Talk:Area of Square

Be careful, this can't be circular. --Cynic (talk) 01:21, 12 March 2009 (UTC)

If someone can check the irrational side length section (especially the limits), I would appreciate it. --Cynic (talk) 21:50, 12 March 2009 (UTC)

Looks impressive. I think it's okay, I need to look at it a bit harder (I'll do that later, not tonight) but I think it's fine. --prime mover (talk) 21:53, 12 March 2009 (UTC)

I'm a touch concerned that the integral proof is circular. It seems to me that the definition of the definite integral as area is based on the area of a rectangle, which in turn is based on the area of a square. Also, I'm not sure if the fundamental theorem of calculus is really relevant for this proof. --Cynic (talk) 23:05, 18 April 2009 (UTC)

You're absolutely right. Even if we adopted a notion of double integral with arbitrary sets and Lebesgue measure, the notion of set measure in Euclidean space depends on the area of a rectangle being the product of the length of its sides. Also, though it hardly matters in light of that, the fundamental theorem of calculus is over kill, one could use Integral of Constant for the same use here. Zelmerszoetrop 23:09, 18 April 2009 (UTC)

I have to agree. I'll leave the decisions about geometry up to you guys (especially Cynic) - you seem to know our way round this area better than me. But I'd say: scrub this or label it "circular" or something. OTOH it is a proof (of sorts) and so is to all intents and purposes "valid". --prime mover (talk) 23:13, 18 April 2009 (UTC)

Tidied and added a note regarding the circularity. --Cynic (talk) 23:32, 18 April 2009 (UTC)

If anyone's still keeping an eye on this, feel free to give my latest proof a once-over. It goes right back to Euclid and I believe it bypasses all issues of circularity. Purists may quibble that we may need to take on board the sections on incommensurability from book X, but once we confirm that lines can be irrational in length (once a unit has been set) then this proof takes on board integer, rational and real without the need for a constructivist algebraic approach. --prime mover 09:59, 14 August 2011 (CDT)

Seems good to me - a touch more straight forward than the case by case version I put together since it essentially moves all the complex stuff to Similar Polygons are composed of Similar Triangles. --Alec (talk) 16:56, 15 August 2011 (CDT)

Exactly. Probably the most profound thing in book VI: "When you scale something up by a linear factor, the area scales up by a square of that factor." Our teacher did a demonstration of this with us in school when I was about 13 or 14 and it was jaw-droppingly revelatory. --prime mover 00:14, 16 August 2011 (CDT)