The connection between two geometrical axioms of H. N. Gupta

Wolfram Schwabh{äuser
<span title="1969-01-01">1969</span> <i title="American Mathematical Society (AMS)"> <a target="_blank" rel="noopener" href="" style="color: black;">Proceedings of the American Mathematical Society</a> </i> &nbsp;
The answer is given here to a problem raised by H. N. Gupta at the end of [l ], namely, if two axioms (E) and (B) are equivalent when added to Euclidean geometry of arbitrary dimension over arbitrary ordered fields. (B) states that, for any points x, y, z such that y is between x and z, there is a right triangle having x and z as endpoints of the hypotenuse and y as foot of the perpendicular drawn from the right angle vertex. (B) holds in the w-dimensional Cartesian space SniS) over an ordered
more &raquo; ... ield % (n ^ 2) if and only if (b) for all au • • • ,an,lE^ with 1^0, the system of the two equations (1) aixi + • • • + anx" = 0, (2) x\+ ■ ■ ■+xl = l-(a\+ ■ ■ -+ al) has a solution with xi, • • • , xnin %. (E) is, essentially, a combination of the axiom of segment construction and the circle axiom. It is analytically expressed (as above) by (e) % is Euclidean (i.e., in $. each positive element is a square). As stated in [l], (E) implies (B) (also for infinite dimension), and the converse holds in Euclidean geometry over Pythagorean ordered fields. It will be shown here that (B) does not imply (E) in general, in fact, that we get a counterexample, satisfying (b), by taking the field O of rationals for % together with any «^5, while, clearly, (e) does not hold for fÇ = O. Thus, the answer to the above problem is negative. First, let the ordered field % and n^2 still be arbitrary.
<span class="external-identifiers"> <a target="_blank" rel="external noopener noreferrer" href="">doi:10.1090/s0002-9939-1969-0246189-1</a> <a target="_blank" rel="external noopener" href="">fatcat:62cla3j4jzajrlstnzkbqcrsfq</a> </span>
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