Why can we prove facts about Euclidean geometry using coordinate method?

Question

It's easy to show that coordinate geometry based on real number axioms satisfies the Euclidean postulates.

But how do we go the other way around?

Say we prove an arbitrary* statement about Euclidean geometry using coordinates. How can we translate it to a proof that solely relies on Euclid's postulates?

By "coordinate geometry" Euclidean space with Euclidean norm is meant specifically.

* there I'm assuming the statement is provable in Euclidean geometry

Answer 1

You can use Euclid's geometric axioms to construct a coordinate system in the plane: choose a point for the origin, a line through it for one axis, a point on that line to define the unit of distance. Construct a perpendicular for the second axis and draw the unit circle to find the unit point on the second axis.

Then show that equations for lines and circles and their intersections follow from Euclid's axioms.

Then any analytic proof using coordinate algebra can be reduced to a proof using Euclid's axioms.

You will need a good set of axioms to make this rigorous. Euclid's have some logical gaps. See Hilbert's .

Answer 2

With a few basic metric assumptions, I think Euclid's postulates can be translated into coordinates.

1. To draw a line between two points. Suppose our points are $(a,b)$ and $(c,d)$. You can just use a straight edge to determine that line.

In coordinates $(y-b)=\frac{d-b}{c-a}x$ determines the line.

2. Extend a line segment arbitrarily in any direction. Similar argument, only you are given slope and y intercept. Implies global application of line's equation.

3. Create a circle of radius r at any center translates to: consider center (a,b) and radius r the equation$(x-a)^2+(y-b)^2=r^2$

4. All right angles are congruent: I think this is equivalent to saying what's perpendicular remains perpendicular after rotations of the coordinate axes or translation of the center. Euclidean vectors can be used here. Might be equivalent to the product of the slopes of perpendicular lines is -1 and remains constant under said translations.

5. Two lines crossing a third will intersect on the side of that third line which the alternate interior angles are less than two right angles. Equivalently, given a line and a point C not on that line, a unique line can be draw through C parallel to the first.Coordinate equiv: Lines of the same slope either have all points in common or no points in common.

So consider the construction of an equilateral triangle given a line segment.

Euclidean approach: Consider Segment AB. Center a circle at A and draw a circle through B (III). Center a circle at centered at B draw through A (III). The intersections of the circles are the same distance from A or B as A and B are to each other. Call one of the two intersections C. Draw AC and CB (I used twice). ACB is a triangle with all sides equal. Call the other point of intersection D.

Theorem: Line CD intersects AB at its midpoint.

Proof: Call the point where CD intersects AB, M. Consider triangles CDA and CDB. CD is congruent to itself. CA, CB, DA, and DB are congruent by construction. This tells us that CDA and CDB are congruent triangles by SSS. This allows application of SAS to triangles CMA and CMB, proving that AM congruent to BM.

In coordinates:

Point A=(a,b) point B=(c,d)

M=( (a+b)/2, (c+d)/2)

More throughly:

$(x-a)^2+(y-b)^2=(x-c)^2+(y-d)^2=(a-c)^2+(b-d)^2$

$x^2-2ax+a^2+y^2-2by+b^2=x^2+c^2-2cx+y^2+d^2-2dy$

$-2ax+a^2-2by+b^2-c^2+2cx-d^2+2dy=0$

$x(-2a+2c)+y(2d-2b)+a^2-c^2=0$

This is an equation of a straight line.

$x^2-2ax+a^2+y^2-2by+b^2 =a^2+c^2-2ac+b^2+d^2-2bd$

$(x-a)^2+(y-b)^2= (a-c)^2+(b-d)^2$

is the equation of a circle.

Plug in the equation for the line into the equation for the circle to solve for points C and D. Use coordinates and the pythagorean theorem to check the lengths work to make the equilateral triangles.

This changes in non-euclidean geometries.

$\vec{r_1}=\sin\phi_1\cos\theta_1 \hat{i} + \sin \phi_1 \sin \theta_1 \hat{j}+ \cos{\phi_1} \hat{k}$

$\vec{r_2}=\sin\phi_2\cos\theta_2 \hat{i} + \sin \phi_2 \sin \theta_2 \hat{j}+ \cos{\phi_2} \hat{k}$

$\cos \alpha = \sin \phi_1 \sin\phi_2 \cos \theta_1\cos\theta_2+\sin \phi_1 \sin \phi_2 \sin \theta_1 \sin \theta_2 + \cos \phi_1\cos \phi_2$

$\cos \alpha = \sin \phi_1\sin \phi_2 \cos (\theta_1-\theta_2) + \cos \phi_1 \cos \phi_2$ Where $\alpha$ is the distance between two points on the unit sphere, $\theta$s are longitude and $\phi$s are the complementary angle of the latitude. .

Here you already see a breakdown in the circle formula for geometry on the surface of a sphere.

Answer 3

You have asked:

But how do we go the other way around?

The answer is: we cannot go the other way around. Euclid's axioms have gaps, and some of Euclid's own theorems are not provable using Euclid's own axioms.

Here is example of such a theorem, namely Euclid's very first construction: Given a line segment $\overline{AB}$, construct an equilateral triangle with one side on $\overline{AB}$.

Here's how Euclid describes that construction. First, construct the circle with center $A$ and radius $\overline{AB}$. Next, construct the circle with center $B$ and radius $\overline{AB}$. Next, let $C$ be a point that lies on each of those two circles. Then $\triangle ABC$ is equilateral.

The gap in Euclid's proof is that he never proves $C$ exists. Nowadays we know how to fill that gap, and to prove that $C$ exists, by using coordinate geometry. But the existence of $C$ is not provable using Euclid's axioms: no argument using only Euclid's axioms implies that there exists a point that lies on each of those two circles.

The gaps in Euclid's arguments, and in his axioms, were unveiled and filled during a lonnnngggggg, slowwwwwwww process that took from Euclid's own time up through the 19th century. In the early 20th century, Hilbert was in a position to write down a single account of how to extend Euclid's axioms to a fuller set of axioms which are sufficient to prove everything that can be proved using coordinate geometry. You will find this all laid out in the book by Hartshorne mentioned in the comments.