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CHAPTER VII 
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Most of this chapter will be devoted to showing why the Pythagorean Theorem cannot be used in the argument against atomism. Without the Pythagorean Theorem many arguments against atomism fail. The less mathematically inclined may wish to skip ahead to the concluding remarks on page 208.
One alleged proof of infinite divisibility appeals indirectly to the Pythagorean theorem.^{(1)} In this section I examine a classic proof of the Pythagorean Theorem and show that that proof depends upon and presumes that extension is infinitely divisible. I also show that, when atomism is presumed, the corrected proof fails to yield the well known formula; a different formula results. Since the original Pythagorean formula itself depends upon infinite divisibility, using it to "prove" infinite divisibility, in effect, begs the question.
One diagram used in demonstrating the proof of the Pythagorean theorem involves 6 overlapping figures in the plane.^{(2)} Figure 26 shows a square with sides of length C that is inscribed inside another square. The sides of the triangles so formed are of length A and B respectively. The area of each triangle is AB/2. The length of a side of the outer square is A + B. To prove the Pythagorean theorem, the total area of the outer square is set equal to the area of the 4 triangles plus the area of the inner square. The proof involves simple algebra.
Pythagorean Theorem Proof (with infinite divisibility)^{(3)}  


This proof of the Pythagorean theorem depends upon a line having no area and a perspective ratio of 1:1. In the proof the area of contiguous triangles and a square is added and set equal to the area of the overall figure. In order for the area of contiguous figures to add to the area of the overall figure, there must be no overlapping. That means that the common border, a shared line, must have no area. But a line is made up of at least two points. And if points have a minimum extension, as would be the case under atomism, then the line they make up must also have a minimum area. Area, after all, is the square of extension. The area of a line cannot be less than the sum of the areas of its points. For the area of a line to actually be zero, the width of the line must be zero. But the width of a line is just its extension in another direction. For the width to be zero, extension must be zero, and that violates the presumption that there is a minimum extension.
Another way to look at this is by making an analogy between extended figures with area and extended figures with length. When a line segment is divided into shorter and shorter segments, the limiting case is a single point. If Infinite divisibility is presumed, then the limiting case has no extension; but if atomism is presumed, then the limiting case is reached after a finite number of divisions and has a minimum extension. Analogously, when an extended figure with area is divided into narrower and narrower widths, the limiting case is a single line. And this line cannot be "narrower" than the minimum atomic extension. In order for the limiting width to be zero, the division process must continue infinitely. Since a line is the limiting case of a plane figure which is being divided, a zero width is obtained only by presuming that this figure can be divided to infinity  by presuming infinite divisibility. In other words, a line whose width (and area) is zero is obtained only by presuming infinite divisibility. As a consequence, proofs of the Pythagorean theorem that do not take into consideration the area of shared points and lines cannot be valid under atomism.
In the atomic case, a line has a small but finite width, and hence a determinate area. This means that adjacent plane figures, figures which share a common boundary line, have overlapping areas, and the above proof of the Pythagorean theorem is therefore invalid in the atomic case; it does not account for these overlapping areas. The Pythagorean theorem is used to prove the incommensurability of the diagonal with the sides of a triangle. Without the Pythagorean theorem, this incommensurability cannot be shown.
In atomic metrics, points have a minimum extension, and consequently, have a minimum length, area, and volume. If a point is extended in one direction, it must also be extended in all other directions. If it weren't, the proposition that there is a minimum extension would be contradicted. Lines, which are made up of points, must also have a minimum length, area, and volume.
As noted in the last chapter, the length of two line segments, say of length A and B, which touch at a point is the sum of the individual lengths less the extension of the common point: A + B  1. The same principle holds for area and adjacent plane figures with a line of intersection as holds for length and lines with a point of intersection. Two plane figures which share a common boundary line have overlapping areas. The sum of the areas of two plane figures is more than the area they jointly cover by exactly the area of the line they share. The true (atomic) area of the combined figure is obtained by adding the areas of the contiguous figures and subtracting the area of their common border line.
Lines drawn in the atomic plane have a minimum area of two atomic units, since it takes two points to determine a line, but may have an area from two up to the "length" of the line; skew lines which pass "between" points are less "dense" than lines which hit the points exactly.
In general, the calculation of the area of a plane figure is not a simple matter. Moreover, determining how many points are shared by adjacent figures is no simple matter. Standard (continuous metric) mensuration formulas are based on lines having no area. Blithely using these standard continuous metric mensuration formulas when "refuting" "atomic" distances presumes infinite divisibility  begs the question  or worse yet, introduces an implicit contradiction.
The length of two lines which meet at a point is:
A + B  1
The area of two figures which share line C is:
D + E  C
For a square with a side whose length is divided into A and B the area is:
(A + B  1)^{2} = A^{2} + B^{2} + 2·A·B  2·A  2·B + 1
The area of a right triangle with sides B and H is computed by the formula:
A = INT((I1)·(H1)/(B1)+1)
This formula does not readily render itself into a simple relation using B and H. We can transform this formula into one which can be compared with the familiar formula in continuous metrics for the area of a triangle. For the purpose of this demonstration we will not need to make the transformation into a general formula that works for all atomic triangles. We will be showing that one particular special case of atomic triangles disproves the Pythagorean theorem, so we will only need an area formula for that special case. If the Pythagorean formula is to hold, it must hold in the special case we will be considering; if it does not hold in that special case, then it cannot hold in general. By limiting our examination to this special case, our task is greatly simplified.
The length of the sides of an atomic triangle is an integer. As such, the length is either even or odd; there are just 4 possible combinations. Let the base of the triangle be B and the height of the triangle be H. We will look at the special case when B is even. This will allow us to divide it by 2 and still have an integer. The advantage to doing this will become apparent.
Since B is even, there are an even number of terms in the sum, and half an even number is also an integer. We can divide the series into two halves and then add the first and last term, the second and the next to the last, etc., ending with adding the two middle terms. To convert these terms for a sequence of integers we would substitute an expression which evaluates to the number of terms when an indexing variable is set to 1. Such a conversion is accomplished by substituting B+1I for I. This allows us to sum to B/2.
INT((I1)·(H1)/(B1)+1) becomes:
INT((I1)·(H1)/(B1)+1)+INT((B+1I1)·(H1)/(B1)+1)
We want to get both INT terms of the expression into a form with as much similarity as possible; we want to combine like terms and remove quantities from under the INT function. As I perform a little algebra, I bold items of interest to make following the steps easier.
INT((I1)·(H1)/(B1)+1)+INT((B+1I1)·(H1)/(B1)+1)
By rearranging we get a '(B1)' which we can divide out.
= INT((I1)·(H1)/(B1)+1)+INT((B1I+1)·(H1)/(B1)+1)
Multiply through by '1/(B1)'.
= INT((I1)·(H1)/(B1)+1)+INT((B1)·(H1)/(B1)+(I+1)·(H1)/(B1)+1)
Cancel out (B1)/(B1).
= INT((I1)·(H1)/(B1)+1)+INT(H1+(I+1)·(H1)/(B1)+1)
Factor out a '1'.
= INT((I1)·(H1)/(B1)+1)+INT(H1(I1)·(H1)/(B1)+1)
Integer values may be removed from the INT function, so we do so.
= INT((I1)·(H1)/(B1))+1+H1+INT((I1)·(H1)/(B1))+1
Rearrange terms.
= INT((I1)·(H1)/(B1))+INT((I1)·(H1)/(B1))+1+H1+1
Simplify.
= INT((I1)·(H1)/(B1))+INT((I1)·(H1)/(B1))+H+1
We have reduced the expression to one which is the sum of the integer part of a number plus the integer part of its negation plus an integer  INT(Z)+INT(Z)+N.
Note that INT(X.0) = X, INT(X.0)=X and INT(X.Y) = X; INT(X.Y)=X1; so, INT(X.0)+INT(X.0)=0 and INT(X.Y)+INT(X.Y)=1.
The INT parts of the above expression is just 0 or 1, depending upon whether the expression (I1)·(H1)/(B1) evaluates to an integer or a number with a fractional part.
We are summing
INT((I1)·(H1)/(B1))+INT((I1)·(H1)/(B1))+H+1
from I=1 to B/2. So that we may compare our result with the familiar continuous formula for the area of a triangle, let us include the 1 with the INT expression and leave the H separate.
The INT parts of the expression evaluated to 0 or 1, so adding the 1 yields an expression which evaluates to 1 or 0. This leaves the H. Summing from I=1 to B/2 gives BH/2 plus the sum of the revised INT expression, which can be interpreted as just the amount by which the area of an atomic triangle is larger than a corresponding continuous triangle. We get:
BH/2 + INT((I1)(H1)/(B1))+INT((I1)(H1)/(B1))+1
Without a loss of generality we can assume that H B. By this hypothesis, H/B is either 1 or something less than 1. Since the final formula must hold for all special cases, it must also hold if H=B. To simplify our demonstration we let H=B.
If H=B then (H1)/(B1) is 1.^{(4)} With this special case we will be able to completely eliminate the INT function from the expression.
INT((I1)·(H1)/(B1))+INT((I1)·(H1)/(B1))+H+1 = INT(I1)+INT((I1))+H+1 = I1(I1)+H+1 = H+1
Pictorially we can show what we are doing; we are "chopping off" part of a triangle, rotating it, and fitting the parts together to make a rectangle as in figure 27.
H+1 = (H+1)B/2
The area is B·H/2 + B/2 (or B·H/2 + H/2, since B=H).
Having worked out a formula for the area of certain atomic triangles, let us consider the proof of the Pythagorean theorem for the special case when B=H and both are even.
Let us now examine the same diagram which purports to demonstrate the proof of the Pythagorean theorem, by using 6 overlapping plane figures, using an atomic metric. The area of the outer square is (A+B1)^{2} = A^{2} + B^{2} + 2·A·B 2·A 2·B +1. The area of a triangle is A·B/2 + B/2.
We will assume the area of the inner square is C^{2}. This assumption is not generally valid because atomic squares not aligned with the axes do not generally have square areas, as Figure 28 demonstrates.
There are some diagonal squares with square areas though. The illustration in Figure 33 below has an offset square with an area of 16, and the inner square in Figure 31 below has an area of 25.
The number of points on the diagonal of a triangle which are in common with a side of the inner square is B. This is true even of triangles with odd lengths. Figure 29 illustrates this.
The triangles overlap at the end points, so four points would be counted twice and must be subtracted. Figure 30 illustrates this. Adding the four triangle areas and subtracting the area of the overlapping points gives 4·(A·B/2 + B/2)  4.
The perimeter of the inner square consists of 4 lines, each of which overlaps with a triangle and has B points. But these lines intersect with each other at the vertices of the square, so the combined area of these 4 lines themselves is 4·B  4. We already have 4·(A·B/2 +B/2)  4 as the area of the four outer triangles, so we can add the area of the inner square, C^{2}, and subtract the area of the common boundary (4B  4), to get the combined total area. This gives us:
4·(A·B/2+B/2)  4 + C^{2}  (4·B  4)
Setting these two areas equal and performing algebra,
Outer square area = Four triangles + Inner square (A + B 1)^{2} = 4(AB/2+B/2) 4 + C^{2} (4B  4) A^{2} + B^{2} + 2AB 2A 2B +1 = 2AB + 2B  4 + C^{2}  4B + 4 A^{2} + B^{2} 2A 2B +1 =  2B + C^{2} A^{2} + B^{2} 2A =  1 + C^{2} A^{2} + B^{2} = + 2A  1 + C^{2} A^{2} + B^{2} = C^{2} +2A 1
Since A=B this reduces to
2A^{2} = C^{2} +2A 1
2A^{2} 2A +1 = C^{2}
The infinitely divisible Pythagorean equivalent would be 2A^{2} = C^{2}. Figure 31 is an illustration with A = B = 4.
Since 2A^{2} 2A +1 = C^{2} is satisfied for A=4 and C=5, we can interpret the fact that the diagonal line has only 4 points while C = 5 as meaning that the 4 points are "sparsely" distributed along the line that, were it aligned with the axis, would be "densely" populated with 5 points. Figure 32 shows a 5×5 square rotated 45 and superimposed upon a diagonal square with an area of 25 points. As you can see, they take up the same space as well as have the same area.
Let us look at atomic Pythagorean theorem demonstration for the simple case of A=3, B=4, and C=4^{(5)}  as shown in figure 33. The length of the outer square side is A + B  1, which is 3 + 4  1 = 6. The area of the outer square is 36. The area of a triangle with sides 3 and 4 is 7. The four triangles touch another at four vertices, so the combined area of the four triangles is 4 times 7 less 4 or 24. The area of the inner square is 16^{(6)}. But the outer triangles and the inner square share a common border which is only 4 points. Adding these areas and subtracting the area of the common border gives 24 + 16  4 = 36.
The Pythagorean formula fails for even this special case. It is unnecessary to go through the even more complicated examples for the other cases, since the damage could not be undone. The necessary condition for the Pythagorean theorem to hold is that the intersection between two lines be of zero area. But, for that to be true, points must have zero extension; for points to have zero extension, infinite divisibility must hold.
Loomis collects and presents 370 "proofs" for the Pythagorean theorem.^{(7)} These proofs all involve either the addition of lengths, or ratios of lengths. None of the proofs which involve the addition of lengths takes into consideration the area of the shared points. And none of the proofs which depend upon ratios makes any provision for integer results. Both conditions fail to yield the Pythagorean formula when atomic extension is considered.
Since the Pythagorean theorem does not hold in the atomic plane, it cannot be appealed to in any argument purporting to disprove atomism. Doing so begs the question.
Is there some a priori way to rule one way out as somehow not primary? That is, can one of the general views, atomism or divisionism, be ruled out by showing it to be flawed in some way? Over the millennia most philosophers have thought so. The pendulum has swung back and forth many times. And each swing usually followed some development in our way of viewing the world.
Prior to its expression atomism held sway in the form of Melanesian Monism. Everything was one. But monism implied that motion was impossible. So monism gave way to pluralism. But pluralism suffered from infinite regress. In a resurgence of monism, extension was claimed to be infinite. But the problem of motion remained, so space was invented. Now the problem of infinite divisibility can be applied to pure extension (without body). But each side continues to find flaws and questionbegging in the other side's arguments. At the same time we find a proliferation of mathematical tools on both sides. Measuring develops right along side counting.
Infinite sets correspond to many; singleton sets correspond to one. Either thesis (one or many) generates its antithesis (many or one), but both are required to synthesize a solution. In physics relativity theory shows the identity of matter and energy, and De Broglie developed an equation relating the wavelength of a particle to its momentum.
Modern developments in the philosophy of science show that any description we make of reality is at best a notyetdisconfirmed model, and while we can continually improve the model by testing and revising it, there is no a priori way to know when the final model is achieved. Moreover, we know that our present model of physics is not the final model because it does not include an account of gravity.
We also now have ample evidence that both points of view (atomism and divisionism) can reliably be used in various circumstances. Although there is as yet very little general familiarity with discrete metrics, modern computer display screens are becoming more and more common, and users who deal with computer graphics are becoming much more accustomed to the discrete metrics involved.
For some things, we use the perspective of infinite divisibility; for others, we use the perspective of atomism. Although incompatible with each other, these perspectives appear to be internally selfconsistent. I make an analogy using model theory. Each point of view is a language capable of describing "stuff" in a certain way. And either point of view may be taken. But as modern philosophy of science has shown, we have no access to the "stuff" except through the language we choose.
It has been said by that great philosopher Anonymous:
There are two kinds of people in the world  those who divide the world up into two kinds of people and those who do not.
That jest illustrates the tension that pervades our thinking. The two kinds illustrates counting and atomism while the opposition illustrates measuring and divisionism. We count and we measure, and the distinction between these pervades even our basic physical theory. Wave mechanics is contrasted with Quantum mechanics. Interference patterns in light are explained by choosing the divisionist perspective. The photoelectric effect is explained by choosing the atomist perspective. In our physical theory there is an equation to relate the two views. In mathematics we can use either to generate the other.
We may, in fact, reject the need to choose between these views, choosing each in its turn according to our needs and its efficacy in the use to which we intend to put it. Heraclitus may have said it first.
From out of all the many particulars comes oneness, and out of oneness comes all the many particulars.^{(8)}