Footnote symbols: - Rasmusen



December 16, 2011

For AMerican Mathematical Monthly?

This is an idea of how to do a simple patch-up of Euclid's Elements, Book I-- in particular, how to get round using his Method of Superposition. I think it's interesting to see how close we can keep to Euclid's simplicity and still have proofs work.

For more than a thousand years, students have learned mathematics from Euclid's elements. Its importance is not so much in the mathematical results as in the idea of the proof from first principles. It is therefore dismaying that even in the limited domain of circles and rectilinear figures in planes Euclid fails to be rigorous. Or, rather, he is rigorous in proving some obvious propositions but loosely assumes the obvious in proving other propositions. It is worth patching up Euclid for two reasons. One is to show how to construct an axiomatic system to prove his propositions. The other is to show how his method can be used rigorously by students learning proofs. The first of these objectives has long ago been tackled, most famously in Hilbert's set of axioms. Hilbert, however, tried to escape from the diagrammatic style which produces the essential flavor of Euclid's geometry. Recently a literature has sprung up trying to axiomatize that diagrammatic approach, keeping the diagrams but adding rigorous rules for drawing them (Barwise, Manders, Mumma, Miller). The systems of Mumma and Miller are valuable, but although they provide foundations for Euclid's approach, they complexify it greatly.

We have a more modest objective: to patch up some of the more obvious and grating problems in Euclid's Book I. This allows at least a rigorous introduction to Euclid, and it will not require changes to Euclid's essential style. Specifically, we will address three problems. (1) Euclid does not specify the size of an angle (except for right angles), which leaves it unclear when two angles can be of the same size. (2) Euclid lacks a postulate saying when circles intersect--- his only postulate dealing with intersections is the famous Postulate 5 about the intersection of non-parallel lines. (3) Very early on, in Proposition 4 of Book I (and thus foundationally for later propositions), Euclid uses the ``method of superposition,'' in a key proof about the congruence of triangles (that Side-Angle-Side defines a triangle). This method, criticized for centuries, takes one figure and superimposes it on another to show their equivalence.

We argue that the method of superposition is unnecessary, provided that we clean up the definition of an angle and add the circle intersection postulate necessary to prove Proposition 1. The method reaches valid results; its problem is that it skips steps. Rather than superimposing figure DEF on figure ABC directly, we will construct a duplicate GHI of DEF on top of ABC with straight-edge and compass, so that the proof is to show that GHI equals DEF and GHI also equals ABC so DEF equals ABC. To do the construction, we will need to build an angle of size equal to an existing angle but in a different location, To build such an angle, we will need to use the intersection of circles.

Barwise, J., & Hammer, E. (1996). Diagrams and the concept of a logical system. In G. Allwen & J. Barwise (Eds.), Logical reasoning with diagrams. New York: Oxford University Press.

Hilbert, D. (2004). David Hilbert’s lectures on the foundations of geometry: 1891–1902. In M. Hallet & U. Majer (Eds.). Berlin: Springer.

Manders, K. (2008). The euclidean diagram. In P. Mancosu (Ed.), Philosophy of mathematical practice. Oxford: Clarendon Press.

Miller, N. (2007). Euclid and his twentieth century rivals: Diagrams in the logic of euclidean geometry. Stanford, California: Center for the Study of Language and Information.

Mumma, J. (2006). Intuition formalized: Ancient and modern methods of proof in elementary euclidean geometry. PhD Dissertation, Carnegie Mellon University. .

Mumma, J. (2008) Review of Euclid and his twentieth century rivals: Diagrams in the logic of euclidean geometry. Philosophia Mathematica, 16(2), 256–264.

Joyce’s superb web edition of the Elements:

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First, let us have a new definition 8a.

Definition 8a: The SIZE OF AN ANGLE BAC is defined relative to some line DE. Extend or contract the lines AB and AC to length DE to create lines AF=DE and AG=DE.

The size of the angle BAC is the line FG.

Note: angle size thus defined is ordinal, not cardinal. When two angles are combined, the size of the combined angle is not the sum of the sizes of each angle. If you want that, you need a different definition:

Definition 8b: The ADDITIVE SIZE of an angle is defined relative to some straight line DE. Extend or contract the straight lines AB and AC to length DE to create straight lines AF and AG. Draw a circle centered at A with radius DE. The additive size of the angle BAC is the length of that part of the circle lying between F and G.

Note: We do not need the additive size of an angle in Book I. It might be useful for talking about the length of arcs. There is a one-to-one mapping between the size of an angle and its additive size. The additive size is not useful in Book I because there is no way to measure the length of an arc. (Or maybe there is and I just haven't seen it, by cutting up circles with other circles.)

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Definition 24 (maybe) : The EXTENSION of a line AB is the long line obtained by producing a straight line indefinitely from point A to B and beyond (maybe not helpful).

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We also need a new postulate.

Postulate 6: If the distance between the centers A and B of two circles is greater than the radius of either but less than the sum of their radii then the two circles have at least one point in common. (they will have 2, but that is unnecessary) (from Joyce’s Prop 1 notes, but modified significantly—adding the second radius condition)

Note: The bit about the radius of either is necessary to avoid circle A having such a large radius as to enclose circle B without intersecting it. This postulate is about the circle as a boundary of a disk.

Use Postulate 6 to prove Prop. 1 too. It will guarantee that there is an intersection of the two circles in that proof.

Maybe another postulate is needed to guarantee the intersection of a long enough straight line sarting at the interior of a circle with the circle itself.

It would be handy to have a proposition 3aa: A circle can be constructed centered at A and with a radius equal to a given line BC. Maybe that could be replaced by just broadening Proposition 3. Proposition 3 says

Proposition 3

To cut off from the greater of two given unequal straight lines a straight line equal to the less.

It SHOULD say

Proposition 3

To cut off from the greater of two given unequal straight lines a straight line equal to the less; or to extend the lesser to be equal to the greater.

I bet Euclid uses it that way all the time.

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We will accept Euclid’s Propositions 1, 2, and 3. Then we will move up Proposition 20 to become Propsition 3a:

Prop 3a/ 20 (The Triangle inequality). In any triangle the sum of any two sides is greater than the remaining one. (Move Prop 20 up to be Prop 3a, since Prop 3b needs it.)

The proof is by contradiction. Suppose side AC of triangle is longer than BC plus AB. Draw a circle centered at A with radius AB and a circle centered at C with radius CB. If AC>AB+BC then the distance between the two centers is greater than the sum of the radii. Thus, they will not intersect (Pos 6). But we know that point B is in both circles, so we have a contradiction. QED.

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We will also move up Proposition 23 to become Proposition 3b. I will modify it, though, to make use of the new definition 8a that says when angles are equal.

Proposition 3b/23 (move up to be Proposition 3b, since Prop. 4 needs it)

To construct a rectilinear angle equal to a given rectilinear angle on a given straight line and at a point on it.

[pic]

Proposition 23 (MODIFIED FOR PROOF) (move up to be Proposition 3b, since Prop. 4 needs it)

To construct a rectilinear angle WITH THE SAME SIZE as a given rectilinear angle on a given straight line and at a point on it.

Proof.

Given angle BAC and line DE, we can construct an angle of the same size with D as the vertex and lying on DE.

Construct a straight line DF equal to AB (Prop I-2) but ending at D. Draw a circle centered on D with radius DF (Pos 3). Label as G the intersection of this circle with the the extension of line DE out beyond E (new postulate 7?).

Draw the line BC (Pos 1). Construct a straight line FH equal to BC but starting at point F (Prop I-2). Draw a circle centered on E with radius FH (Pos 3).

This F-circle will intersect our earlier D-circle. That is because BA=AC by definition of an angle, and AB+AC> BC by the triangle inequality (Prop 20, provable as Prop 1a), so 2AB > BC. The D-circle has the radius DF=AB and the E-circle has radius FH=BC. The D-circle's center and the F-circle's center are distance DF=AB apart, which is less than the sum AB+BC of the radii but greater than either AB or BC, so the two circles intersect (Pos 6). Label that point I.

Draw the lines DI and FI (Pos 1). Note that FI =BC, by construction of the F-circle. Also DF=DI, because both are on the circle centered at D, and DF=AB by construction. Thus, angle FDI has the size FI, which equals BC, the size of the angle BAC (Def of angle size). So we've constructed the desired angle size with vertex at D.

Next we will move the angle to align with DE. Draw a line GJ the length of IF but starting at G (Prop 2). Draw a circle with radius GJ centered at G. The G-circle and the D-circle will intersect. By Postulate 6 that is true, because D and G are distance DG=AB=AC apart and their radii are DG and GL=BC, where by the triangle inequality (Prop 20) AB+BC>AC. Label the intersection (either intersection) K. Since K is on both circles, it is distance DG=AB from D and distance GL=BC from G. Since GD =AB, the reference length for the angle BAC, the size of the angle GDK is GL=BC, the same as that of the angle BAC. Q.E.F.

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Proposition 4

If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, then they also have the base equal to the base, the triangle equals the triangle, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides.

Euclid starts thus: ``Let ABC and DEF be two triangles having the two sides AB and AC equal to the two sides DE and DF respectively, namely AB equal to DE and AC equal to DF, and the angle BAC equal to the angle EDF.''

Make AB our unit length for angle size, and measure angle BAC by cutting the extension of line AC to length AG = AB. Then BG is the size of the angle BAC (Def 8a).

Use Proposition 3b to construct an angle of the same size as BAG but with the A vertex moved to D and the AB side moved to lie on top of DE. There will be two ways to construct the angle: choose the one which puts the AC side on the same side as DF. Label the third vertex of the angle H.

The size of the angle FDE is found by labelling as point I (which will turn out to equal H) the distance AB on the extension of the line from D to F. The size of angle FDE is then IE. This equals the size of angles BAC, so IE=BC, and since the size of BAC is the same as that of HDE by construction, IE=BC =HE.

Suppose I and H are not the same point. Then we have a triangle HIE with two sides HE and IE and one side IH. But since HE=IE, we would have HE + IE=IH, whereas the Triangle inequality requires that HE+IE>IH. So I and H are the same.

In that case the points of the transferred angle HDE are the same as those of the original angle HDE.

Draw a line JH of length GC starting from J (Prop 2). Draw a circle with radius JH centered at J. It will cut the extension of DH at point F, because DH =AG and where it cuts adds the radius length GC to DH, resulting in length DH+GC= AG+GC= AC (cn 2) and AC equals DF by assumption.

We now have BAC superimposed on DEF and must show that the angles and sides are the same.

Draw a line EL of length BC on the extension of EF starting at E. (First draw a length KE=BC starting at E. Then draw a circle of radius BC centered on E. The intersection with DF will be point L.) We need to show that L=F.

Consider the point L. If it is part of a triangle DEF with opposite angle of size BF=EH and side ED, a line DL from L to D must pass through H. If L=F, that is the case. If L is not equal to F, then the angle will be bigger or smaller. Thus, L=F, and we have proved that LE=FE and since LE=BC it follows that FE=BC. (I struggled with this last paragraph-- it may be wrong, tho it is the key to everything).

Still to be proven is that the angle DFE equals ACB and DEF equals ABC. This will be tedious. It can work back from the first part of the proof, showing that if the opposite side and the two angle sides are the same in both triangles, so must the angle be.

(This last part of the proof would also work for the Side-Side-Side proposition 8.)

This is a lot of steps to get the same idea as Superposition, but it is doing it right instead of waving hands. And it only needs to be done once, I hope-- in Book I only Propsition 8 also uses it. (and ther’s some prposition about semicricles in Book III).

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