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Let $l$ be a straight line, and $P$ be a point not on it. We do not assume the parallel postulate. Then from one of the equivalent versions of the absolute geometry, it is not clear that we can draw a line $l'$ parallel to $l$ while containing $P$.

However, let us proceed as follows. I can draw $l''$ perpendicular to $l$ passing $P$, and then draw $l'$ perpendicular to $l''$ at $P$. By the alternate angle theorem (more precisely, its converse, as it is usually stated) which does not require the parallel axiom, we have $l'\parallel l$.

Where was I possibly wrong?

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    $\begingroup$ There is nothing wrong: that a parallel to a line passing through a given point exists is a theorem in absolute geometry. Uniqueness of the parallel requires the parallel postulate. See also here: math.stackexchange.com/questions/2379549/… $\endgroup$ – Aretino Feb 10 '18 at 20:59
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I would modify the wording of @Aretino's comment:

"that a [line not having common points] to a line passing through a given point exists is a theorem in absolute geometry."

A line having no common points to another line is called an ultra parallel line. With this the sentence above reads:

"That an ultra parallel line to a line passing through a given point exists is a theorem in absolute geometry."

In the Euclidean geometry the following concepts are equivalent: ultra parallel, equidistant, parallel.

In the hyperbolic geometry parallel or asimptotically parallel and equidistant and ultra parallel are different concepts.

Consider the following figure:

enter image description here

In absolute geometry: If we drop a perpendicular to $l$ from $P$ and then another perpendicular to this latter line then we will get a line $l'$ that will be ultraparallel to $l$. That is, a line not having common points to $l$.

Since this is an absolute theorem, it is true both in Euclidean and hyperbolic geometries.

In hyperbolic geometry: $l'$ constructed the way given above will be simply an ultra parallel line.

In Euclidean geometry the said line will be equidistant to $l$.

How to construct the asimpptotitically parallel lines in hyperbolic geometry? The following is the Bolyai construction:

Drop a perpendicular from $P$ to $l$ and construct $l'$, a perpendicular line at $P$ to $PA$. Then take an arbitrary point $B\not =A$ on $l$. Drop a perpendicular from $B$ to $l'$. Draw a circle of radius $AB$ around $P$. Construct the straights joining $P$ and the intersection points of the circle and $b$.

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