Is the Euclidean postulate a theorem?

[binis]

TL;DR 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.

 

[mathwonk]

that is a nice observation. unfortunately one cannot prove that parallelism satisfies transitivity without using the postulate referred to.

 

[binis]

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?

 

[mathwonk]

yes. you used the equivalent assumption (axiom) that if alpha is perpendicular to beta, then the only line y parallel to line alpha, and passing through point A on beta, must also be perpendicular to beta.

I.e. one can indeed prove that any two lines making the same angle with a third line, are parallel to each other. (This is proposition 28, Book 1, of Euclid's Elements.) But the converse, which you are using, that any two parallel lines must make the same angle with a third line they both meet, is not provable without the 5th postulate. (This is proposition 29, Book 1 Euclid.) Of course just because Euclid uses the 5th postulate to prove People. 29 does not prove that it could not be proved without using it, but that is in fact true, if harder to show. For that one has to construct a "non euclidean model" of geometry.

This problem baffled people for ages until it was discovered that there is another geometry, also satisfying all axioms except the 5th, and where the 5th is false. For the simplest rough sort of example, think of "table top geometry" where one can easily find two lines through a common point, and not either of them meeting a third line, simply because the table is not big enough. Of course this seems flawed because the table is not very large, but one can extend such a table so that the two lines still do not meet, by making the extended surface curved like the ruffles on a skirt.

There are many good books on this subject (neutral geometry). Here are some free notes which discuss your question in the first few pages. I.e. uniqueness of perpendiculars does not imply uniqueness of parallels.

https://www.math.ust.hk/~mabfchen/Math4221/Neutral Geometry.pdf

 

[binis]

that any two parallel lines must make the same angle with a third line they both meet, is not provable without the 5th postulate.

Thanks a lot,but this caused me a new query:the educative way for the "division of a given line segment in equal parts" is based on the 5th postulate? If so, then it is incorrect?

 

[mathwonk]

I apologixze that I cannot answer all questions on this subject since they are so numerous. Please enjoy one of the good books on the topic. I think you will enjoy the study. Euclid is the first recommended book.

 

[wrobel]

It is quite banal: if you take a set of propositions as axioms then another set of propositions are the theorems and converse
5 Euclidean postulate can not be deduced from the other Euclidean axioms because the Lobachevski plane exists

 

[binis]

1)Draw the unique vertical line β from the point A to the line α. 2)Then draw the unique line γ vertical to β at the point A.These are well known as the two theorems of perpendicularity and they have proven by arcs.
The line γ is parallel to α and it is unique due to the second theorem.It is the unique vertical to a unique vertical.It is unique and it is parallel. Where is the misunderstanding?

 

[binis]

I think that this is a misleading.If an aspect dominates for ages it does not make it right.Till last century scientists was thinked that flies have four legs because Aristotle have written it.

 

[jbriggs444]

1)Draw the unique vertical line β from the point A to the line α. 2)Then draw the unique line γ vertical to β at the point A.These are well known as the two theorems of perpendicularity and they have proven by arcs.
The line γ is parallel to α and it is unique due to the second theorem.It is the unique vertical to a unique vertical.It is unique and it is parallel. Where is the misunderstanding?

As I understand it, you claim to have proved that given a line $\alpha$ and a point $A$ not on that line, you can construct a line through the point which is parallel to the original line.

You assert that this line is unique. And it is -- the line you construct is the only line that your construction produces from a given $A$ and $\alpha$.

But it is not necessarily the only line parallel to $\alpha$ that passes through point A.

 

[Svein]

First: Define "parallel". Then try to prove that it can be deduced form the other axioms (hint: Using "obvious" results that are not directly proven from the other four axioms do not count as proofs).

 

[binis]

As I understand it, you claim to have proved that given a line $\alpha$ and a point $A$ not on that line, you can construct a line through the point which is parallel to the original line.
You assert that this line is unique. And it is -- the line you construct is the only line that your construction produces from a given $A$ and $\alpha$.
But it is not necessarily the only line parallel to $\alpha$ that passes through point A.

Can we draw another line through the A which is parallel to the α?

 

[jbriggs444]

Can we draw another line through the A which is parallel to the α?

For some geometries, yes.

The simple example is a geometry on a surface with constant negative intrinsic curvature -- a sort of saddle shape. On this surface, consider two lines that are locally "parallel" in the sense that they point in the same direction but do not touch. As you follow these lines toward infinity, you will find that they diverge from each other in both directions.

In this geometry you will find that there is a small range of angles for which the lines will still diverge in both directions. [Along with a critical angle at which they will converge without meeting in the one direction and at which they will converge without meeting in the other].

 

[binis]

You assert that this line is unique. And it is -- the line you construct is the only line that your construction produces from a given $A$ and $\alpha$.

For my construction it is.Could be other constructions produce another line?

 

[jbriggs444]

For my construction it is.Could be other constructions produce another line?

Yes. I already spoke about existence. Now you want constructibility.

Perform your construction -- drop a perpendicular from point $A$ to line $\alpha$. Set your compass to the distance between $A$ and the intersection point of the perpendicular with $\alpha$. Use the compass to mark of a point this distance down line $\alpha$ from the point of intersection. Draw a perpendicular to $\alpha$ from this point and mark off a point on this perpendicular at the same distance as was set before (and on the same side of $\alpha$ as before). Take the perpendicular to the perpendicular at this point and you have a line that is locally "parallel" to $\alpha$.

For Euclidean geometry, the new construction produces the same line as the old. For non-Euclidean geometries, such is not assured.

 

[binis]

Take the perpendicular to the perpendicular at this point and you have a line that is locally "parallel" to $\alpha$.

The line is parallel but is not passing through A.

 

[jbriggs444]

The line is parallel but is not passing through A.

You are right. I got caught up in the mechanics of drawing and forgot the goal. But that is a minor fix.

At the last step, instead of drawing a line perpendicular to the perpendicular, draw a line passing through A.

You can construct a whole family of parallel lines based on where you put the second perpendicular.

 

[binis]

At the last step, instead of drawing a line perpendicular to the perpendicular, draw a line passing through A.

How can you prove that this line is parallel to α?

 

[jbriggs444]

How can you prove that this line is parallel to α?

With reference to Euclid's axioms, I am not sure. However, some informal handwaving makes it clear that it must be parallel in the sense of having no intersection with $\alpha$.

The new construction is sandwiched between two lines that you claim are parallel to $\alpha$ and intersects each of them at a point not on $\alpha$. It cannot intersect $\alpha$ without intersecting at least one of them again which would contradict the properties of unique lines.

So if the new construction is different from the old, it must nonetheless still construct a parallel. [And if it is the same as the old, you already agree that it is parallel]

 

[Svein]

For two thousand years, many attempts were made to prove the parallel postulate using Euclid's first four postulates. The main reason that such a proof was so highly sought after was that, unlike the first four postulates, the parallel postulate is not self-evident. If the order the postulates were listed in the Elements is significant, it indicates that Euclid included this postulate only when he realized he could not prove it or proceed without it.[10] Many attempts were made to prove the fifth postulate from the other four, many of them being accepted as proofs for long periods until the mistake was found. Invariably the mistake was assuming some 'obvious' property which turned out to be equivalent to the fifth postulate (https://en.wikipedia.org/wiki/Parallel_postulate)

 

[binis]

With reference to Euclid's axioms, I am not sure. However, some informal handwaving makes it clear that it must be parallel in the sense of having no intersection with $\alpha$.
View attachment 262767
So if the new construction is different from the old, it must nonetheless still construct a parallel. [And if it is the same as the old, you already agree that it is parallel]

You constructed a parallelogram so yes, indeed the "new" line is parallel to α. But it is not a new line.It is the perpendicular.The three lines is the same, the original.
It seems to me that we are all confused playing a word game.So I will keep claiming my statement: the 5th postulate is a theorem.

 

[binis]

For two thousand years, many attempts were made to prove the parallel postulate using Euclid's first four postulates. The main reason that such a proof was so highly sought after was that, unlike the first four postulates, the parallel postulate is not self-evident. Invariably the mistake was assuming some 'obvious' property which turned out to be equivalent to the fifth postulate (https://en.wikipedia.org/wiki/Parallel_postulate)

If a belief is remaining for ages does not make it true.Once upon a time people believed that the Earth is flat. Mathematical logic (ML) is a branch of mathematics.For the ML any rational proposition is either true or false.Using the ML,since we can construct only one line,the proposision of the parallel postulate is true.

 

[jbriggs444]

You constructed a parallelogram so yes, indeed the "new" line is parallel to α. But it is not a new line.It is the perpendicular.The three lines is the same, the original.
It seems to me that we are all confused playing a word game.So I will keep claiming my statement: the 5th postulate is a theorem.

You have failed to understand how locally straight lines project on the surface of a saddle function. The drawings I have produced are accurate depictions and conform to the four axioms but not the fifth.

For the ML any rational proposition is either true or false

"Truth" is more nuanced than you imagine.

A proposition may be provable, disprovable or neither. "Truth" is relative to a model in which the axioms may or may not hold. Provability does not always imply truth. Truth does not always imply provability.

There are models in which the four axioms hold but in which the fifth does not.

But you are in good company. Marilyn vos Savant failed to grok this aspect of mathematics also. See https://dms.umontreal.ca/~andrew/PDF/VS.pdf

 

[binis]

You have failed to understand how locally straight lines project on the surface of a saddle function. The drawings I have produced are accurate depictions and conform to the four axioms but not the fifth.

I am sorry I failed to imagine a saddle.We were studied a plane defined by α and A.

 

[jbriggs444]

I am sorry I failed to imagine a saddle.We were studied a plane defined by α and A.

We are talking about an abstract space characterized by a set of axioms. The fact that you picture a flat plane in your mind is not relevant.

A plane fits the axioms. But so does a saddle or potato chip shape. There is a picture here.

 

[binis]

The fact that you picture a flat plane in your mind is not relevant.

May I have a short imagination.

 

[binis]

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.

 

[Infrared]

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]

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?

 

[pbuk]

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 existence 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]

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.

 

[binis]

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]

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.