Mathematical recreations and essays . it unsuccessfully through infinite space, just as a curve pursuesits asymptote and never catches it. Froduss Froof of the Fostidate. Proclus, after criticisingPtolemys demonstration, gave a proof of his own, but in thecourse of it he assumed that if two intersecting straight linesbe produced far enough the distance between a point on one ofthem and the other line can be made greater than any assignedfinite length, and that if two parallel straight lines be producedindefinitely the perpendicular from a point on one of them tothe other remains finite. On the
Mathematical recreations and essays . it unsuccessfully through infinite space, just as a curve pursuesits asymptote and never catches it. Froduss Froof of the Fostidate. Proclus, after criticisingPtolemys demonstration, gave a proof of his own, but in thecourse of it he assumed that if two intersecting straight linesbe produced far enough the distance between a point on one ofthem and the other line can be made greater than any assignedfinite length, and that if two parallel straight lines be producedindefinitely the perpendicular from a point on one of them tothe other remains finite. On these assumptions the postulatecan be proved. But just as we cannot assume that two con-verging lines (ex. gr. a curve and its asymptote) will ultimatelymeet, so we must not assume that the distance between twodiverging lines will be ultimately infinite. Walliss Froof of the Fostulate. I will give next a demonstra-tion offered by J. Wallis, Savilian Professor of Geometry, in alecture delivered at Oxford on July 11,1663. The substance of. CH. XV] THE PARALLEL POSTULATE 315 his argument may be put thus*. It is desired to prove that iftwo lines AB, CD meet a transversal HACK, so that the sumof the angles BAG, ACD is less than two right angles, then ABand CD must (if produced) meet. One of the angles BAG, ACDmust be acute; suppose it is BAG. He first showed that, inthis case, from any point J5 in we can draw a line BEwhich will cut AG m E, so that the sum of the angles BEG,EGD is equal to two right angles; hence the angle BE A isequal to the angle DC A. Then if we take the triangle BAE(drawn on ^^ as base and with B as vertex) and constructa similar triangle on AG as base, he proved that its vertexmust be at a finite distance from AG, must lie on AB produced,and must lie on GD (produced if necessary). Hence AB andGD when produced must meet. The proof is ingenious, but it rests on the assumption thatit is possible to construct a triangle on any specified scalesimilar to a given tr
Size: 2239px × 1116px
Photo credit: © The Reading Room / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookcentury1900, bookdecade1920, booksubjec, booksubjectastrology
