Is the Euclidean postulate a theorem?

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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.

Main Question or Discussion Point

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.
 
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  • #2
mathwonk
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that is a nice observation. unfortunately one cannot prove that parallelism satisfies transitivity without using the postulate referred to.
 
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  • #3
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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?
 
  • #4
mathwonk
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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
 
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  • #5
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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?
 
  • #6
mathwonk
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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.
 
  • #7
wrobel
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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
 
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  • #8
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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?
 
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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.
 
  • #10
jbriggs444
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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.
 
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  • #11
Svein
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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).
 
  • #12
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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?
 
  • #13
jbriggs444
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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].
 
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  • #14
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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?
 
  • #15
jbriggs444
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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.

1589455966433.png
 
  • #16
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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.
 
  • #17
jbriggs444
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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.

1589457054931.png

You can construct a whole family of parallel lines based on where you put the second perpendicular.
 
  • #18
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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?
 
  • #19
jbriggs444
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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##.
1589461468063.png


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]
 
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  • #20
Svein
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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 realised 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)
 
  • #21
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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.
 
  • #22
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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.
 
  • #23
jbriggs444
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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
 
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  • #24
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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.
 
  • #25
jbriggs444
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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.
 

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