Return to Vignettes of Ancient Mathematics
Return to Introduction to the method of the Method
Return to Survey of Infinitary Arguments

We go through the theorem, with the additions due to the new reading of the palimpsest the Method, by Reviel Netz and Ken Saito (Sciamus, 2001).  I shall then make some observations.
Let there be a cylinder enclosed in a square based prism.  The base of the prism is a, b, g, d.
Take a plane from the top of the cylinder tangent to the prism and cut the prism and cylinder so that it cuts through the base of the cylinder at its tangents to the prism, as shown, ae = ed.

 
The result is a segment of the cylinder, which Archimedes calls a hoof.
On the base of the hoof, draw a parabola hze, where z is on the tangent of the cylinder and prism.
We will now take an arbitrary ordinate of the parabola, ls, and parallel to the diameter of the parabola kz, we draw mlxn with perpendiculars at to the top of the hoof (cylindrical section) and at n to the top of the prism.

 

It is fairly easy to see that (since they are similar triangles):
the triangle in the prism on mn : the triangle in the cylindrical section on mx
           ≈ mn2 : mx2.
But, by the basic property of a parabola (note hk = kz)
mn : nl  ≈ hk2 : ls2
Hence,
mn : nl  ≈ hk2 : ls2  ≈ mn2 : ls2
It follows that,
mn : ml
mn2 : mn2 - ls2 since a : b = c : d => a : a-b = c : c-d
mn2 : mn2 - mk2
mn2 : mx2 since mn2= (the radius)2= mx2 + mk2
Hence, 
mn : ml mn2 : mx2
That is, where all are distance ls from kz:
the straight line parallel to kz : the line in the parabola ≈ the triangle in the prism : the triangle in the cylindrical segment

Let the triangles in the prism be A1, ..., An, ...
Let the lines in the rectangle dh be B1, ..., Bn, ...
Let the lines in the parabola be C1, ..., Cn, ...
Let the triangles in the cylindrical segment be D1, ..., Dn, ...

Since Ai = Aj and Bi = Bj, it trivially follows that

Ai : Aj ≈ Bi : Bj
Archimedes has also shown that: Bi : Ci ≈ Ai : Di

We now know that Archimedes claims:
1.   The triangles in the prism are equal in number to the lines in rectangle dh  (the A's and B's are equinumerous)
2.   The triangles in the prism are equal in number to the triangles in the cylindrical segment. (the A's and C's are equinumerous)
3.   The lines in the parabola are equal in number to the lines in the rectangle dh. (the B's and D's are equinumerous)
 

From these it follows by the Conoids and Spheroids 1,

(B1 +...+ Bn, ...) : (C1 +...+ Cn, ...) 
≈ (A1 +...+ An, ...) : (D1 +...+ Dn, ...)


Or

all the lines in rectangle dh parallel to kz : all the lines in the parabola ≈ all the triangles in the prism : all triangles in the section of the cylinder
Or, if we think of the magnitudes as 'filled up' by the slices',
the triangle based prism : the cylindrical section or hoof
≈ the rectanble dh : the parabola ≈ 3 : 2
Hence, since the prism on the square is quadruple the triangle based prism, the whole prism on the square : the hoof ≈ 12 : 2 = 6 : 1.

Several points about the argument:
Archimedes has no difficulty in saying that

  1. The n-dimensional figure is composed of the n-1-dimensional figures.
  2. The (infinite) n-1-dimensional figures in one figure are equal in number to the m-1-dimensional figures in another figure.
  3. The lemma appealed to, which is mentioned in the introduction to the Method, from Conoids and Spheroids 1 is only proved for the finite case, and probably could only be proved for the finite case.  Yet, he has no difficulty extending it to the infinite case.
From these three claims it is reasonable to infer that Archimedes is not troubled by infinitary arguments in the way that some ancient philosophers were (e.g., Aristotle).