# Is the Euclidean postulate a theorem?

• B
The fact that you picture a flat plane in your mind is not relevant.
May I have a short imagination.

For the ML any rational proposition is either right or wrong.The proposition:"From a given point outside a line you can draw many lines parallels to the line" is wrong.

jbriggs444
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The proposition:"From a given point outside a line you can draw many lines parallels to the line" is wrong.
No, it is perfectly consistent with your other axioms, as explained earlier in the thread. For example, take the hyperbolic plane (set of points in the plane with positive ##y##-coordinate with the metric ##\frac{dx^2+dy^2}{y^2}).## The lines (geodesics) are vertical lines together with semi-circles perpendicular to the ##x##-axis. It satisfies your other axioms. It's not hard to see that for any line and point off the line, there are infinitely many lines through the point not intersecting the circle. For example, the top half of the circle ##x^2+y^2=1## and the vertical line ##x=0## are both lines passing through ##(0,1)## that do not intersect the line ##x=2.##

I think you're going too far in claiming "any rational proposition is either right or wrong". For example, suppose your only axiom was "There exists a unique line passing through any two given points". Then you'd be able to prove very little, and certainly not every 'rational proposition' would be either right or wrong.

binis
take the hyperbolic plane
With all my respect,you undoubtfully know that we are reffering to a regular plane. Jbriggs444 is talking about a saddle. You are talking about an hyperbolic plane. I am talking about a regular plane. I do not want to imagine something else. To keep on discussing,we at least must agree to one point.
the top half of the circle
You undoubtfully know that we are all talking about straight lines. The half circle is a curvylinear line.
[/QUOTE]

Infrared
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That is all fine, but the other four axioms do not distinguish hyperbolic space from the Euclidean plane, so there cannot be a proof of the parallel postulate just from these other axioms (since then any such proof would be just as valid in the hyperbolic plane, where the parallel postulate is false).

It follows that if you want to prove the parallel postulate, you must assume something other than Euclid's four other axioms. What else are you taking as axiom?

jbriggs444 and binis
pbuk
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With all my respect,you undoubtfully know that we are reffering to a regular plane.
There is a misunderstanding here, let me see if I can clear it up.

I am afraid that only you are referring to a 'regular plane', because nobody else understands what that means. The other posters on this thread are referring to all 2-dimensional geometries where Euclid's first four axioms hold. Such geometries are referred to as geometries of constant curvature and can be divided into three groups:
• Euclidean geometry, which is defined as the geometry of constant curvature with the axiom "given any straight line and a point not on it, there exists one and only one straight line which passes through that point and never intersects the first line, no matter how far they are extended", in other words the parallel postulate is true
• Elliptic geometries, which are defined as geometries of constant curvature with the axiom "given any straight line and a point not on it, there does not exist any straight line which passes through that point and never intersects the first line, when they are both extended without bound", so the parallel postulate does not hold
• Hyperbolic geometries, which are defined as geometries of constant curvature with the axiom "given any straight line and a point not on it, there may exist more than one straight line which passes through that point and never intersects the first line, no matter how far they are extended", so again the parallel postulate does not hold.
Elliptic and hyperbolic geometries are together referred to as non-Euclidean geometries.

As you are claiming a proof of the parallel postulate, then you must be referring to Euclidean geometry (because in non-Euclidean geometry the parallel postulate is false). You must also be inferring some other definition of Euclidean geometry, otherwise the parallel postulate is an axiom not a theorem and doesn't need to be proved. From your post
Nevertheless,you can prove it otherwise:draw the unique vertical line β from the point A to the line α.Then draw the unique line γ vertical to β at the point A.The line γ is parallel to α and it is the unique.Do I use any axiom?
you have assumed that you can draw a unique line from a point A intersecting a line β at right angles. This is an alternative axiom that defines Euclidean geometry, and your proof is correct.

It is also possible to define Euclidean geometry with many other axioms instead of the parallel postulate including the equidistance postulate, Playfair's axiom, Proclus' axiom, the triangle postulate, and the Pythagorean theorem. In each case it is possible to prove the parallel postulate using that axiom together with Euclid's first four axioms.

In conclusion, as in many misunderstandings, it could be said that you are right, and so is everybody else.

You are right that in what you call a 'regular' plane (perhaps 'flat' would be a more commonly understood term), which is properly called a 2-dimensional Euclidean geometry the parallel postulate is provably true. You have presented an example of such a proof using the assumption of the existance of a unique perpendicular, which is provably true in Euclidean geometry.

Everybody else is right that you cannot prove the parallel postulate using only the first four axioms because non-Euclidean geometries exist where the first four postulates are axioms and the parallel postulate is provably false.

binis
I am afraid that only you are referring to a 'regular plane', because nobody else understands what that means.
A 'regular' plane is defined (and described) by a triangle (three intersected straight lines). Does n't?

jbriggs444
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A 'regular' plane is defined (and described) by a triangle (three intersected straight lines). Does n't?
No. It is not.

If you have an existing space on which your axioms apply then yes, three points are typically accepted as defining a "plane" within that space.

But if you are trying to define the space within which you are working, three points is not adequate to select the sub-space you are interested in.

In particular, one can have the same three points on a flat plane, on a sphere or on a saddle shape embedded within Euclidean 3-space. A two-dimensional sub-space (sphere, plane or saddle) is not uniquely selected by the three points.

etotheipi
three points are typically accepted as defining a "plane" within that space.
the same three points on a flat plane, on a sphere or on a saddle shape embedded within Euclidean 3-space. A two-dimensional sub-space (sphere, plane or saddle) is not uniquely selected by the three points.
I didn't say three points. I said 3 straight lines.

lavinia
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Summary:: Applying the transitive property of the parallelism to the Euclidean postulate you can prove it.Therefore it is not a postulate but a theorem.

Consider a point A outside of a line α. Α and α define a plane.Let us suppose that more than one lines parallels to α are passing through A. Then these lines are also parallels to each other; wrong because they all have common point A.
I think that in some sense, two perpendiculars through two separate points on a line must be parallel without the parallel postulate, But as @mathwonk explained there may be others. In Lobachesky geometry there are infinitely many parallels. This is the other possibility.

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you have assumed that you can draw a unique line from a point A intersecting a line β at right angles. This is an alternative axiom that defines Euclidean geometry, and your proof is correct.
No,this is not an axiom.In my school textbook is a theorem proven by arcs,not by a postulate.

I think that in some sense, two perpendiculars through two separate points on a line must be parallel without the parallel postulate,
Yes.This is proven by arcs.
But as @mathwonk explained there may be others.
No.You cannot draw others.Read former posts.

jbriggs444
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I didn't say three points. I said 3 straight lines.
So? Note that "straight" is an undefined term. You are not allowed to use it.

binis
So? Note that "straight" is an undefined term. You are not allowed to use it.
What is a triangle?

jbriggs444
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What is a triangle?
A geometric figure consisting of three points and the three line segments connecting them pairwise. The word "straight" does not appear in this definition. If you wish to use the term "straight", you must be prepared to define it first.

Often there is a restriction that the three points not be co-linear. Optionally, one accepts degenerate triangles where this restriction is not obeyed.

Getting back to the subject matter at hand, a "triangle" in the full space may not match a "triangle" in a lower dimensional subspace embedded therein. Nor do "lines" in the full space necessarily match "lines" in the subspace.

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A geometric figure consisting of three points and the three line segments connecting them pairwise. The word "straight" does not appear in this definition. If you wish to use the term "straight", you must be prepared to define it first.
I think that the definitions of the point, the (straight) line etc. are the first four postulates.If you know any definition please share it with us.Can the three line segments be curves?

jbriggs444
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I think that the definitions of the point, the (straight) line etc. are the first four postulates.
Those are more characterizations than definitions. The terms "point" and "line" are taken as undefined. The four postulates establish some relationships between the terms.
Can the three line segments be curves?
You have not provided a definition for "curve".

But I think I know what you mean. The answer is yes. A "straight" line in a sub-space can be a "curved" line in the full space. For instance, if we restrict our attention to the surface of a sphere, a great circle path is "straight" -- it is the shortest path between two points.

Infrared and pbuk
pbuk
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No,this is not an axiom.In my school textbook is a theorem proven by arcs,not by a postulate.
<sigh> In that case your 'proof by arcs' (I do not know what that is) must either be an axiom or a theorem proved by some other axiom in addition to the first four postulates. You must understand that noone else here has access to your school textbook so you need to put more effort in to explaining the terms that you use. Perhaps you could start by considering how your school textbook treats the parallel postulate: is it an axiom? If so, you cannot use any theorems derived from it in order to 'prove' it.

You keep stating that things that are true in Euclidean geometry are always true, despite being provided with many explanations and counter-examples. If you are not going to make the effort to understand what others are posting there is not much point in continuing this thread.

Have you tried learning about non-Euclidean geometry? These references at MathWorld and Wikipedia may help or you could just search for it.

Infrared and binis
The answer is yes.
So a cycle is a triangle.
a sphere, a great circle path is "straight" -- it is the shortest path between two points.
What is a sphere? What is circle? What does it mean "shortest"?

<sigh> In that case your 'proof by arcs' (I do not know what that is) must either be an axiom or a theorem proved by some other axiom in addition to the first four postulates. You must understand that noone else here has access to your school textbook so you need to put more effort in to explaining the terms that you use. Perhaps you could start by considering how your school textbook treats the parallel postulate: is it an axiom? If so, you cannot use any theorems derived from it in order to 'prove' it.
I had been surprised when I was reading my old textbook. Inside this,the two theorems of perpendicularity are proven by arcs.After that,in the next pages the parallel postulate is presented as an axiom.

jbriggs444
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So a cycle is a triangle.
I did not say that. Please use sarcasm more carefully.
What is a sphere? What is circle? What does it mean "shortest"?
A sphere is the set of all points equidistant from a chosen point -- in three dimensions.
A circle is the set of all points equidistant from a chosen point -- in two dimensions.

Shortest is difficult to define tersely. One way to approach it is to start with a distance measure on the space and the idea of a parameterized curve. A parameterized curve is a set of points that, loosely speaking, one can draw without picking up a pencil from the paper. The parameter is any numeric measure that smoothly increases from one end of the curve to the other. The length of the curve is the path integral of the distance measure from one end to the other. The "shortest path" from A to B is the set of points in the parameterized curve that starts at A and ends at B and which has the smallest length. [I am not sure that this is the standard approach. I've never been exposed to a formal exposition of the notion of path length].

Edit: Wikipedia uses pretty much the same approach. They call it a differentiable function rather than a parameterized curve. Six of one, half dozen of the other.

There.

I had definitions available for the terms I used. You have yet to present definitions for the terms you use.

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binis
I did not say that. Please use sarcasm more carefully.
I did not sarcasm.Your definition for the triangle also applies to the cycle.
You have yet to present definitions for the terms you use.
A straight line is the shortest line connecting two points.

jbriggs444
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I did not sarcasm.Your definition for the triangle also applies to the cycle.A straight line is the shortest line connecting two points.
What is a "cycle"?

In Euclidean geometry, there is only one line connecting two points. So it is automatically the shortest. That means that the adjective "straight" by your definition conveys no information.

What is a "cycle"?
A cycle is the set of all points equidistant from a chosen point -- in two dimensions.

jbriggs444
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A cycle is the set of all points equidistant from a chosen point -- in two dimensions.
Ahh. I would have called that a "circle". A language difficulty, it seems.

So you have in mind that if we pick out three points on it then the equator on a sphere counts as a degenerate "triangle" on the surface of that sphere. Yes, it seems to fit the definition.

Similarly, I expect that you consider a line segment to be a triangle.