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Playfair's axiom states:

Through any point in the plane, there is at most one straight line parallel to a given straight line.

This axiom is equivalent to the parallel postulate.

Intuitively, we know that if the point $P$ does not lie on the given straight line $\mathcal{l}$, there is exactly one straight line through $P$ that is parallel to $\mathcal{l}$. But how do we prove this? From Playfair's axiom, we know that there is at most one such straight line. But how do we conclude, from the first four postulates, that there is at least one?

Note: My question is different from Parallel postulate from Playfair's axiom. I have provided an answer to my own question.

Siddhartha
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  • What is the definition of "parallel" here? That the lines do not intersect? In that case it might simply be impossible to show existence: think of geometry on a sphere, where every two straight lines (i.e. great circles) intersect. Or is there another axiom (from the first 4) that does not hold on the sphere? – Vincent Dec 27 '19 at 11:56
  • @Vincent: spherical geometry doesn't satisfy what we normally considered to be "Euclid's postulates". If it was that easy, people would have discovered non-Euclidean geometry long ago. – calcstudent Dec 27 '19 at 12:03
  • @calcstudent I really like that "If it was that easy" argument, it makes a lot of sense. Still I find it hard to see what axioms are violated. Of course right-angles are all equal, I cannot even begin to imagine what non-equal right angles would look like. Similar with "through every point you can draw a line". Ehm I forgot the other two, but can you tell me what goes wrong? Or is it just that people were reluctant to consider a great circle a straight line as it looks so much more like a circle (which, obviously, is non-straight)? – Vincent Dec 27 '19 at 12:07
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    @Vincent: I will just have to put the wikipedia article here for details https://en.wikipedia.org/wiki/Spherical_geometry – calcstudent Dec 27 '19 at 12:09
  • @calcstudent thank you! Wikipedia explains it quite well! – Vincent Dec 27 '19 at 12:11
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    Does this answer your question? Parallel postulate from Playfair's axiom – calcstudent Dec 27 '19 at 12:26
  • @calcstudent Really, no. I already know and have stated that Playfair's axiom is equivalent to the parallel postulate. – Siddhartha Dec 27 '19 at 12:29
  • @Thomas: you asked for a proof that there is at least one parallel line from the first 4 postulates. That proof written in there is a more detailed version of my (now deleted) answer, which is why I delete it. – calcstudent Dec 27 '19 at 12:30

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Let $O$ be any point on $\mathcal{l}$. Join $OP$. Now draw a line $\mathcal{m}$ through $P$ such that the alternate interior angles, where $OP$ is a transversal to $\mathcal{l}$ and $\mathcal{m}$, are equal. Then, by Proposition 1.27 of Euclid's Elements, which does not rely on the parallel postulate, $m\parallel l$. So, there is at least one such line.

Siddhartha
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