What is Unity?
Table of Contents
When we seek the infinite among numbers we find it in unity.
That which is ever divisible is derived from indivisibles.
The vacuum is found inseparably connected with the plenum.
The views commonly held concerning the nature of these matters are so reversed that even the circumference of a circle turns out to be an infinite straight line.
Start with a straight line divided into unequal parts which bear to each other any ratio whatever.
Draw a circle such that 2 straight lines drawn from the ends of the given line to any point on the circumference will bear to each other the same ratio as the 2 parts of the given line.
This makes those lines which are drawn from the same terminal points homologous.
Let AB represent the given straight line divided into any two unequal parts by the point C.
The problem is to draw a circle such that 2 straight lines drawn from the terminal points, A and B, to any point on the circumference will bear to each other the same ratio as the part AC bears to BC, so that lines drawn from the same terminal points are homologous.
Around C as center draw a circle having the shorter part CB of the given line, as radius.
Through A draw a straight line AD which shall be tangent to the circle at D and indefinitely prolonged toward E.
Draw the radius CD which will be perpendicular to AE.
At B erect a perpendicular to AB.
This perpendicular will intersect AE at some point since the angle at A is acute.
Call this point of intersection E. From it, draw a perpendicular to AE which will intersect AB prolonged in F.
The 2 straight lines FE and FC are equal.
For if we join E and C, we shall have 2 triangles, DEC and BEC.
The 2 sides of the one, DE and EC, are equal to the 2 sides of the other, BE and EC.
Both DE and EB are tangents to the circle DB while the bases DC and CB are likewise equal.
Hence the 2 angles, DEC and BEC, will be equal.
Since the angle BCE differs from a right angle by the angle CEB, and the angle CEF also differs from a right angle by the angle CED, and since these differences are equal, it follows that the angle FCE is equal to CEF.
Consequently, the sides FE and FC are equal.
If we draw a circle with F as center and FE as radius it will pass through the point C.
Let CEG be such a circle.
This is the circle sought, for if we draw lines from the terminal points A and B to any point on its circumference they will bear to each other Fig 8 (46) the same ratio as the 2 portions AC and BC which meet at the point C.
This is manifest in the case of the 2 lines AE and BE, meeting at the point E, because the angle E of the triangle AEB is bisected by the line CE.
Therefore AC:CB = AE:BE.
The same may be proved of the two lines AG and BG terminating in the point G.
For since the triangles AFE and EFB are similar, we have AF:FE = EF:FB, or AF:FC = CF:FB, and dividendo AC:CF = CB:BF, or AC:FG = CB:BF; also componendo we have both AB:BG = CB:BF and AG:GB = CF:FB = AE:EB = AC:BC. Q.E.D.
Take now any other point in the circumference, say H, where the two lines AH and BH intersect; in like manner we shall have AC:CB = AH:HB. Prolong HB until it meets the circumference at I and join IF; and since we have already found that AB:BG = CB:BF it follows that the rectangle AB. BF is equal to the rectangle CB. BG or IB. BH. Hence AB:BH = IB:BF. But the angles at B are equal and therefore AH:HB = IF:FB = EF:FB = AE:EB.
It is impossible for lines which maintain this same ratio and which are drawn from the terminal points, A and B, to meet at any point either inside or outside the circle, CEG.
For suppose this were possible; let AL and BL be two such lines intersecting at the point L outside the circle: prolong LB till it meets the circumference at M and join MF. If AL:BL = AC:BC = MF:FB, then we shall have two triangles ALB and MFB which have the sides about the two angles proportional, the angles at the vertex, B, equal, and the two remaining angles, FMB and LAB, less than right angles (because the right angle at M has for its base the entire diameter CG and not merely a part BF: and the other angle at the point A is acute because the line AL, the homologue of AC, is greater than BL, the homologue of BC). From this it follows that the triangles ABL and MBF are similar and therefore AB:BL = MB:BF, making the rectangle AB. BF = MB. BL; but it has been demonstrated that the rectangle AB. BF is equal to CB. BG; whence it would follow that the rectangle MB. BL is equal to the (47) rectangle CB. BG which is impossible; therefore the intersection cannot fall outside the circle. And in like manner we can show that it cannot fall inside; hence all these intersections fall on the circumference.
It is impossible to resolve a line into an infinite number of points. It is easy to divide it into its finite parts.
This I will do under the following condition which I am sure, Simplicio, you will not deny me, namely, that you will not require me to separate the points, one from the other, and show them to you, one by one, on this paper; for I should be content that you, without separating the four or six parts of a line from one another, should show me the marked divisions or at most that you should fold them at angles forming a square or a hexagon: for, then, I am certain you would consider the division distinctly and actually accomplished.
If now the change which takes place when you bend a line at angles so as to form now a square, now an octagon, now a polygon of forty, a hundred or a thousand angles, is sufficient to bring into actuality the four, eight, forty, hundred, and thousand parts which, according to you, existed at first only potentially in the straight line, may I not say, with equal right, that, when I have bent the straight line into a polygon having an infinite number of sides, i.e., into a circle, I have reduced to actuality that infinite number of parts which you claimed, while it was straight, were contained in it only potentially?
Nor can one deny that the division into an infinite number of points is just as truly accomplished as the one into four parts when the square is formed or into a thousand parts when the millagon is formed; for in such a division the same conditions are satisfied as in the case of a polygon of a thousand or a hundred thousand sides.
Such a polygon laid upon a straight line touches it with one of its sides, i. e. , with one of its hundred thousand parts; while the circle which is a polygon of an infinite number of sides (48) touches the same straight line with one of its sides which is a single point different from all its neighbors and therefore separate and distinct in no less degree than is one side of a polygon from the other sides.
And just as a polygon, when rolled along a plane, marks out upon this plane, by the successive contacts of its sides, a straight line equal to its perimeter, so the circle rolled upon such a plane also traces by its infinite succession of contacts a straight line equal in length to its own circumference.
I am willing, Simplicio, at the outset, to grant to the Peripatetics the truth of their opinion that a continuous quantity [il continuo] is divisible only into parts which are still further divisible so that however far the division and subdivision be continued no end will be reached; but I am not so certain that they will concede to me that none of these divisions of theirs can be a final one, as is surely the fact, because there always remains “another”; the final and ultimate division is rather one which resolves a continuous quantity into an infinite number of indivisible quantities, a result which I grant can never be reached by successive division into an ever-increasing number of parts.
But if they employ the method which I propose for separating [93] and resolving the whole of infinity [tutta la infinità], at a single stroke (an artifice which surely ought not to be denied me), I think that they would be contented to admit that a continuous quantity is built up out of absolutely indivisible atoms, especially since this method, perhaps better than any other, enables us to avoid many intricate labyrinths, such as cohesion in solids, already mentioned, and the question of expansion and contraction, without forcing upon us the objectionable admission of empty spaces [in solids] which carries with it the penetrability of bodies.
Both of these objections, it appears to me, are avoided if we accept the above-mentioned view of indivisible constituents.