the following attempted proof of the parallel postulate is similar to Proclus\'
ID: 3406247 • Letter: T
Question
the following attempted proof of the parallel postulate is similar to Proclus' but the flaw is different; detect the flaw with the help of exercise 1. Given P not on line l, pq perpendicular to l at q, and line m perpendicular to pq at p. let n be any line through p distinct from m and pq. We must show that n meets l. Let px be a ray of n between pq and a ray of m emanting from p and let y be the foot of the perpendicular from x to pq. As x recedes endlessly from p, py increases indefinitely. Hence, y eventually reaches a position y' on ray pq such that py'>pq. Let x' be the corresponding position reached by x on the line n. New x' and y' are on the same side of l because line x'y' is parallel to l. But y' and p are on oppsite sides of l. Hence, x' and p are on opposite sides of l, so that segment px' meets l.
Explanation / Answer
Given P not on line l, pq perpendicular to l at q, and line m perpendicular to pq at p. let n be any line through p distinct from m and pq. We must show that n meets l. Let px be a ray of n between pq and a ray of m emanting from p and let y be the foot of the perpendicular from x to pq.
Till this point it is fine. Also As x recedes endlessly from p, py increases indefinitely is true. But this increase is not bounded is not accomplished. Because a sequence may increase indefinitely yet it may be bounded by a fixed number and may not cross that number at any time. For example, 1/2, 3/4, 7/8, 9/16, 31/32, 63/64,.. is increasing but it never crosses the point 1. Thus saying that y eventually reaches a position y' on ray pq such that py' >pq requires justification.
Hence the argument is not complete. It is assuming that the distance between the two lines l and m is constant and also assuming that it can be crossed eventually.
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