CHAPTER VIII.. The Delphic Problem, or Duplication of the

THE DELPHIC PROBLEM, OR DUPLICATION OF THE CUBE.

A. Have you seen a printed paper sent from Paris, containing the duplication of the cube, written in French?

B. Yes. It was I that writ it, and sent it thither to be printed, on purpose to see what objections would be made to it by our professors of algebra here.

A. Then you have also seen the confutations of it by algebra.

B. I have seen some of them; and have one by me. For there was but one that was rightly calculated, and that is it which I have kept.

A. Your demonstration then is confuted though but by one.

B. That does not follow. For though an arithmetical calculation be true in numbers, yet the same may be, or rather must be false, if the units be not constantly the same.

A. Is their calculation so inconstant, or rather so foolish as you make it?

B. Yes. For the same number is sometimes so many lines, sometimes so many planes, and sometimes so many solids; as you shall plainly see, if you will take the pains to examine first a demonstration I have to prove the said duplication, and after that, the algebraic calculation which is pretended to confute it. And not only that this one is false, but also any other arithmetical account used in geometry, unless the numbers be always so many lines, or always so many superficies, or always so many solids.

A. Let me see the geometrical demonstration.

B. There it is. Read it.

TO FIND A CUBE DOUBLE TO A CUBE GIVEN:

Let the side of the cube given be V D. Produce V D to A, till A D be double to D V. Then make the square of A D, namely A B C D. Divide A B and C D in the middle at E and F. Draw E F. Draw also A C cutting E F in I. Then in the sides B C and A D take B R and A S, each of them equal to A I or I C.

Lastly, divide S D in the middle at T, and upon the centre T, with the distance T V, describe a semi-circle cutting A D in Y, and D C in X.

I say the cube of D X is double to the cube of D V. For the three lines D Y, D X, D V are in continual proportion. And continuing the semi-circle V X Y till it cut the line R S, drawn and produced in Z, the line S Z will be equal to D X. And drawing X Z it will pass through T. And the four lines T V, T X, T Y and T Z will be equal. And therefore joining Y X and Y Z, the figure V X Y Z will be a rectangle.

Delphic Problem. Vol. VII. Eng. p. 60. ]

Produce C D to P so as D P be equal to A D. Now if Y Z produced fall on P, there will be three rectangle equiangled triangles, D P Y, D Y X, and D X V; and consequently four continual proportionals, D P, D Y, D X, and D V, whereof D X is the least of the means. And therefore the cube of D X will be double to the cube of D V.

A. That is true; and the cube of D Y will be double to the cube of D X; and the cube of D P double to the cube of D Y. But that Y Z produced, falls upon P, is the thing they deny, and which you ought to demonstrate.

B. If Y Z produced fall not on P, then draw P Y, and from V let fall a perpendicular upon P Y, suppose at u.. Divide P V in the midst at a., and join a u.; which done a u. will be equal to a. V or a. P. For because V u. P is a right angle, the point u. will be in the semi-circle whereof P V is the diameter.

Therefore drawing V R, the angle u. V R will be a right angle.

A. Why so?

B. Because T V and T Y are equal; and T D, T S equal; S Y will also be equal to D V. And because D P and R S are equal and parallel, R Y will be equal and parallel to P V. And therefore V R and P Y that join them will be equal and parallel. And the angles P u. V, R V u. will be alternate, and consequently equal. But P u. V is a right angle; therefore also R V u. will be a right angle.

A. Hitherto all is evident. Proceed.

B. From the point Y raise a perpendicular cutting V R wheresoever in t., and then (because P Y and V R are parallel) the angle Y t V will be a right angle. And the figure u Y t V a rectangle, and u t equal to Y V. But Y V is equal to Z X; and therefore Z X is equal to u t. And u t must pass through the point T (for the diameters of any rectangle divide each other in the middle), therefore Z and u are the same point, and X and t the same point. Therefore Y Z produced falls upon P. And D X is the lesser of the two means between A D and D V. And the cube of D X double to the cube of D V, which was to be demonstated

A. I cannot imagine what fault there can be in this demonstration, and yet there is one thing which seems a little strange to me. And it is this. You take B R, which is half the diagonal, and which is the sine of forty-five degrees, and which is also the mean proportional between the two extremes; and yet you bring none of these proprieties into your demonstration. So that though you argue from the construction, yet you do not argue from the cause. And this perhaps your adversaries will object, at least, against the art of your demonstration, or enquire by what luck you pitched upon half the diagonal for your foundation.

B. I see you let nothing pass. But for answer you must know, that if a man argue from the negative of the truth, though he know not that it is the truth which is denied, yet he will fall at last, after many consequences, into one absurdity or another. For though false do often produce truth, yet it produces also absurdity, as it hath done here. But truth produceth nothing but truth. Therefore in demonstrations that tend to absurdity, it is no good logic to require all along the operation of the cause.

A. Have you drawn from hence no corollaries?

B. No. I leave that for others that will; unless you take this for a corollary, that, what arithmetical calculation soever contradicts it, is false.

A. Let me see now the algebraical demonstration against it.

B. Here it is:

Let A B or A D be equal to 2 Then D F or D V is equal to 1 And B R or A S is equal to the square root of 2 And D Y equal to 3 want the square root of 2 The cube of A B is equal to 8 The cube of D Y is equal to 45 want the square root of 1682 that is almost equal to 4 For 45 want the square root of 1681 is equal to 4

Therefore D Y is a little less then the greater of the two means between A D and D V.

A. There is I see some little difference between this arithmetical and your geometrical demonstration. And though it be insensible, yet if his calculation be true, yours must needs be false, which I am sure cannot be.

B. His calculation is so true, that there is never a proposition in it false, till he come to the conclusion, that the cube of D Y is equal to 45, want the square root of 1682. But that, and the rest, is false.

A. I shall easily see that A D is certainly 2, whereof D V is 1, and A V is certainly 3, whereof D V is 1.

B. Right.

A. And B R is without doubt the square root of 2.

B. Why, what is 2?

A. 2 is the line A D as being double to D V which is 1.

B. And so, the line B R is the square root of the line A D.

A. Out upon it, it is absurd. Why do you grant it to be true in arithmetic?

B. In arithmetic the numbers consist of so many units, and are never considered there as nothings. And therefore every one line has some latitude, and if you allow to B I, the semi-diagonal, the same latitude you do to A B, or to B R, you will quickly see the square of half the diagonal to be equal to twice the square of half A B.

A. Well, but then your demonstration is not confuted; for the point Y will have latitude enough to take in that little difference which is between the root of 1681 and the root of 1682. This putting off an unit sometimes for one line, sometimes for one square, must needs mar the reckoning. Again he says, the cube of A B is equal to 8; but seeing A B is 2, the cube of A B must be just equal to four of its own sides; so that the unit which was before sometimes a line, sometimes a square, is now a cube.

B. It can be no otherwise when you so apply arithmetic to geometry, as to number the lines of a plane, or the planes of a cube.

A. In the next place, I find that the cube of D Y is equal to 45, want the square root of 1682. What is that 45? Lines, or squares, or cubes?

B. Cubes; cubes of D V.

A. Then if you add to 45 cubes of D V the square root of 1682, the sum will be 45 cubes of D V; and if you add to the cube of D Y the same root of 1682, the sum will be the cube of D Y, plus the square root of 1682, and these two sums must be equal.

B. They must so.

A. But the square root of 1682, being a line, adds nothing to a cube; therefore the cube alone of D Y, which he says is equal almost to 4 cubes of D V, is equal to 45 cubes of the same D V.

B. All these impossibilities do necessarily follow the confounding of arithmetic and geometry.

A. I pray you let me see the operation by which the cube of D Y (that is, the cube of 3, want the root of 2) is found equal to 45, want the square root of 1682.

B. Here it is.

A DETECTION OF THE ABSURD USE OF ARITHMETIC AS IT IS NOW APPLIED TO GEOMETRY.

3————√2 3————√2 ——————— —√18 + 2 9—√18 ————————— 9—√72 + 2 3——√2 —————————————— ——√162 + 12——√8 27——√648 + 6 ——————————————————— 27—√658—√162 + 18—√8

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A. Why for two roots of 18 do you put the root of 72.

B. Because 2 roots of 18 are equal to one root of four times 18, which is 72.

A. Next we have, that the root of 2 multiplied into 2 makes the root of 8. How is that true?

B. Does it not make 2 roots of two? And is not B R the root of 2, and 2 B R equal to the diagonal? And is not the square of the diagonal equal to 8 squares of D V?

A. It is true. But here the root of 8 is put for the cube of the root of 2. Can a line be equal to a cube?

B. No. But here we are in arithmetic again, and 8 is a cubic number.

A. How does the root of 2 multiplied into the root of 72 make 12?

B. Because it makes the root of 2 times 72, that is to say the root of 144 which is 12.

A. How does 9 roots of 2 make the root of 162?

B. Because it makes the root of 2 squares of 9, that is the root of 162.

A. How does 3 roots of 72 make the root of 648?

B. Because it makes the root of 9 times 72, that is of 648.

A. For the total sum I see 27 and 18, which make 45. Therefore the root of 648 together with the root of 162 and of 8, which are to be deducted from 45, ought to be equal to the root of 1682.

B. So they are. For 648 multiplied by 162 makes 104976, of which the double root is 648 and 648 and 162 added together make 810 Therefore the root of 648, added to the root of 162, makes the root of 1458 Again 1458 into 8 is 11664. The double root whereof is 216 The sum of 1458 and 8 added together 1466 The sum of 1466 and 216 is 1682, and the root, the root of 1682

A. I see the calculation in numbers is right, though false in lines. The reason whereof can be no other than some difference between multiplying numbers into lines or planes, and multiplying lines into the same lines or planes.

B. The difference is manifest. For when you multiply a number into lines, the product is lines; as the number 2 multiplied into 3 lines is no more than 3 lines 2 times told. But if you multiply lines into lines you make planes, and if you multiply lines into planes you make solid bodies. In geometry there are but three dimensions, lengths, superficies, and body. In arithmetic there is but one, and that is number or length which you will. And though there be some numbers called plane, other solids, others plano-solid, others square, others cubic, others square-square, others quadrato-cubic, others cubi-cubic, &c., yet are all these but one dimension, namely number, or a file of things numbered.

A. But seeing this way of calculation by numbers is so apparently false, what is the reason this calculation came so near the truth?

B. It is because in arithmetic units are not nothings, and therefore have breadth. And therefore many lines set together make a superficies though their breadth be insensible. And the greater the number is into which you divide your line, the less sensible will be your error.

A. Archimedes, to find a straight line equal to the circumference of a circle, used this way of extracting roots. And it is the way also by which the table of sines, secants, and tangents have been calculated. Are they all out?

B. As for Archimedes, there is no man that does more admire him than I do: but there is no man that cannot err. His reasoning is good. But he, as all other geometricians before and after him, have had two principles that cross one another when they are applied to one and the same science. One is, that a point is no part of a line, which is true in geometry, where a part of a line when it is called a point, is not reckoned; another is, that a unit is part of a number; which is also true; but when they reckon by arithmetic in geometry, there a unit is sometimes part of a line, sometimes a part of a square, and sometimes part of a cube. As for the table of sines, secants, and tangents, I am not the first that find fault with them. Yet I deny not but they are true enough for the reckoning of acres in a map of land.

A. What a deal of labour has been lost by them that being professors of geometry have read nothing else to their auditors but such stuff as this you have here seen. And some of them have written great books of it in strange characters, such as in troublesome times, a man would suspect to be a cypher.

B. I think you have seen enough to satisfy you, that what I have written heretofore concerning the quadrature of the circle, and of other figures made in imitation of the parabola, has not been yet confuted.

A. I see you have wrested out of the hands of our antagonists this weapon of algebra, so as they can never make use of it again. Which I consider as a thing of much more consequence to the science of geometry, than either of the duplication of the cube, or the finding of two mean proportionals, or the quadrature of a circle, or all these problems put together.

FINIS.

DECAMERON PHYSIOLOGICUM;

OR,

TEN DIALOGUES OF NATURAL PHILOSOPHY.

THOMAS HOBBES

OF MALMESBURY.

TO WHICH IS ADDED

THE PROPORTION OF A STRAIGHT LINE TO HALF THE ARC OF A QUADRANT,

BY THE SAME AUTHOR.

DECAMERON PHYSIOLOGICUM.