----- Mensaje de "Papirfoldning.dk" <h...@papirfoldning.dk>
Thanks a lot, Hans, for sharing such an interesting discovery!
Since O1-O7 are independent axioms, and O4 is Euclidean (you can
construct a line through a point, perpendicular to another line), it
seems that Donovan's postulates cannot construct all Euclidean
constructs, despite his claim.
I believe they actually can.
On one hand, the manouvre described in O4 may be regarded as a
particular case of O5, when the first point (the one you are moving,
say P) belongs to the given line (say, l) thus, you are folding
perpendicularly to l. So, you just need to choose ANY point of l to
perform O4.
So, in most of the theoretical settings I can imagine, you can do it
by using the other axioms and something else.
If your theory lets you use the axiom of choice (most of the formal
settings for constructions do NOT), you are done.
If you are using the axioms with an initial set consisting in two
points (as it is customary), then you can construct two perpendicular
lines (namely, Y=0 and X=1/2, in cartesian notation, if you started
with (0,0) and (0,1)). At least one of those lines intersects the
original line l and you are done.
On the other hand, it is well known that the first five axioms O1-O5
give the Euclidean constructions (that is, using compass and ruler),
if you have (0,0) and (0,1) as initial set.
So, the "only" thing Johnson was missing is O6.
All the best!
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JOSE IGNACIO ROYO PRIETO
Matematika Aplikatua Saila
University of the Basque Country UPV/EHU
Pza. Ingeniero Torres Quevedo, 1
48013 Bilbao
SPAIN
Phone: 00 34 946013987
FAX: 00 34 946014244
E-mail: joseignacio.r...@ehu.eus
Web: http://www.ehu.eus/joseroyo
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