Solution 26: Desargues' Theorem in the Affine Plane

Index

26 (a)

Let's first prove part I of Desargues' Theorem. Recall:

Desargues' Theorem (I). Suppose that A, B and C are distinct noncollinear points with

l (A, A') || l (B, B') || l (C,C') and l (A,B) || l (A', B') and l (A,C) || l (A', C'). Then l (B,C) || l (B',C').

So, restating this theorem a bit for F2, let's suppose that u = A, v = B and w = C are noncollinear points in F2. We then need to show that if (u,u',v',v ) and ( u,u',w',w ) are parallelograms, ( v,v',w',w ) is a parallelogram as well, where u' = A', v' = B', and w' = C'.

Now for a proof. By Theorem 1 we know that v = v' - u' + u and w = w' - u' + u. So we see that u = v - v' + u'. So w = w' - u' + v - v' + u' = w' + v - v'. So ( v, v',w',w ) is a parallelogram by Theorem 1.

Now let's prove the second part of Desargues' Theorem. Recall:

Desargues' Theorem (II). Suppose that A, B and C are distinct noncollinear point with nonempty and l (A,B) || l(A',B') and l(A,C) || l(A',C'). Then l(B,C) || l(B',C').

Again, looking at this in F2, let's suppose that and l (u,v) || l(u',v') and l(u,w) || l(u',w'). Then we need to show that l (v,w) || l (v',w').

By theorem 2 we know that u' = u + t ( z - u ) for some nonzero scalar t. If we set x = v + t ( z - v ) then x is on l (v,z) and l (u,v) || l(u',x). This follows by the converse of theorem 3. So x = v'. In addition w' = w + t ( z - w) so it follows that w' - v' = w - v + t (z - w - z + v) = ( 1 - t )(w - v). So, w' - v' = (1 - t )( w - v ). So by theorem 3, l (v,w) || l (v',w').

Thus, Parts I and II of Desargues Theorem hold in the affine plane over F.

26 (b).

Let's first show that the Moulton plane is a linear space (2 distinct points are contained in a unique line). Our points are of the form ( x,y ) such that x and y are both real numbers. We need to show that any two points ( x1, y1 ) and ( x2, y2 ) lie on one and only one line. We have seen that two points determine a unique line in the real coordinate plane (recall problem 22), so for any two points that lie on a vertical, horizontal, or negative slope line, it follows that they will be contained by one and only one line. We also know that if two given points lie below the x-axis, then they will lie on a positively sloped line the same as they would in the Euclidean plane (although the line is not the typical line, as it changes slope at the x-axis). The same holds true for two points above the x-axis. However if one of our points lies below the x-axis and the other above, it is a bit trickier.

Let's look at our first point ( x1, y1 ). If this point lies below the x-axis and our second point (x2, y2 ) lies above, then we can determine the "line" between them by finding its intersection with the x-axis. Let's call this intersection point ( x, 0 ). The broken line is made up of 2 Euclidean lines with slope m and 2m, where the value of m has already been fixed.

So let's first look at the part of the broken line below the x-axis. Now 2m = -y1 / (x - x1). Thus, the slope of the line from ( x1, y1 ) to ( x, 0 ) is m = -y1 / 2(x - x1). The slope of the broken line above the x-axis is m = -y2 / (x - x2).

So m = -y1 / 2(x - x1) = -y2 / (x - x2). Solving this for x, we find that:

x = (2y2x1 - y1x2 ) / (2 y2 - y1 )

So there is a unique x - intercept for the broken line through any two points. From this, it is easy to show that a unique (broken) line exists (we can find the equation for the Euclidean lines below and above the x - axis). Thus, there exists a unique broken line between any two points such that one is above and one is below the x-axis and x1 and x2 are distinct. So for any two points on the Moulton plane, there is a unique line between them.

Now, for axiom A1 of an affine plane (given a line and a point, there is a unique line through the point that is parallel to the given line). This axiom is easily verified as holding true for the Moulton plane, as any two horizontal lines are parallel, any two vertical lines are parallel and any two lines with the same negative slope are parallel. In addition, any two broken lines are also parallel. Finaly, two lines of different "types" are never parallel. Thus, the parallel axiom holds in this plane.

Finally, it is trivial to see that axiom A2 holds (there exist 3 points not all contained in a line) true for the Moulton plane.

Now, to show that this affine plane is not Desarguesian, let's look at the figure in Problem 26 corresponding to the plane. By the first part of Desargues Theorem, if

l (A, A') || l (B, B') || l C,C') and l (A,B) || l (A', B') and l (A,C) || l (A', C'). then we need l (B,C) || l (B',C').

But in our figure we see that l (B,C) intersects l (B',C') at point X. Therefore, the Moulton plane is not Desarguesian!