by Jonathan Tennenbaum
Contrary to British-authored mythologies, the intense interest on the part of Greek geometers from Pythagoras to Eratosthenes, in so-called “unsolvable problems” of geometry, had nothing to do with an idle fascination in mathematical puzzles. At issue, in the investigation of such problems as doubling a cube, trisecting an arbitrary angle, constructing a regular 7-sided polygon etc. was nothing less than the notion of Natural Law, as a higher principle subsuming an ordered series of {sets of physical principles}, each embodying a higher per-capita power of Mankind over the Universe.
In fact, there is no absolute “unsolvability” of the above-mentioned and other problems, except relative to a given, fixed set of principles of construction such as the ruler-and-compass constructions of so-called Euclidean geometry. Archimedes, Nichomedes and Eratosthenes and others already developed a whole array of “solutions” based on introducing addition principles, embodied in higher-order curves, constructions in higher dimensionalities, and the use of various physical machanisms and instruments.
The issue was not, that a given problem were “unsolvable” in some absolute sense, but rather that: 1) it could not be solved in the terms in which it had been posed, i.e. in terms of a certain implicitly circumscribed set of principles; 2) it {could} be solved with the help of the discovery and introduction of one or more {new} principles, lying {outside} the given domain, but demonstrated to be physically valid; 3) that the arrays of principles, arising this way, are implicitly ordered by a notion of Man’s increasing {power} over the Universe.
A simple illustration is the realization that a straight line, by and of itself, could never generate a surface. Nor could a surface ever transform itself into a solid. In both cases, a process of (rotational) extension is required, acting upon the line or the plane “from the outside”, and which already embodies the principle of the higher domain. This realization was the basis for the classical differentiation of geometrical problems, between so-called linear, plane and solid species; and for the notion, that the lower-order domain is always derived from the higher one, and not vice-versa.
However, by the time of Plato the Greek geometers had already conclusively established, that the actual ordering of “powers” is {not} that of simple dimensionality in visual space. The former lies beyond the reach of visual geometry per se, but actually determines the characteristics of action as reflected in visual space.
For example, the problem of {trisecting} any given angle in a plane, only {appears} to be a “two-dimensional” or “plane” problem. In actuality, as demonstrated by Nichomedes and others, it belongs to the same domain of “power” as the doubling of a cube!
That relationship is the original focus of the following discussion. It is key to understanding, why attempts by Cardan and others in the 16th century an afterwards, to derive an algebraic formula for the solution of a arbitrary cubic equation, inevitably ran into a devastating anomaly in the emergence of so-called “impossible”, “imaginary quantities”. The second installment of this discussion, will examine Cardan’s anomaly through the eyes of Abraham Kaester, setting the stage for Gauss’ subsequent discovery of the complex domain.
Trisecting an Angle
Since the origin of all angles is rotation, we reference all constructions to a circle whose center (marked “O”) is the vertex of the given angle, and whose radius is taken as “1”. The angle itself corresponds to the circular arc of rotation between the two points P and Q on the circle, where the sides (rays) of the angle intersect the circle.
Nothing is simper in visual geometry, than to double, triple, or multiply a given angle a whole number of times. We have only to set our compass to PQ and mark off a succession of equal arcs on the corresponding circle, starting at Q. Relative to OP, the rays OR, OS, OT, joining the center O to the endpoints of those successive arcs, represent double, triple, quaduple etc. the original angle between OP and OQ.
But what about the {inversion} of that process: to {divide} a given angle into a whole number of equal parts? Bisecting an angle is easily accomplished with the aid of ruler and compass, but the problem of dividing an arbitrary angle into {3} equal parts, presents a whole different kettle of fish. Centuries of attempts to develop a general solution within the domain of the ruler-and-compass constructions of plane geometry, ended in failure. Why? Key to this is the species-relationship to the doubling of the cube, known to the Greek geometers around the time of Plato.
Evidently, dividing the angle into three or any other number of equal angles, is equivalent to dividing the corresponding {circular arc} into the same number of equal arcs.
Now, it is a relatively easy matter to divide a straight line segment (hypothetical “zero curvature”) into three or any other number of equal segments, by ruler-and-compass constructions. Someone might, accordingly, attempt the following “solution” for trisecting a circular arc: First trisect the {chord} of the given arc; then {project} the division-points from the center onto the circular arc (see Figure 1a).
This attempt fails, for reasons Cusa and Leonardo emphasized in their discussions of the {distortion} introduced by any projection between a linear and curved surface. The projected images of the three equal segments on the chord onto the circle, are no longer equal as arcs.
Figure 1 |
Conversely, if we project the division-points of an already trisected arc onto the chord joining the arc’s endpoints (by drawing radial lines from the circle’s center to the division-points on the circular arc) we get an {unequal} division of the chord (see Figure 1b). Furthermore, the lengths of the resulting segments on the chord, taken in and of themselves, do not manifest any simple proportionality; rather — as it turns out — the attempt to express the convoluted relationships in algebraic form, inevitably leads to what are called “equations of the third (cubic) degree”.
Turn that around in your mind. Might it not be the case, that the appearance of complicated combinatorial-algebraic relationships among ordinary “scalar” magnitudes (whole numbers, “real numbers” measuring the lengths of straight line segments etc.) reflects the fact, that we are dealing, not with self-evident realities, but rather with “shadows”, cast from a higher physical-geometrical domain? Yes indeed, as Gauss demonstrated most conclusively in his work on biquadratic residues! The same principle underlies Gauss’ “fundamental theorem of algebra”, and its pre-history in the celebrated, centuries-long controversy over “imaginary numbers” appearing in the solution of the cubic equation.
That is where we are headed right now, along a trajectory defined by the ancient problem of trisecting a given angle.
Take a closer look, first, at the {inverse process} — {tripling} an angle — and at the lawful functional relationships, which are generated among selected “shadows” cast by that process.
To triple a given angle POQ, mark off points R and S from Q on the circle, such that the arcs QR and RS are equal to PQ. The rays OP, OQ, OR and OS are equally spaced, so that the angle between OR and OP is {double} that between OQ and OP, and the angle between OS and OP is {triple} the original angle.
For purposes of illustration it is best to represent P as the right-hand endpoint on the horizontal diameter of the circle. Rather than only considering a fixed angle, imagine the point Q as moving along the circle, starting at P (angle 0) and going around in a counterclockwise direction. Call the size of the angle POQ, “alpha”. What happens to the points R and S, as the alpha grows?
Clearly, for alpha = 0 all three points Q, R, S coincide with the position P. As alpha grows, Q moves at a proportional rate along the circle, while R runs ahead at twice, S at three times that rate. When Q reaches 90 degrees, R will have reached the position opposite to P at 180 degrees, and S will be at 270 degrees (see Figure 2).
Figure 2 |
Next, investigate the lawful relationships among the “shadows” cast by that process under various sorts of projections.
The simplest and most characteristic case, is perpendicular projection onto the horizontal diameter (axis) of the circle (the line through O and P).
Denote by q, r, s the perpendicular projections of the points Q, R, S, respectively onto the horizontal axis (see Figure 3a).
Figure 3 |
How do q, r, s move, as the point Q runs at a uniform rate along the circle starting at P?
Focus first on q. Imagine one end of a string is attached to Q on the circle (the latter oriented in a vertical plane), while the other end is attached to a lead bob, so that the string hangs vertically downward. Q’s projection q is the point where that vertical crosses the horizontal axis through O and P. (This works when Q lies on the upper half of the circle; when Q is on the lower half, we have to project “upward” by extending the direction of the string until we reach the axis.) Note, that the motion of the point q is not uniform like that of Q; rather, q starts very slowly (Q near P), and then accelerates, reaching maximum speed as the angle alpha approaches 90 degrees, and then slowing down again as Q approachs the point opposite P on the left of the circle (180 degrees). At that point, q reverses direction and repeats the process in reverse, as Q runs back to P along the lower half of the circle, and so forth (see Figure 4).
Figure 4 |
This will be familiar to anyone who knows the so-called trigonometric functions (better termed “circular functions”); The position of the “shadow” point q relative to O, corresponds to the so-called {cosine} of the angle alpha; and q’s motion has the form of a simple “harmonic” vibration whose frequency is the number of revolutions of the moving point Q around the circle, per unit time.
More entertaining, is to watch the {simultaneous} interrelated motions of q, r and s, as the latter two oscillate along the axis with frequencies {twice} and {three times} that of q! (see Figure 5).
Figure 5 |
Does there exist a mutual relationship among the positions of the three “shadows” q, r, s, taken by themselves, which remains valid and invariant throughout the process?
To answer that, “freeze” the motions for a moment, and pose the question again as follows: what direct relationships among q, r, and s flow from the circumstance, that the corresponding points on the circle — Q, R and S — were generated from P by rotation through the angles {alpha}, {2 x alpha} and {3 x alpha} respectively?
(For the purpose of illustration, it is best to take the case, where Q lies in the upper right-hand quadrant of the circle with the angle POQ less than 45 degrees. The other cases are essentially equivalent.)
Note the three {right triangles} OQq, ORr, OSs, each of which have hypotenus equal to unity (the radius of the circle), and whose angles at O are {alpha}, {2 x alpha} and {3 x alpha} respectively. How are those three triangles related? In effect, the successive rotations through the angle {alpha} on the circle, imply transformations of the triangle OQq into ORr and then into OSs. But generally speaking, the triangles are neither congruent, nor similar to each other in shape (see Figure 3a).
Noting the doubly and triply self-reflexive character of the action involved — i.e. an action applied to itself, and then to the result of that — it should occur to us, that the points R and S bear the same relationship to the axis OQ, as the points Q and R do to the axis OP.
This remark suggests that we consider the perpendicular projections or “shadows” of R and S on the axis OQ — call these r’ and s’, respectively — as well as the projections r and s on original axis OP (see Figure 3b [partial construction]).
Figure 3 |
Note, that the right triangles ORr’ and OSs’, arising from the new projection, are congruent to OQq and ORr respectively. In fact, the latter are carried into the former by exactly the same circular rotation through {alpha}, that carries Q into R and R into S. For the same reason, Or’ = Oq and Os’ = Or.
These observations now provide the key to unravelling the relationships between the “shadows” q, r and s.
The construction required is a bit laborious, but worth going through in detail, while keeping an eye out for the underlying bounding principle. For behind the following, nested chain of similarity relationships, lurks Gauss’ complex domain.
Algebraic Equations Arise Through Projection of Rotational Action
The point of departure for unravelling these relationships is the first triangle, OQq, and in particular its base and height — the segments Oq and Qq — whose lengths we shall call X and Y, respectively (see Figure 3a). The two are linked to each other through Pythagorus’ relationship: the sum of the squares of X and Y is the square of the radius of the circle, which we took as 1.
Now proceed as follows, concentrating first on q and r.
First project the point r’, lying on the axis OQ, down onto the original axis OP, obtaining r”. This is, so to speak, “the shadow of a shadow” (see Figure 3b).
We can obtain the position of r” quite easily, because the result of projecting from a straight line (in this case OQ) onto another straight line (OP) is to transform the distances along the line by a {constant factor}, as measured from the point of intersection of the two lines.
In our present case, we can determine the factor involved by comparing the length OQ, with the projected length Oq. OQ being of length 1 (Q lies on the circle), the factor is just the value of the length Oq, namely the quantity we have called X.
Since Or” is the projection of Or’, its length is X times that of Or’. The latter, as we already noted, is equal to Oq, whose length is X. So the length of Or” is X times X, or {X squared}!
So much for r”. How do we get from there, to r? We have to compare the {direct} projection of R from the circle down to the axis OP — which gives us r — with the “double” projection, from R onto r’ on OQ, and then from r’ to r” on OP.
The difference between the two arises from the fact, that the first step in the “double” projection occurs at an angle, which is “skewed” relative to the vertical direction of the other projection. I am talking about the angle betwen the vertical line Rr, and the segment Rr’. What is that angle? With a bit of reflection, you can see it is none other than {alpha}. For, Rr’ is perpendicular to the line OQ, which itself is “rotated” by alpha relative to the horizontal line OP. (To put it a different way, the triangle ORr’ is the result of rotating the triangle OQq around O by the angle alpha. In that process, the directionality of each of the triangle’s sides is changed by the same amount.)
The result of the “skew” projection that generated r’, is that its ensuring projection onto the horizontal axis will lie a certain distance to the right of the direct projection r. By how much?
Draw the perpendicular line segment from r’ to the vertical line Rr, and let r* denote the endpoint of that segment. Then r’r* will be parallel, and equal in length, to the segment between r” and r on the horizontal axis. (see Figure 3b)
Now observe that the triangle Rr’r* is {similar} to the original triangle OQq. Indeed: by construction Rr’r* has a right triangle at r*, while the angle at the vertex R, as we just saw, is alpha.
(Note also, that Rr’r* is rotated by 90 degrees relative to OQq. This, as we shall see later, reflects the action of the complex number “i”, lurking in the background of this whole construction!)
The similarity means that the the sides of Rr’r* are proportional to the corresponding sides of OQq, by a common factor of proportionality. Comparing the hypotenuses of the two, note that the hypotenus of OQq — the segment OQ — has length 1; while the hypotenus of Rr’r* — the segment Rr’ — is equal to Qq, the length of which we designated “Y”. The ratio is thus 1:Y, i.e. the factor of proportionality is Y.
In the similarity relationship between the triangles Rr’r* and OQq, length we are looking for — namely r’r* — corresponds to the side Qq of the triangle OQq, which again has length Y. So, to get the length r’r*, apply factor of proportionality Y to length Y. The result is YY, {Y squared}!
To get from r” to r, we thus have to move to the {left} by a distance {Y squared}. On the other hand, we found the length of Or” to be {X squared}.
Our conclusion: {r is located at distance [X squared – Y squared] from O along the horizontal axis}.
Here X stands for the length Oq, Y for the length Qq. Remember, that X and Y are linked to each other, as we noted above, by Pythagorus’ relationship XX + YY = 1 (Q lies on the unit circle). From this YY = 1 – XX, so that XX – YY = XX – (1 – XX) = 2XX – 1.
The result is to express the position of r {directly} as a function of q: The distance Or is equal to 2XX – 1, i.e. twice the {square} of the distance Oq, minus one.
Don’t miss the remarkable implication: the process of {doubling} the angle, by self-reflexively applying the rotation to itself, results in a {quadratic} relationship — i.e. one involving a {square} power — among the scalar “shadows” generated by that process!
Note also certain “topological” features of the relationship of X to 2XX – 1, that reflect the different rates of motion of Q and R along the circle, as the angle alpha grows. (Remember, X measures the segment Oq, not alpha directly). For example, X = 1 corresponds to the case alpha = 0, when P, Q, and R coincide. Sure enough, for this value of X, 2XX – 1 is also equal to 1. On the other hand, X = 0 corresponds to the case where Q lies at the top of the circle (alpha is 90 degrees); in this case, 2XX-1 = -1 and, sure enough, R lies {opposite} to P, at an a angle of 2 x alpha = 180 degrees.
Those skillful in geometry, will find no great difficulty applying entirely analogous methods, to determine the position of the {third shadow}, s — first in terms of r, and then in terms of q. It turns out that the distance Os is equal to 4XXX – 3X. Thus, {tripling} an angle results in {cubic} relationship among the corresponding “shadows” — i.e. one involving the {third power} of X. Hence, by inversion, the implicit relationship between {trisecting an angle} and constructing the {cube root} of a given quantity, which is the general form of the classical problem of doubling a cube.
In the following installment we shall derive the cubic relationships for the trisection of an angle from an improved and simplified standpoint, and then turn to the celebrated paradox of “Cardan’s formula”, as seen through the eyes of Gauss’ predecessor and teacher, Abraham Kaestner.
From Cardan’s Paradox to the Complex Domain, Part II
by Jonathan Tennenbaum
“In this place it will please us to speak of the great advantages of opening up the fountain of Transcendental Magnitudes , and discovering the reasons, why certain problems are neither plane, nor solid, nor of any other degree, but surpass all algebraic equations.” (Gottfried Wilhelm Leibniz, 1686).
We began the first installment of this series, by recalling the physical-geometrical “powers” investigated by the Greeks, and the unexpected emergence of higher “powers” in connection with what appeared to be a straightforward problem of plane geometry: to divide a given angle or circular arc into two, three or a larger number of equal parts. In the following discussion we shall nail down that relationship, in particular deriving the cubic equation corresponding to the trisection of a given angle, and thereby revealing the inseparable relationship with the doubling of a cube.
The same relationship will become clear, when we later examine the implications of the self-similar spiral, itself a reflection of Gauss’s complex domain. On such a spiral, tripling a given angle of rotation, translates into taking the cube (third power) in terms of the corresponding ratio of radial distances. However, the self-similar spiral itself provides neither the means to trisect an arbitrary angle, nor to construct a cube of a given volume. However we shall see later, that both those ancient problems — and countless others — can easily be solved the using the higher principle or “power” embodied in the catenary .
The `Complex’ Composition of Angles
In the last installment we found an algebraic relationship between the scalar “shadows” generated by the doubling of an angle, in terms of the corresponding arcs on a circle of unit radius. We identified the center of the circle as “O” and called the right-hand endpoint of a chosen diameter (taken as the horizontal) “P”; also, we denoted by Q and R the points on the circle, corresponding to the angles {alpha} and {2 x alpha}, respectively, measured as rotations around O relative to the horizontal ray OP. Finally, we denoted the perpendicular projections of Q and R onto the horizontal diameter q and r respectively.
Our analysis showed, that the distance Or is equal to twice the square of the distance Oq, minus one. In other words: the position of r relative to O is given by 2XX – 1, where X measures the corresponding position of q. (See discussion on the meaning of negative values of these parameters, toward the end of this installment.)
Now I want to take on the case of tripling the angle {alpha}!
Recalling last week’s discussion, let S denote the point on the circle, corresponding to the angle {3 x alpha}, and let s be the projection of S onto the horizontal diameter. How is s related to q and r?
As a bit of reflection shows, the answer is already implied by what we did last time, to analyze the relationships for doubling an angle. Reworking the essential steps here again, if possible with an actual physical model or corresponding animated diagram as reference, should cause the principle involved to “leap out” at the reader (see Figure 6).
Figure 6 |
We started from the right triangle OQq, whose horizontal and vertical sides we called X and Y, respectively. We noted that the point R arises from Q, by rotating Q through the angle {alpha}. Applying that rotation to the whole triangle OQq, yields a congruent triangle ORr’, where r’ marks the perpendicular projection of R onto the axis OQ. At the same time, that rotation generates a new right triangle: ORr, whose third vertex r is the projection of R onto the original horizontal axis OP. Our analysis of the relationship of the triangles, showed that the horizontal side of the latter triangle, Or, had length equal to XX – YY. (Using Pythagorus’ relation between X and Y, we found XX – YY to be equivalent to 2XX – 1.)
Now, to get the point S corresponding to tripling the original angle, it is enough to rotate R — itself obtained by doubling the angle alpha — through the same angle once again. Apply that rotation to the whole right triangle ORr. Observe, that the resulting relationships have essentially the same form as the earlier case, when we rotated the triangle OQq through {alpha}, to obtain the result of doubling the angle. The only difference is, that the angle of the triangle ORr at O is not {alpha}, but {2 x alpha}. But our earlier analysis did not really depend on any special assumptions concerning the shape of the right triangle being rotated, but only the angle of rotation itself ({alpha} and the parameters X and Y connected with alpha).
Accordingly, suppose we have an arbitrary right triangle ORr with hypotenus 1 (i.e. R lying on the given circle) and its side Or lying along the horizontal axis OP. Call its angle at O “{beta}”, and the lengths of its horizontal and vertical sides “A” and “B” respectively (see Figure 7a).
Now rotate ORr by the angle {alpha} around O. The vertex R is carried to a point S, the which, relative to the original point P, corresponds to an angle of {alpha} + {beta}. Imagine a weighted string attached to R and hanging down vertically as the triangle ORr rotates. Observe how the angle, between that vertical and the triangle’s side Rr, grows, as the rotation progresses; in fact, that angle will be equal to the angle of rotation itself (see Figure 7b).
Figure 7 |
After completing the rotation of the triangle ORr through the full angle {alpha}, R is carried to the point we called S, and r to a point s’, corresponding to the projection of S onto the axis OQ. The vertical string attached to R (now at position S) hits the horizontal diameter at the point s, creating the new right triangle OSs (see Figure 8).
Figure 8 |
We can now unfold relationships entirely analogous to the ones we found for the doubling of the angle {alpha}, but which now apply to the generalized case of the sum {alpha} + {beta}. To wit:
Let s” and s* be the perpendicular projections of s’ onto the horizontal diameter and onto the vertical line Ss, respectively. As we noted in last week, the first projection changes lengths by a factor X; since the length of the segment Os’ is the same as that of Or, which we called “A”, the projection of Os’ — i.e. Os” — will have length X x A.
Observe, in addition, that in virtue of the process of rotation which generated an angle {alpha} between the vertical at S and the line Ss’, the right triangle Ss”s* will be similar to the original triangle OQq. At the same time, the hypotenus of that triangle, Ss’, is congruent to Rr, whose length we called “B”. Since the original triangle’s hypotenus is 1, the factor of proportionality must be B. As a result, the horizontal side s’s**, which corresponds to the vertical side Qq of OQq, has length B x {length of Qq} = B x Y. On the other hand, that distance is the same as the gap between s and s” on the horizontal axis. Since s” lies to the right of s, we must subtract that distance from Os” — whose length we just found to be X x A) — in order to obtain the length Os.
The result of this chain of relationships is: length Os = {X x A} – {Y x B} That was the horizontal side of the triangle OSs. With a little extra effort, we can also find its vertical side. The latter, Ss, is divided by s* into the two segments Ss* and s*s. The first of them, which coincides with the vertical side of the triangle Os’s*, is proportional to the horizontal side of the original triangle OQq (length X), by factor of proportionality B. So, length of Ss* = B x X The second segment s*s, is parallel to and equal in length to the vertical segment s’s”. Note, that the points O, s’, s” form a right triangle which is slightly smaller than, but similar to the original triangle OQq. The factor of proportionality is the hypothenus of the former triangle, namely Os’, which is equal in length to Or = A. As a result, the length of the side s’s”is A times the corresponding side of the original triangle, namely Qq = Y. So, length of s*s = length of s’s” = A x Y.
Putting these results together, we find: length of Ss = length Ss* + length s*s = {B x X} + {A x Y}. Summing up: the lengths of the horizontal and vertical sides of the right triangle, generated by the angle {alpha} + {beta}, are {X x A} – {Y x B} and {X x B} + {Y x A} respectively, where X,Y and A, B are the sides of the right triangles corresponding to {alpha} and {beta} (see Figure 7b).
Notice, that the horizontal and vertical sides of the triangle for the sum or composition of the two angles {alpha} and {beta}, each involve all of the four values X, Y, A and B. This “complex” interwining of parameters is merely the algebraic “shadow” of the physical process of combining two rotations .
The Cubic Equation for the Trisection of an Angle
At this point we can easily derive the relationships resulting from tripling an angle, and invert these to obtain the third degree algebraic equation corresponding to trisecting an arbitrary angle.
First, what happens when we apply our “composition formula” to doubling a given angle? In this case {beta} is the same as {alpha}, A = X, B = Y, and the horizontal and vertical sides of the triangle for {2 x alpha} come out as {X x X} – {Y x Y} and {X x Y} + {Y x X} respectively. The first one, XX – YY, we had before; and now we have 2XY as the second side.
Now take that triangle, and rotate it by alpha again. In this case {beta} is the double of {alpha}, and A = XX – YY, B = 2XY, as we just found. The result of the composition formula is now a bit more “hairy”, but lawfully so. For our present purposes we only need the horizontal component, which expresses the position of the “shadow”-point s: {X x (XX – YY)} – {Y x (2XY)} = XXX – XYY – 2YXY = XXX – 3XYY. Recalling Pythagorus’ relation XX + YY = 1, YY = 1 – XX, we can express the latter magnitude in terms of X alone: XXX – 3XYY = XXX – 3X(1 – XX) = XXX – 3X – 3XXX = 4XXX – 3X This is the result I announced at the end of last week’s discussion. The position of the “shadow”-point s, resulting from tripling the angle {alpha}, is related to that of the point q, corresponding to {alpha}, as follows: The length Os is equal (in scalar magnitude) to 4 times the cube of Oq, minus three times Oq.
Thus, tripling an angle, is reflected in an essentially cubic or third-power algebraic relationship among the corresponding “shadows”!
Recall the relationships for doubling an angle, involved a square or second-power relationship among the shadows. Thus, the linear, plane and solid geometrical “powers” of Classical geometry, seem to be subsumed within the process of successive transformations of rotation by once, twice, three times an arbitrary angle.
But, what is to prevent us from applying the composition of rotations once again, to obtain analogous relationships for 4, 5 or any other whole-number multiple of an angle? Evidently, each time we apply the rotation {alpha} we increase by one the dimensionality or degree of the corresponding algebraic relationship in the domain of the “shadows”. In this sense, the circular rotation subsumes and transcends all those algebraic dimensionalities. And by the way, didn’t Nicolaus of Cusa refer to the circle as reflecting a higher principle, bounding the linear (algebraic) world of the regular polygons? The latter constitute, of course, special cases of the whole-number division and multiplication of angles.
But returning to our cubic relationship, two remarks are in order.
First, someone who has not been utterly brainwashed by high school or college algebra courses, might rightfully object to subtracting what appears to be a simple one-dimensional magnitude — 3 times the length Oq, i.e. 3X — from the cubic or three-dimensional magnitude 4XXX. Such a subtraction would be plainly absurd; evidently sort of error or foul play has occurred!
Looking a bit closer at what we did, however, reveals, that the multiplier “3” in the above expression does not signify a simple linear magnitude. Indeed, if you check back, you will find that this “3” originated from the “1” in Pythagorus’ relationship XX + YY = 1. That 1, however, signifies the square of the hypotenus of the right triangle OQq, i.e. a two-dimensional magnitude . Thus, the magnitude “3X” is actually a magnitude of “cubic” or third order, while being at the same time proportional to X. (Much more could be said about this matter, under the rubric of the devastating fallacies arising from belief in the supposed self-evidence of “simple numbers”.)
Secondly, we implicitly assumed, in our analysis of the relationships among q, r, and s, that the points Q, R, S all lay in the same, upper right quadrant of the circle; we also spoke of “lengths” always as positive magnitudes. Whereas, as the angle {alpha} grows, the points R and especially S, race ahead of Q, and can come to lie on opposite sides and different quadrants of the circle. In these cases, the layout of the triangles and projections, upon which our derivation depended, change (see Figure 4). At the same time, note that both 2XX-1 and 4XXX-3X can take on nominally negative values, as for example when X = 1/2. What is the significance of those negative values?
Gauss himself, as well as Lazard Carnot in his famous book on the “Geometry of Position”, devoted careful attention to this question, which is closely related to the analysis situs origin, not only of the negative numbers, but also of the so-called “imaginary” numbers. The following should suffice to identify the essential point:
Real physical magnitudes — as opposed to mathematical fictions — are never “indifferent,” but are invariably associated, at least implicitly, with a notion of directionality or orientation in the Universe. “Negative numbers” arise very simply, in connection with the notion of reversal of direction or orientation. Indeed, when for example the point “r” (corresponding to the angle {2 x alpha}) crosses over to the left of the midpoint O — the which occurs at the moment {alpha} hits 45 degrees, and X becomes less than the corresponding value, namely 1/sqrt(2) — the “length” Or reverses its direction. Exactly at that point, indeed, the value of 2XX-1 hits zero and becomes negative .
Thus, the so-called “rules” of algebraic operations with negative numbers, are no mere conventions or arbitrary inventions of so-called “pure mathematics”; on the contrary, they are determined by the geometrical characteristics of rotational action . When those characteristics are taken into account, and when the differentiation of positive and negative values for the “lengths” Oq, Or, Os etc. correspond to the distinction between “right and left” relative to the chosen origin O, then the expressions 2XX – 1 and 4XXX – 3X etc. turn out to be valid for all values of the angle {alpha}.
To explore these relationships, graph the cubic function represented by 4XXX – 3X. Note, that the horizontal coordinate X of the graph corresponds to the position of the point “q”, whereas the vertical coordinate (with value 4XXX – 3X) corresponds to the position of the point “s”. (Keep this separate from the representation of motion on the circle, to avoid confusion!) Imagine the overall form that curve must have, to represent the relative motions of q and s, as {alpha} increases. Next, explore the graph numerically, by calculating the value of 4XXX – 3X for a variety of values of X between -1 and 1, noting the points of reversal of direction and their significance in terms of the interrelation of Q, S and q, s.
So far we have been focussing on tripling a given angle. What about trisecting an angle? For that case, the point S is given, and we have to find the point Q, such that S is the result of tripling the rotation from P to Q. This is evidently equivalent to determining a value of X, such that 4XXX – 3X is equal to the length Os (taking account of +/- sign), where s is the projection of the given point S onto the horizontal diameter. For, once we have the value X, we know the position of Q’s projection q on the diameter; then Q can be constructed as one of the two points of intersection of the perpendicular at q with the circle.
Thus, trisecting an arbitrary angle corresponds to solving the cubic equation 4XXX – 3X = c, where c (corresponding to the position of s on the axis) can assume any value between -1 and 1.
In terms of the graph of 4XXX – 3X, this means finding the intersection-point between that cubic curve, and a horizontal line at height “c” parallel to the X-axis. But, wait a minute! Given the “looping” form of the curve, there will be not just one, but in general three different points of intersection! What do they signify? How could there be more than one solution to trisecting an angle? And what about the doubling of a cube, which corresponds to the cubic equation XXX – 2 = 0? Could there exist three different cubes, having the same volume?
Part III takes us from the birth of “algebra” in the famous “Hisab al-jabr w’al maqalaba” of the 9th-Century Baghdad astronomer ibn Al-Khwarizmi, to Cardan’s paradox and its discussion by Kaestner.
From Cardan’s Paradox to the Complex Domain, Part III
by Jonathan Tennenbaum
As we have emphasized, the subject of Gauss’ “Fundamental Theorem of Algebra” — ostensibly the solution of algebraic equations of arbitrary degree — goes back to very long before the emergence of what came to be known as “algebra”, to the discussions emanating from the Pythagoreans and continued in the circles of Plato, on the general notion of physical magnitude . It is exactly that line of development, which culminated via Gauss’ breakthroughs in Riemann’s conception of magnitude as a self-developing multiply-extended manifold, where “extension” signifies the differential action of generation and integration of a new principle of physical action into the ongoing process.
The very nature of Riemannian physical action is such, that it generates an increasing density of discontinuities or other sorts of anomalies relative to any attempted formal representation “projection” of the action involved.
Look at algebra from this standpoint, and all fearful mysteries dissolve into pure fun. That’s what we shall pursue now, in examining the devastating anomalies which developed within algebra itself in connection with the celebrated “Cardan’s rule” for the solution of cubic equations, and which was played a central role in the disputes which culminated to Gauss’ 1799 refutation of Euler and Lagrange on the so-called “imaginary numbers” and the heredity defects of formal-algebraic method.
The origin of “algebra”
According to available accounts, the term “algebra” derives from the Arabic word “al-jabr”, signifying “completion” or “healing”. The term became current through a famous Arabic mathematical treatise, the “Hisab al-jabr w’al maqalaba”, composed by the astronomer and mathematician Abu Ja’far Muhammad ibn Al-Khwarizmi (approx. 780-850). Al-Khwarizmi worked together with Al-Kindi and others at the “House of Wisdom” in Bagdad, founded by the son of Harun al-Rashid as a center of learning. There, alongside original investigations and writings, classical Greek scientific manuscripts were collected and translated into Arabic. The “Hisab al-jabr w’al maqalaba”, later became widely used in Europe in Latin translation, transmitting the Indian (decimal) system of arithmetic and now-familiar methods for the rearrangement and solution of equations, which Al-Khwarizmi called “completion” (al-jabr) and “balancing” (al-muqabala).
Of particular historical influence was Al-Khwarizmi’s method of solving quadratic equations, such as XX + 10X = 39 . He did not express this with letters, as later became commonplace, but posed the problem instead this way: “a square plus ten of its roots is 39 units. Find such a square.”
How? Draw any square to represent, in hypothetical manner, the square we are looking for, with unknown side (“root”) X. Then “ten of those roots”, constitutes corresponds to a rectangular area with sides 10 and X. Added together the square and rectangular areas are supposed to give a total area of “39”; but there is no evident way to combine the two areas, in such a way, that the value of the square will be evident. Al-Khwarizmi proposes the following “remedy” (al-jabr!):
Divide the rectangle into 4 equal rectangles, by cutting the it parallel to the “X” side, into 4 rectangles of sides X and 10/4 (= 5/2, or 2 1/2). Now, arrange these four rectangles alongside the four sides of the square XX (you have to draw this to follow the argument!). The resulting figure is nearly a square — all that is missing, is four “corners”! “Fix” the defect by adding four squares, each 5/2 by 5/2. The result is a single, bigger square, whose sides are 5/2 + X + 5/2 = X + 10/2 = X + 5. The area of the big square will be the SQUARE of X + 5. On the other hand, the area of each of the four supplementary squares, is (5/2)(5/2) = 25/4, so the total area supplied was 4 time 25/4, i.e. 25.
Adding that same amount to the right side of the original equation, Al-Khwarizmi finds: the area of the big square, namely {X + 5 SQUARED} is 39 + the added area of the four “corners”, i.e. 39 + 25 = 64. Thus the square of X + 5 is 64, and thus X + 5 and X = 8 – 5 = 3. In fact, we can easily verfiy that the value X = 3 does indeed solve the equation XX + 10X = 39. The square Al-Khwarizmi demanded is the 3 x 3 square of area 9.
Some people may recognize in Al-Khwarizmi’s method for “completing the square” the origin of the famous formula for the solution of a general quadratic equation, which students are mechanically drilled in, without ever encountering the simple geometrical idea behind it.
(Indeed, a typical feature of the sadistic “New Math” educational reforms, pushed through in the 1960s, was to systematically suppress the geometrical underpinning of algebra. This included obfuscating the commonplace “rules” for what Abraham Kaestner called “Buchstabenrechnung” — calculation with letters representing unknown or hypothetical magnitudes –, the which became a prominent technical feature of the development of algebra in the 15th century and afterwards. Consider, for example the formula (A + B) x (C + D) = AC + AD + BC + BD.
The “New Math” typically presents this as a deduction from the so-called “associative law” of addition and multiplication. But the origin of the formula is geometrical : it simply describes the division of the rectangular area with sides A+B and C+D, by corresponding perpendiculars to those sides, drawn at the points that divide them into lengths A,B and C,D respectively. The result is four rectangles of sides A,C; A,D; B,C and B,D respectively. Similarly, the equation:
(A + B) squared = A squared + 2AB + B squared describes the special case of the division of a square of side A+B into two squares, with sides A and B respectively, and two rectangles with sides A,B. There do arise some interesting subtleties and paradoxes, when the magnitudes involved take on negative, imaginary or some other species of values, and are no longer assumed to be simple lengths, as we shall discuss below. But by depriving students of even the simplest, visual-geometrical image of these relationships, the perpetrators of the “New Math” fraud also blocked the pathway to the deeper physical principles that underlie both algebra and the geometry of visual space. In fact, Bertrand Russell’s “New Math” represents nothing but a revival of the worst features of Euler and Lagrange’s formalist method — exactly the method Gauss refuted in his first paper on the “Fundamental Theorem of Algebra”!)
Applied to a general quadratic equation of the form
XX + BX + C = 0 where B and C represent any arbitrary choice of parameters, Al-Khwarizmi’s approach yields the following result. First, we make XX + BX into a square, by adding B/2 squared = BB/4, exactly as we did for the particular case of XX + 10X above. The general case takes the form:
XX + BX + BB/4 = (X + B/2) squared
Accordingly, add BB/4 to both sides of the quadratic equation, and apply “al muqabala” — the art of “balancing” or shifting the components of an equation between the sides, in order to “box in” the value of the unknown X. To wit:
XX + BX + C = 0
XX + BX + (BB/4) + C = BB/4
(X + B/2) squared + C = BB/4
(X + B/2) squared = BB/4 – C
X + B/2 = square root of ( BB/4 – C )
X = – B/2 + SQRT( BB/4 – C )
The last in this chain of relationships is essentially nothing but the famous formula for the solution to the general quadratic equation XX + BX + C = 0 (see footnote), and the precursor to Cardan’s attempted solution to the general cubic equation.
Al-Khwarizmi’s quadratic species
But, beware the algebraicist’s sleight-of-hand! A number of subtleties and paradoxes lie buried beneath the apparently routine procedure we just went through, to “solve” the quadratic equation.
Al-Khwarizmi distinguished between at least four different species of “quadratic” problems. For example:
1) XX + 10X = 2
2) XX = 10X + 2
3) XX + 2 = 10X
4) XX + 10X + 2 = 0 each representing a distinct sort of geometrical relationship.
In the first case, we seek “a square which, when combined with the rectangle whose sides are the square’s root and 10, respectively, gives a total area of 2”. We discussed a problem of this sort already above (with 39 instead of 2).
In the second case we seek “a square whose area is 2 units more than the area of the rectangle whose sides are the square’s root and 10, respectively.” At first glance it might not be clear at all, from the geometrical picture, how to apply the method of “completing the square”. On the other hand, changing the “balance” by shifting the rectangle 10X to the other side of the equation, we can put it in a form apparently similar to the first case. In fact, from the standpoint of determining the unknown X, XX = 10X + 2 is equivalent to XX – 10X = 2.
Compared with Al-Khwarizmi’s construction for the first case, we run into a significant difference: this time we have to take away the area of the rectangle 10X from the square, rather than adding it. After cutting the rectangle into four equal subrectangles with sides 10/4 and X and trying to fit them in an analogous way into the square XX — in order to “take them away” — we find that they overlap at the corners (make the drawing, to see what I am getting at!) Thus, taking away the four areas from the square would mean to remove each of the four corners twice . But how can one “take away” an area twice, from the same place? After we have removed it once, it is gone ; and to remove the same area again would mean taking something from nothing !
Al-Kharizmi scrupulously avoided such “impossible” operations. In fact, if you add in four extra little squares of sides 10/4, one at a time, at the appropriate moments, interspersed between removing the four rectangles one at a time, you can circumvent the difficulty. The result of a suitably ordered sequence of adding the four square areas and taking away the four rectangles is a smaller square, whose side is X – 10/2 = X – 5. From this point on, the solution proceeds entirely analogously to the first case.
The third case is more tangled-up still, if we hold to a simplistic visual-geometrical conception. Indeed, when we try to rearrange the equation XX + 2 = 10X in order to be able to “complete the square”, we get XX – 10X = – 2.
The equation demands, explicitly, that we take away a bigger area from a smaller , to get a negative area — clearly an “impossible” proposition! Here again, the difficulty can be circumvented by adding the four squares (10/4 x 10/4) to both sides of the original equation, before proceeding further.
The fourth case, XX + 10X + 2 is truly “impossible” in Al- Khwarizmi’s sense. Areas are by their very nature positive magnitudes . The sum of a square (XX) and a rectangle (10X) cannot be less than zero. Hence XX + 10X + 2 cannot be less than 2, and certainly never equal to zero.
All of this indicates quite clearly, that, to the extent the famous “general formula” for the solution of a quadratic equation XX + BX + C = 0 is valid at all, it must presuppose the existence of a different domain , distinct from that of simple visual geometry, but coherent with the latter.
Kaestner on “negational magnitudes”
At first glance, that domain is distinguished by the introduction of negative numbers , which have no self-evident interpretation in terms of lengths, areas and volumes, and were long branded as “impossible” for the indicated reasons. In fact, this issue was first fully cleared up by Gauss himself, in the context of his conceptualization of the physical principle underlying the “imaginary numbers”.
In his 1758 “Anfangsgrnde der Mathematik”, which was a standard reference for the teaching of mathematics at the time Gauss began his studies at the Carolineum in Braunschweig, Abraham Kaestner introduced negative numbers in the following manner:
“Opposing magnitudes are magnitudes of the sort, that arise through consideration of such conditions, in which one magnitude reduces another — for example assets and liabilities, forward and backward motion, etc. One of the magnitudes, whichever one chooses, is called positive or affirmative; the opposite is called negative or negational.”
Kaestner proceeds to develop the arithmetic of such “opposing magnitudes” as a new domain subsuming simple “positive” magnitudes together with their “opposites”. Keastner demonstrates, among other things, why the product of two “negational magnitudes” must necessarily yield a “positive magnitude”. For example, multiplying by -1 transforms a magnitude into its opposite, so that (-1) x 2 is the opposite of 2, i.e. -2; while (-1) x (-2) is the opposite of the opposite of 2, i.e. 2 itself. As Kaestner emphasizes, these relationships apply only to such magnitudes, as admit of a unique opposite , as for example in the case of forward and backward motion along a pathway.
Relative to such a domain of “opposing magnitudes” — as typified by the so-called “real number line” — Al-Khwarizmi’s four species of equations can all be subsumed under a common form. Indeed, in the new domain the species 1) – 4) above are equivalent to:
1) XX + 10X – 2 = 0
2) XX – 10X – 2 = 0
3) XX – 10X + 2 = 0
4) XX + 10X + 2 = 0 respectively, all of which fall under the form XX + BX + C = 0, where B and C can take on both positive and negative values. At first glance, it would seem that the formula derived above, namely X = – B/2 + SQRT( BB/4 – C ) indeed provides a general solution to the quadratic equation, subsuming each of the four cases distinguished by Al-Khwarizmi. The operations he regarded as “impossible” or absurd from the standpoint of lengths, areas and volumes — such as “subtracting a larger from a smaller” — have a perfectly determinate meaning in the domain of forward and backward motion along a line. For example, subtracting a forward motion of 5 from a forward motion of 2, results in a backward motion of 3, i.e. -3, and so forth. The chain of relationships, by which we derived the general solution to the quadratic equation XX + BX + C, are valid irrespective of the values of B and C.
In particular, the general formula provides a solution to the equation XX + 10X + 2 = 0, which Al Khwarizmi regarded as “impossible”: X = -10/2 + SQRT( 100/4 – 2) = – 5 + SQRT (23) = (approx) -2.0416848, a negative magnitude.
The fallacy of closure
Have we invented a “perfect system”? A closed domain in which the algebraicist, sitting in a room with no windows, can solve all problems at the blackboard, without any need to heed the real universe outside?
On the contrary! In fact, our derivation of the “general solution” glossed over an even more devastating paradox, than the emergence of the negative numbers which Al-Khwarizmi and many of his successors treated as “impossible”. Look back at the next-to-last step in the “general solution”:
(X + B/2) squared = BB/4 – C
X + B/2 = square root of ( BB/4 – C )
Contrary to common misconceptions, the algebraic expression, “square root of,” constitutes a question , not an answer ! We are asked to find a magnitude, whose square is the given magnitude. Just because the algebraicist has invented a formal symbol in place of answering the question, does not make it any less a question! Although there exists a simple ruler-and-compass construction for determining the square root of any positive quantity, the analogous problem for cube roots leads, as we indicated in the preceeding installments, beyond the limits of ruler-and-compass geometry.
What, however, if the magnitude, whose square root is demanded, turns out to be negative ? According to Kaestner’s argument, the required square root could be neither negative nor positive. For, the square of a positive quantity is naturally positive, while result of multiplying a negative magnitude by itself, for the reasons Kaestner indicated, is likewise positive. Thus, extracting the square root of a negative magnitude is “impossible” in the domain of the so-called “real numbers”.
On the other hand, our formula for the solution of the general quadratic equation, demands that we find the square root of BB/4 – C. What if the latter magnitude were to turn out to be negative — as for example in the case, where B and C are both equal to 2, and BB/4 – C = 4/4 – 2 = -1 ? Apparently, to solve the equation XX + 2X + 2 = 0, we would have to determine the value of SQRT( -1) !
To get a notion of the problem involved, explore the values of XX + 2X + 2 for a variety of positive and negative values of X. The values turn out all to be positive, with a minimum value of 1, reached for X = -1. In fact, plotting XX + 2X + 2 (on the “Y-axis”) against the value of X (on the “X-axis”), yields a parabola lying entirely in the positive region above the X-axis; whereas XX + 2X + 2 = 0 would imply a crossing of the X-axis by the corresponding curve, which evidently never occurs.
Another “impossible” problem
A useful, related sort of case was discussed by Cardan and his associates in the context of their attempt to grapple with “imaginary numbers”.
Consider for example the following simple geometrical problem: “Given a line segment of length 2, divide the segment into two subsegments of lengths X and W, such that the product of their lengths is equal to 2.”
A bit of reflection shows, that the task is “impossible”! If we divide the segment exactly in half, then the two segments will each have length 1, and the product will be 1, not 2. If, on the other hand, if the division is not equal, then one of the two segments (X for example) will be shorter than 1, and so XW will be less than W, which in turn cannot be longer than the total length 2.
The paradox becomes even clearer, when we look at the proposed problem in terms of a division of the square of side 2, implied by any given division of the sides into subsegments of lengths X and W. Drawing perpendiculars at the points of division, the square is divided into the two squares XX, WW plus two X-by-W rectangles. The total area of the original square is 2×2 = 4. If, on the other hand, we demand that XW = 2, then the sum of the two rectangles would already be 4, and would thus completely fill up the square! But then there would be no room left inside for the smaller squares XX and WW — unless, by some extraordinary circumstance, the two areas XX and WW were somehow to cancel out , i.e. XX + WW = 0 or WW = – XX. But since the areas of squares are always positive, the latter is clearly impossible .
This “impossible” geometrical problem leads directly to a quadratic equation. In fact, we can restate the problem in algebraic form as a combination of two simultaneous equations:
X + W = 2 and XW = 2
The first equation implies W = X – 2, and consequently
XW = X x (2 – X) = 2X – XX.
In terms of X, the requirement that XW = 2 now becomes 2X – XX = 2 or, by “rebalancing” the equation: XX – 2X + 2 = 0.
Again, graphing the values of XX – 2X + 2 yields a parabolic curve lying entirely above the X-axis; the minimum value, reached at X = 1, is 1. On the other hand, if we apply our “general formula” for the solution of the quadratic equation to this case, we obtain the paradoxical value X = 1 + SQRT ( -1 ).
At this point, the formal algebraicist, like Euler and Lagrange, might seek refuge in the following comforting thought:
“After all, doesn’t the appearance of the “impossible” square roots of negative magnitudes in the formula for the solution of a quadratic equation, coincide with the case, where the equation has no real solution? Our formula gives a solution in exactly those cases, in which a solution really exists. ‘Imaginary’ magnitudes like SQRT( – 1) mean nothing, but merely signal the impossibility of solving the corresponding equation. So, our algebraic world is closed and perfect.”
If that’s what you think, you are in for a rather unpleasant surprise! Just wait for the next installment, on Cardan’s formula. Notes:
The solution to the quadratic equation XX + BX + C = 0 is commonly presented in a slightly different, but entirely equivalent form: X = ( -B + SQRT( BB – 4C ) ) / 2 .
In the case of the equation AXX + BX + C = 0, in which the square term appears multipled by an arbitrary coefficient A, the solution takes the form: X = ( -B + SQRT( BB – 4AC ) ) / 2A.
Both formulae are simply alternative manifestations of the principle of “completing the square”, and add nothing of substance to the above discussion.