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April 30th, 2009, 04:42 PM   #1
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An inquiry in Geometry

Draw two squares size a and b side-by-side to each other such that b > 2a and the a-square is on the left bottom corner of the b-square. Now draw another square size a on the top right corner of the b-square in the same manner. Use this to Algebraically and Geometrically study the square (a + b). Now, thus if the nature of Algebra and Geometry are consistence then (a + b)^2 = 2a^2 + 2b^2 - (b - 2a)^2 which will lead to the conclusion that 3a = 2b. But the picture drawn have other variation as validate by ours eyes.

A brief description of the problem:
The fabication of R^2 had, or will, correctly avoided or bybass this problem as each "unit square" in the process are equal and every other distance and length is a dependent of this fabication. Indeed, the primary problem here is the existence of two units, that is, while the nature of Algebra was orginally design to compute, work with, realise with pure numbers, it said nothing of length and units. Thus, the application of Algebra to this problem require a setting of fabicated background, that is, one way to avoid such complication is to rewrite a and b in term of a singular unit. Hencefore, since there is no known system of such is known, one in actuality work with what one have and thusfore have two equations, one is from the geometric-algebra intepretation and the other is from the fabication involving a and b, say, for a pre-theory, ab=lk.

Edit: Forgot to mention, this is Physical Construction Mathematics an part of Applied Nature Mathematics. It is rather, a and b above are Algebraic-physical variable or whatever one wish to label it to increase the understanding and comprehension by it viewer in general and for latter generation as well. I would not likely to call it so-and-so name variable.
MyNameIsVu is offline  
May 1st, 2009, 01:43 AM   #2
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Did you mean b < 2a? If b > 2a (and a is positive), it's necessarily incorrect to form an equation that implies 3a = 2b, i.e., b = (3/2)a.
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