and it's usually first studied in depth at or near the graduate/post-grad level.

From here, proving that the area of any square is what it should be can be done in a few ways, but you'll need to use infinite sums/limits somewhere if you want more than something like "half-open rational-side-length squares have the area they should if they have area at all" (taking assumption 5 to mean "$[0,1)\times[0,1)$ has area $1$").

Mark S.
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    My goodness! What an answer! Look back to the original question and ask yourself if this is what the poser was asking, and had the background (Lebesgue measure indeed!) to understand and appreciate. I'm surprised that Gödel and the foundations of math wasn't part of your answer. – David G. Stork Mar 17 '20 at 18:21
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    @DavidG.Stork, if OP asked a very deep question, so be it... even if it looks naïve. – vonbrand Mar 17 '20 at 20:16
  • @vonbrand: Not a chance. It is not a very deep question at all, and the fact he (reputation 113) refers to engineering (not Lesbegue integration), writes "I agree with David here," and accepted my answer proves it. It also proves that certain answers are not appropriate for him. (My goodness... all this digital ink spilled for "Why is the area of a square $A = s^2$?"!! – David G. Stork Mar 17 '20 at 20:35