Ramble, Part 13: Completing the Cube
Feb. 3rd, 2007 10:03 am![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
stoutfellowIn this installment, I'm going to discuss the methods Tartaglia and the rest developed to solve cubic equations. I'll give some of the mathematical details, but I want to lay particular stress on a couple of methodological and philosophical issues that are involved. More under the cut.
The first item that deserves mention is the fact that this work, in the mid-sixteenth century, was carried out without the benefit of modern algebraic symbolism. Allied to this is the continued dominance of geometry, which made the solution possible even with that handicap.
Let me begin by recalling the solution of the quadratic equation x2+mx=n. Since we're thinking geometrically, we have to set up the equation this way in order to ensure that m and n are magnitudes (and hence necessarily positive). (Equations like x2=mx+n and x2+n=mx have to be handled separately.) Consider a square, with side a; cut a square notch, with side b, in one corner, and let x=a-b. The mutilated square, whose area is a2-b2, can be dissected into three parts, a square with side x and two rectangles with sides x and b. This dissection yields the equation x2+2bx=a2-b2. Applying this to the original equation, we see that if we can solve the pair of equations 2b=m and a2-b2=n for a and b, then we can recover x as a-b.
Let's extend this to the cubic equation x3+mx=n. Picture a cube with side a, and this time cut a cube of side b from, say, the upper right corner in front. Again, set x=a-b. We can dissect this solid (with volume a3-b3) into four pieces: a cube with side x, and three slabs with sides x, a, and b. To see this, start with the cube of side x. Place one of the slabs upright on the right side of the cube, so that it covers the right side and extends an additional distance b above it. Put a second slab on top, covering the top and extending forward beyond it. The third slab lies on its side in front of the cube, covering the front and extending beyond it to the right. Remembering that x=a-b, this dissection tells us that x3+3abx=a3-b3. So, if we can solve the equations 3ab=m and a3-b3=n, we can then set x equal to a-b. If we take the second of these equations, multiply it by 27a3, and substitute in from the first equation (27a3b3=(3ab)3=m3), we get 27a6-m3=27a3n. This is a quadratic equation in z=a3; we solve it for z, set a equal to the cube root of z, find b, and then compute x.
Thus, despite not having algebraic notation available, the Italian mathematicians were able to use geometry to solve cubic equations. The spatial visualization this needed is considerable, but it did lead them to a solution (which merely fiddling with the algebraic equation, even assuming modern notation, might not have).
The other interesting point is this. Suppose a cubic equation has three real roots. If you apply Tartaglia's method, you get a quadratic equation in z with no real roots; z must be complex, and hence so are a and b. The remarkable thing is that a-b still winds up being real. It's not clear whether Tartaglia or Cardano ever thought about this, but not much later Rafael Bombelli confronted it squarely, taking what the formulas said seriously. What makes this interesting is not just the question of the reality of complex numbers, but the fact that they were necessary to deal with an equation in which complex numbers did not appear overtly at all - and all of whose roots turned out to be real!
I distinguished, earlier, between "numbers" and "magnitudes", and spoke of the effort to bring them together under a single heading. With Bombelli's discussion, a new player enters the fray: the complex number. What "numbers" are (including or excluding magnitudes) became an important and vexing question for several centuries, as we will see.
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The first item that deserves mention is the fact that this work, in the mid-sixteenth century, was carried out without the benefit of modern algebraic symbolism. Allied to this is the continued dominance of geometry, which made the solution possible even with that handicap.
Let me begin by recalling the solution of the quadratic equation x2+mx=n. Since we're thinking geometrically, we have to set up the equation this way in order to ensure that m and n are magnitudes (and hence necessarily positive). (Equations like x2=mx+n and x2+n=mx have to be handled separately.) Consider a square, with side a; cut a square notch, with side b, in one corner, and let x=a-b. The mutilated square, whose area is a2-b2, can be dissected into three parts, a square with side x and two rectangles with sides x and b. This dissection yields the equation x2+2bx=a2-b2. Applying this to the original equation, we see that if we can solve the pair of equations 2b=m and a2-b2=n for a and b, then we can recover x as a-b.
Let's extend this to the cubic equation x3+mx=n. Picture a cube with side a, and this time cut a cube of side b from, say, the upper right corner in front. Again, set x=a-b. We can dissect this solid (with volume a3-b3) into four pieces: a cube with side x, and three slabs with sides x, a, and b. To see this, start with the cube of side x. Place one of the slabs upright on the right side of the cube, so that it covers the right side and extends an additional distance b above it. Put a second slab on top, covering the top and extending forward beyond it. The third slab lies on its side in front of the cube, covering the front and extending beyond it to the right. Remembering that x=a-b, this dissection tells us that x3+3abx=a3-b3. So, if we can solve the equations 3ab=m and a3-b3=n, we can then set x equal to a-b. If we take the second of these equations, multiply it by 27a3, and substitute in from the first equation (27a3b3=(3ab)3=m3), we get 27a6-m3=27a3n. This is a quadratic equation in z=a3; we solve it for z, set a equal to the cube root of z, find b, and then compute x.
Thus, despite not having algebraic notation available, the Italian mathematicians were able to use geometry to solve cubic equations. The spatial visualization this needed is considerable, but it did lead them to a solution (which merely fiddling with the algebraic equation, even assuming modern notation, might not have).
The other interesting point is this. Suppose a cubic equation has three real roots. If you apply Tartaglia's method, you get a quadratic equation in z with no real roots; z must be complex, and hence so are a and b. The remarkable thing is that a-b still winds up being real. It's not clear whether Tartaglia or Cardano ever thought about this, but not much later Rafael Bombelli confronted it squarely, taking what the formulas said seriously. What makes this interesting is not just the question of the reality of complex numbers, but the fact that they were necessary to deal with an equation in which complex numbers did not appear overtly at all - and all of whose roots turned out to be real!
I distinguished, earlier, between "numbers" and "magnitudes", and spoke of the effort to bring them together under a single heading. With Bombelli's discussion, a new player enters the fray: the complex number. What "numbers" are (including or excluding magnitudes) became an important and vexing question for several centuries, as we will see.
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Ramble Contents
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no subject
Date: 2007-02-03 07:40 pm (UTC)Since, in the construction, a and b are explicitly real, did Tartaglia simply not address such problems? *tries one* OK, this would lead to a different form of the cubic, with terms on the RHS that are on the left above. And I suppose in that case, critical signs are changed above so that complex intermediate steps are avoided? (will try that later - the sun is shining, time for a walk.)
no subject
Date: 2007-02-03 08:44 pm (UTC)If the three roots u,v,w are real and the x^2 term is zero, they must satisfy u+v+w=0, so either two of them are negative and one positive or two are positive and one negative. If we write it in the form x^3+px+q=0, p is uv+uw+vw and q is -uvw; in either case p and q wind up with the same sign, so (forcing coefficients to be positive) it's either x^3+mx+n=0 or x^3=mx+n.
no subject
Date: 2007-02-04 03:59 am (UTC)Antidisirregardless, I seem to be getting complex intermediate results, as advertised. I suppose that means that the geometrical construction can't be physically realized in the 3 real root case?
So what, we wonders, did Tartaglia do? Not solve such terrible equations?
no subject
Date: 2007-02-04 10:21 am (UTC)I'll look into it tomorrow.
no subject
Date: 2007-02-06 01:05 pm (UTC)He does rather gingerly discuss complex solutions of quadratic equations, but he's even leery of negative numbers, so... (Rather than say the equivalent of "-4 is a solution", he says "4 is a false solution", for instance.)
no subject
Date: 2007-02-03 11:51 pm (UTC)