ind he flaw in the following attempted proof of the parallel postulate by Wolfga
ID: 3110334 • Letter: I
Question
ind he flaw in the following attempted proof of the parallel postulate by Wolfgang Bolyai (Hungarian, 17 1856) (see Fig. 3). Given any point P not on a line construct a line l parallel to l through P in the usual way: a perpendicular toland construct perpendicular to PO. Let l" be any line through P distinct from l To see that intersects l, pick a point A on between Extend P beyond to a point B so that OB. Now Then the foot of the perpendicular from A and extend AR passing through them. Since l is R A, B, and Care not collinear, so there exists that MR RC a unique circle the perpendicular bisector of the of this and l is the perpendicular bisector of the land must in the center of the circle.Explanation / Answer
PB is not the last ray between rays PA and PQ that intersects l, and hence all rays between PA and PQ meet l"
The "hence" here is asserted without proof. In fact, it does not follow. All we have shown is that given any point B there is a point C beyond it.
For example, given any real number b less than 2, there is another real number less than 2 but greater than b. Does it follow that all real numbers are less than 2?
Similarly, given any ray through P intersecting l at B, we have shown there is another ray through P intersecting l at C. Logically, it does not follow that all rays therefore intersect l.
There's not need for a counterexample; we just need to show the reasoning is flawed.
Like many of these attempted proofs that popped up throughout history, it's just a clever way to subtly hide the assumption of the parallel postulate by cloaking it in "geometric intuition". The "hence" is an appeal to the imagination; it is hard to imagine that such a ray between PA and PQ can exist without intersecting ll, but that's not proof that it doesn't exist
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