User:Addemf/sandbox/Technical Reasoning/Examples in Axiomatic Geometry

Our previous study of sets served two purposes. One was to know enough set theory to be able to use it in the study of other topics.

The other was that it provided an opportunity for some relatively simple proofs. These can serve as examples later on in the study of logic.

Before studying logic, it would be nice to see proofs in at least one other setting. For the sake of variety, let's see some proofs in the setting of geometry.

The Search for Axioms
Recall from a previous lesson that we saw a proof that the interior angles of a triangle sum to 180°.

After this proof, we reflected on the fact that we must accept some statements as fundamental. These statements, which we accept without proof, are called "axioms".

We ended with the question "Which statements will we accept as axioms?" Of course in some sense we must pick axioms which are designed to give us the geometry that we know we want.

One of the most important results from Euclidean geometry is the Pythagorean theorem. Let us see a proof of this theorem, and try to work backwards from this highly desirable result.

By thinking about what it is that we want from a theory, we are guided in trying to organize it and choose principles which deliver the thing we want.

The Pythagorean Theorem
Here is a beautiful and intuitive picture proof of the Pythagorean theorem.

The way this works is:

1. Start from any right triangle, call it $$\Delta ABC$$ where C is the vertex at which the sides make a right angle. Make four copies of the original triangle in the following way:
 * 1. Extend $$\overline{CB}$$ by a length equal to AC. Call the end of this extension C'.
 * 2. Through $$C'$$ and perpendicular to $$\overleftrightarrow{CB}$$, draw a segment of length equal to CB. Call the end of this segment $$B'$$.
 * 3. Triangle $$\Delta BC'B'$$ is now congruent to $$\Delta ABC$$.
 * 4. Repeat the above procedure on triangle $$\Delta BC'B'$$ to generate a new triangle.


 * 5. Repeat the procedure again on the new triangle.

2. This creates four triangles all congruent to the original triangle. One can prove that the outer edges of these triangles now form a square ringing the entire figure.

Show that the square surrounding the figure constructed above has area
 * $$(a+b)^2$$

Note that this course is designed to increase in rigor. At this initial point in the course, proofs are meant to be compelling but not extremely rigorous or formal. Just give a convincing explanation.

Notice that the figure also has an inner square, with sides formed from the hypotenuses of all of the copied triangles. Argue that this square has area $$c^2$$.

Moreover notice that, because the four triangles are congruent then they have the same area. Also notice that the area of the entire figure is just the sum of the inner square and these four triangles.

Therefore argue that the area of the square is also equal to
 * $$c^2+2ab$$

Use the two exercises above to prove the Pythagorean theorem
 * $$a^2+b^2=c^2$$

Certainly the proof above assumes the possibility of extending a line segment by a given length. That seems like a good candidate for an axiom, since it sounds quite fundamental and probably cannot be proved from any simpler statements.

Here are other assumptions made by the proof.


 * Given a point P and line $$\ell$$, there is always a line through P and perpendicular to $$\ell$$.
 * Through any two points is a line segment connecting them.
 * If the segments $$\overline{WX},\overline{XY}, \overline{YZ},\overline{YZ'}$$ are all of the same length, at right angles to each other, and have the same "orientation" (i.e. every angle formed is, so-to-speak, a "left turn"), then $$Z=Z'$$.
 * The length of segments on a line are additive: If $$\overline{ABC}$$ is a segment, then its length, AC is the same as the sum of the component lengths, $$AB+BC$$.
 * The area formulas for triangles.
 * Area is additive, in a way analogous to how length is additive.

Not all of these seem like equally good candidates for being axioms. They all seem relatively compelling, I would argue -- but some of them, like the third bullet point above, seems much too complex to be an axiom. Axioms should be relatively simple, fundamental, and not provable from other simpler propositions. That third bullet-point certainly feels as though it should be provable if we chose the right collection of other, simpler axioms.

The ancient Greeks were aware of proofs like this one and others. In an effort to bring all of these various facts into a single, unified, and organized body, Euclid proposed a collection of axioms from which all of geometry could be proved.

Euclid wrote down only a few axioms, which were


 * 1) To draw a straight line from any point to any point.
 * 2) To produce (extend) a finite straight line continuously in a straight line.
 * 3) To describe a circle with any centre and distance (radius).
 * 4) That all right angles are equal to one another.
 * 5) The parallel postulate: That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which the angles are less than two right angles.

This last one is quite complicated, so above is an animation demonstrating its meaning. First two lines are shown and then two points, one on each. The segment between them is constructed, and then the angles on one side of the segment. Because these angles sum to less than 180° then the lines intersect on this side.

More information about Euclid's work can be found at the Wikipedia page | Euclidean geometry.

Hilbert-style Axioms
Ultimately it turns out that Euclid's axioms are not quite adequate for the level of rigor and completeness that we would like. In more recent times, a more precise and complete set of axioms were laid out by various mathematicians.

One of the more famous of these was | Hilbert's axioms. This consisted of 20 axioms, and in order not to get too distracted, we won't try to discuss all of them or get too deep into the study of geometry.

However, it will be instructive to at least prove a few of the most basic results of geometry by means of Hilbert-styled axioms.

As a general note about language, we will talk about points being "on" or "in" a line. Equivalently we may say that that the line "runs through" a point, or "contains" a point. This will be defined after we have precisely defined the word "line".

Here is a first axiom:

For any two points there is exactly one line through them.

Notice that this axiom refers to objects which we have not defined.

As we have said repeatedly, statements cannot have infinitely many proofs, and we must choose some foundational statements for axioms.

Well the same, too, must be true of definitions as well! While we can define some concepts in terms of other concepts, eventually we must have some concepts which are not defined.

Our first undefined concept is that of a point. We will talk about sets of points, we will prove things about them, but we will not try to state what they are. We take the idea of a point to be foundational and not in need of further definition.

In fact, we will define lines in terms of distances between points.

In order to that, we first have to define the concept of distance. This will come shortly.

We next define the plane as the set of all points, and we write this as $$\mathfrak P$$.

We will now define distance to be a relationship between points and a nonnegative real number. We denote the distance between points A and B by the expression $$d(A,B)$$.

In the following axiom we show that the distance must also satisfy certain properties that we would recognize, from anything worthy of the name "distance".

For any two points, $$A,B\in \mathfrak P$$, the distance from any point to itself is zero.
 * $$d(A,A)=0$$

Also, distance is commutative, which is to say, the distance from A to B is the same as the distance from B to A.
 * $$d(A,B)=d(B,A)$$

Prove that if B is between A and C then it also follows that B is between C and A.

Note that, at this point, proofs are meant to be increasing in their rigor and therefore, in this proof, we do not want to rely on things which "should intuitively be true". Rather, we want the proof to come very formally and precisely from the exact statements of definitions and axioms.

However, also note that we will be assuming that numbers work in the way that we are familiar with. For example we will assume that $$1+2=2+1$$ without need for justification. Later in the course we will actually investigate axiom systems for numbers, but for now we assume their familiar properties.

Read this as saying "The line through A and B is defined to be the set of all points, C, such that either A is between C and B, or C is between A and B, or B is between A and C."

At this point we have finally defined all of the objects referred to in the Two Point Axiom! With all of this ground cleared, we are actually in a position to state several more axioms, and the prove another lemma.

There exist some three distinct points which are not all collinear.

For any three distinct collinear points, $$A,B,C\in\mathfrak P$$, precisely one is between the other two.

Prove that for any three distinct colinear points $$A,B,C\in\mathfrak P$$, precisely one of the following holds,


 * $$[ABC]$$
 * $$[ACB]$$
 * $$[BAC]$$

If $$B,D\in\mathfrak P$$ then there are points $$A,C,E\in\mathfrak P$$ such that $$[ABCDE]$$.

Prove, using any of the axioms and definitions above, that there are infinitely many points.

Use the axioms and definitions above to prove that, for any $$A,B\in\mathfrak P$$, we have
 * $$d(A,B)=0$$ if and only if $$A=B$$

Now you do it!

With inspiration taken from how we defined a line, write a reasonable definition of a ray.

Recall that, for any two distinct points $$A,B\in\mathfrak P$$, the ray from A through B is supposed to be effectively half of the line through A and B. In particular, it is suppose to be the half of the line which contains both of these points, and is "cut" at A.

This is a half-formal definition of a ray. Your challenge is to re-state this, fully formally.

Likewise, define the segment from A to B.

We will write the ray from A through B by $$\overrightarrow{AB}$$. We will write the segment from A to B with $$\overline{AB}$$.

Once you have defined these formally, then prove the following. For any two distinct points $$A,B\in\mathfrak P$$,


 * $$\overrightarrow{AB}\cap\overrightarrow{BA} = \overline{AB}$$

and
 * $$\overrightarrow{AB}\cup\overrightarrow{BA} = \overleftrightarrow{AB}$$