History of Mathematics: Archimedes' Work of The Method
Archimedes first proved it by a mechanical way and then confirmed by a geometrical way, both of the solutions were confirmed by the method of exhaustion.
In his work of The Method, Archimedes assumed that figure are consist of their indivisible cross-sections, second, by the use of the law of the lever, assumed the balancing of cross-sections of a given figure against corresponding cross-sections of a known figure. He obtained the right answer by the following method: to inscribe the given solid within a solid within a solid of the known center of gravity and then, using his law of the lever, to ‘weigh’ the former against the latter about an imagined fulcrum. He assumed that there is no change by assuming that the triangle is located at its center of gravity.
To prove this property, he assumed the following lemma: that the excess by which the greater of unequal areas exceeds the less can, by being added to itself, be made to exceed any given finite area. We can observe that Archimedes takes the segment and the triangle to be made up of parallel lines indefinitely close together. In reality, they are made up of indefinitely narrow strips, but the width being the same for the elements of the triangle and segment respectively, divides out. And also that the weight of each element in both is proportional to the area. He also did not mention, that moment, in effect assumes that the sum of the moments of each particle of a figure, acting where it is, is equal to the moment of the whole figure applied as one mass at its center of gravity. Neither mechanical principles nor ‘indivisible’ cross-sections could appear in a formal mathematical argument, moreover, his ‘mechanical’ derivation depends on other things, including on Democritus’ idea that a plane figure is the sum of an infinite number of parallel lines just as a solid is the sum of an infinite number of parallel planes, which might be the reasons for why Archimedes considered his ‘mechanical’ derivation did not furnish an actual demonstration.
The geometrical proof of the result on the segment of a parabola is based on Eudoxus’s method of exhaustion. The idea is to construct rectilinear figures inside the parabolic segment whose total area differs from that of the segment by less than any given value. The figures he used for this purpose are triangles. At each stage, the area of the tringles constitutes more than half the area of the parabolic segments in which they are inscribed. If this process is continued, Eudoxus’ theorem suggests that the segments left over will have a combined area less than any given value. Hence the polygon formed by the triangles will ‘exhaust’ the original parabolic segment. For this purpose it is necessary to find, in terms of the triangle with the same base and height, the area added to the inscribed figure by doubling the number of sides other than the base of the segment.
Archimedes used several geometrical theorems to in his Quadrature of the Parabola. He used the properties of similar triangles to show that the sides of the smaller triangle are half of the sides of the larger triangle. He also used Euclid’s theorem on the ratio of triangles on the same base and between the same parallel. It then can be shown that the total areas of the triangles created at each stage are one-fourth of the area of the triangles constructed in the previous stage.
And that the more steps are taken, the more closely the sum of the areas approaches the area of the parabolic segment. Therefore, to complete the proof, it is sufficient for him to find the sum of the geometric series where a is the area of the original triangle.
Archimedes’ work is very original as he did not use Euclid’s formula for the sum of a geometric progression for Elements, but instead gave the sum in the form:
He found this find by noting that, then by subtracting equals and rearranging gives the desired result. He completed the argument through a double reduction ad absurdum, with begin with the assumption that K= is not equal to the area B of the segment. Then by considering two cases, KB, he reached two contradictions, therefore it showed that K=B.
Sources:
- J. Fauvel and J. Gray, The History of Mathematics, pp. 153-154, 167-171
- T. Heath, A History of Greek Mathematics, Vol.2, pp. 27-30, 85-91
- V. Katz, A History of Mathematics, pp. 104-105, 108-109
