Why Yes, As a Matter of Fact You Can Prove the Parallel Postulate, and All the Others Too
I can’t believe I got this far in mathematics without noticing that not only can the parallel postulate be proved, but that it’s been proved for hundreds of years. For the last hundred or so, it’s been completely provable, along with the rest of Euclidean geometry, in Zermelo-Frankel set theory. That is, since ZF was invented. You don’t even need the Axiom of Choice. It’s provable with naive set theory too, or just with the Peano postulates and basic algebra.
Are you surprised? I was. What about non-Euclidean geometry? It turns out we can prove that too, and it’s all consistent (assuming ZF is consistent). How about two thousand years of mathematicians trying (and failing) to prove the parallel postulate? Did they just miss it? In one sense, yes, but in one sense no. Let’s dig a little deeper.
It’s still true that you can’t prove the parallel postulate from Euclid’s first four axioms. Nor is it possible to prove it from the first 19 of Hilbert’s fixed up versions of Euclid’s axioms. But are those the axioms we should be using? Maybe not. Most of modern math starts from the nine axioms of ZFC set theory, and for plane geometry we don’t even need the controversial ninth axiom (choice). The first eight will do nicely. Not only can you use these to prove the parallel postulate. You can prove Euclid’s other four axioms too. And you can precisely define Euclid’s undefined terms point, line, and plane. It’s that last point that really makes ZF a stronger foundation for geometry than Euclid or Hilbert gave us.
I’m not actually going to go all the way back to “If o is a member of A, the notation o ? A is used.” Let’s more comfortably start with the real numbers and high school algebra. This can all be built on top of ZF, and if you want to see those steps, I recommend Paul Halmos’s Naive Set Theory.
First define a point as an ordered pair of real numbers like (1.5, 3). Next define a line as the points that satisfy a linear equation like \(Ax + By + C = 0\). (I’m using the standard form to support vertical lines which the more familiar point-slope form can’t do.) Finally define a plane as all ordered pairs of real numbers. That’s the three basic undefined terms in the standard presentation of plane geometry. It doesn’t matter whether you built the real numbers starting from ZFC or the Peano postulates. Once you have them, it’s easy to define points, lines, and planes. And then from there you use algebra to prove the standard axioms. For example, here’s the promised proof of the parallel postulate:
Assume you have a line that satisfies \(Ax + By + C = 0\) where A, B, and C are constants. Also assume you have a point \((x_1, y_1)\) that is not on the line.
Now consider the line \(Ax + By + K = 0\). Plug in the point \((x_1, y_1)\) to get \(Ax_1 + By_1 + K = 0\). Rearranging terms we find \(K = -Ax_1 – By_1\). So now consider the line \(Ax + By – Ax_1 – By_1 = 0\). Clearly it includes the point \((x_1, y_1)\). A little less obviously it does not include any point that lies on the original line \(Ax + By + C = 0\). To see this suppose to the contrary there is a point \((x_2, y_2)\) that lies on both lines. Then in that case we have
\(Ax_2 + By_2 + C = 0\)
\(Ax_2 + By_2 + K = 0\)
Subtract these two equations to get
\(C – K = 0 \Rightarrow C = K\)
but then the two lines are the same, and \((x_1, y_1)\) also lies on the original line which is a contradiction. Therefore there cannot be any point which lies on both \(Ax + By + C = 0\) and \(Ax + By – Ax_1 – By_1 = 0\). This proves there is at least one parallel line through any given point in plane not on the original line.
We still need to show this line is unique. In order for two lines not to intersect, that is for two of the line equations not to have any solution in common, the ratios A/B or B/A must be the same and the constants C different. (To see this, treat these as two equations in two variables x and y and see the conditions under which the equations don’t have a solution.) So for the specific point \(x_1, y_1\) the line through that point that is parallel to the original line \(Ax + By + K = 0\) must satisfy the equation \(Ax + By + K2 = 0\) where \(K2 \neq K\). And since \(x_1\), \(y_1\), \(A\), and \(B\) are all fixed at this point*, then \(K2\) is fixed too. QED.
* Actually, not quite. You can divide through by either \(A\) or \(B\) (whichever isn’t 0) to get a ratio that is fixed, even if \(A\) and \(B\) can vary in sync. This doesn’t materially affect the proof.
What about non-Euclidean geometry? These use different algebraic definitions of point, line and plane to represent the surface of a sphere or a saddle instead. For instance, in spherical a geometry a “line” is the set of points satisfying the equation \(Ax + By + Cz = 0\). In a Poincaré disk, a “line” is the set of points satisfying the equation \((x – a)^2 + (y – b)^2 = r^2\). Better yet, don’t confuse everyone by calling these things points, lines, and planes. Call them antipodes, great circles, spherical surfaces, chords, arcs, or whatever they naturally look like to regular human beings instead. (Is there a word for a generalized “line” that covers both Euclidean and non-Euclidean lines?) This is all much clearer than trying to guess what shape a revised parallel postulate is really describing.
The real trick here is not the proof. It’s eliminating the undefinedness of the terms point, line, and plane. Once they’re defined, geometry sits on much stronger foundations. I expect some mathematicians will object that I’ve only constructed the standard Cartesian model of Euclidean geometry. Fair enough. I have. There are other models of other geometries. But this is the geometry pretty much everyone studies and uses until a very small fraction of students encounter general relativity much later in their education. It makes more sense to me to fully construct the model we want from well understood properties of regular math than to present multiple postulates by appealing to intuition and some hand-waving about undefined terms, and then try to see how many models fit which permutations of those postulates.
Why isn’t this the standard way to present plane geometry? Maybe because high school geometry is still too wedded to more than two millennia of Euclid. The ancient Greeks were brilliant at geometry, but they were terrible at algebra and arithmetic. They didn’t even have zero, much less real numbers and Cartesian coordinates. But once you have those things geometry makes a lot more sense. We tend to teach analytic geometry after shape-based geometry, but maybe we shouldn’t. If you start by using ordinary real numbers and Algebra I to define point, line, and plane, plane geometry makes a lot more sense.
