Download Equivalents to the Euclidean Parallel Postulate In this section we

Equivalents to the Euclidean Parallel Postulate
In this section we work within neutral geometry to prove that a number
of different statements are equivalent to the Euclidean Parallel
Postulate (EPP). This has historical importance. We noted before that
Euclid’s Fifth Postulate was quite different in tone and complexity
from his first four postulates, and that Euclid himself put off using the
fifth postulate for as long as possible in his development of geometry.
The history of mathematics is rife with efforts to either prove the fifth
postulate from the first four, or to find a simpler postulate from which
the fifth postulate could be proved. These efforts gave rise to a
number of candidates that, in the end, proved to be no simpler than the
fifth postulate and in fact turned out to be logically equivalent, at least
in the context of neutral geometry. Euclid’s fifth postulate indeed
captured something fundamental about geometry.
We begin with the converse of the Alternate Interior Angles Theorem,
as follows:
If two parallel lines n and m are cut by a transversal l , then
alternate interior angles are congruent.
We noted earlier that this statement was equivalent to the Euclidean
Parallel Postulate. We now prove that.
Theorem: The converse of the Alternate Interior Angles Theorem is
equivalent to the Euclidean Parallel Postulate.
~ We first assume the converse of the Alternate Interior Angles
Theorem and prove that through a given line l and a point P not on the
line, there is exactly one line through P parallel to l. We have already
proved that one such line must exist: Drop a perpendicular from P to l;
call the foot of that perpendicular Q. Let
. Now, through
point P use the protractor postulate to construct a line m perpendicular
to t. The Alternate Interior Angles Theorem guarantees that since the
alternate interior angles are all right angles, l and m are parallel. So it
all comes down to proving that there is only one such line.
To this end, suppose there were another line mN through P parallel to l.
By the converse to the Alternate Interior Angles Theorem, alternate
interior angles formed by l, mN, and t must be equal, and since t z l we
must have mN z t as well. But then m and mN would both be
perpendiculars to t though point P, which would violate the uniqueness
of rays forming a right angle with t through a given point (Protractor
Postulate). Thus, m = mN.
To prove the converse, assume that the EPP holds and that l and m are
parallel lines cut by a transversal t at points Q and P, respectively. We
prove that alternate interior angles are congruent. Suppose for
contradiction that alternate interior angles p1 (at point P) and p2 (at
point Q) are not congruent, and WLOG that :(p1) > :(p2). Using the
protractor postulate, create ray
on the other side of t from p2 such
that :(pRPQ) = :(p2). Then from the above theorem, since pRPQ
and p2 are alternate interior angles, it follows that lines l and
are
parallel. But that means there are two lines through P parallel to line l,
contradicting the parallel postulate. So p1 and p2 are congruent. €
Theorem: Euclid’s Postulate V is equivalent to the Euclidean Parallel
Postulate.
~ First we assume EPP and prove from it Postulate V. Suppose l and
m are two lines cut by a transversal t at points P and Q respectively in
such a way that the interior angles on one side of t have measures
adding to less than 180. Using the Protractor Postulate, construct
through point P a line n such that the interior angles formed by n and m
(on the same side of t as before) add up to 180.
Now a Corollary to the Alternate Interior Angles Theorem, states that
if two lines are cut by a transversal such that the interior angles on one
side of the transversal are supplementary, the lines are parallel. Thus n
and m are parallel, and by EPP n is the only line through P that is
parallel to m. Thus l is not parallel to m, and must meet. They must
meet on the side of t we have been working on or the triangle they
form would have angle measure greater than 180.
To prove the converse, let l’ be a line and P a point not on the line.
We have demonstrated how we can create a line through P parallel to
l’ by using perpendiculars, so we will let l be such a line, created by
dropping a perpendicular from P to l’ with foot Q, then forming line l
perpendicular to
at point P. We now prove there can be no other
lines through P parallel to l’. Suppose m is such a line. We know that
m cannot be perpendicular to
at P since l is, and so the angle it
makes with t must have measure less than 90 on one side of t. On that
side, the sum of the interior angles formed by m and l’ add to less than
180, and so l’ and m must meet on that side. Thus l’ and m cannot be
parallel. €
Theorem: The following statements are each equivalent to the
Euclidean Parallel Postulate (EPP):
1.
If l and l’ are parallel lines and
is a line such that t
intersects l, then t also intersects l’.
2.
If l and l’ are parallel lines and t is a transversal such that
, then
.
3.
If l, m, n, and k are lines such that
then either m = n or
4.
,
.
If l is parallel to m and m is parallel to n then either l = n or
.
~ To prove statement 1 from EPP, let l and l’ be parallel lines and
be a line such that t intersects l at point P. By EPP, there can
only be one line through point P that is parallel to l’, so t must intersect
l’.
To prove the converse, let l’ be a line and P a point not on l’. We have
demonstrated how we can create a line through P parallel to l’ by using
perpendiculars, so we will let l be such a line. We now prove there can
be no other lines through P parallel to l’. Let
be a line through
point P. By statement 1, since t intersects l, it must also intersect l’,
and so cannot be parallel to l’. Thus there can only be one line through
P which is parallel to l’.
To prove statement 2 from EPP, let l and l’ be parallel lines and t a
transversal such that
.
The EPP is equivalent to the converse of the Alternate Interior Angle
Theorem, so since l and l’ are parallel, alternate interior angles p1 and
p2 must be congruent, and since p1 is a right angle, p2 must be a right
angle, so that
.
To prove the converse, let l’ be a line and P a point not on l’. We
have demonstrated how we can create a line through P parallel to l’ by
using perpendiculars, so we will let l be such a line, created by
dropping a perpendicular from P to l’ with foot Q, then forming line l
perpendicular to
at point P. We now prove there can be no other
lines through P parallel to l’. Suppose m is a line through P parallel to
l’. Now
is a transversal for m and l’ that is perpendicular to l’ at
Q, and so by statement 2 it must also be perpendicular to m at P. But
this makes l and m both perpendiculars to
at P, contradicting the
Protractor Postulate. Thus there cannot be a second parallel m.
To prove statement 3 from EPP, we note that we have already proved
that statements 1 and 2 above are equivalent to EPP, so we can use
them. So let l, m, n, and k be lines such that
, and suppose that
. In particular,
since l and k are parallel and m intersects k, it must also intersect l by
statement 1. Moreover, by statement 2, since
as well.
But then, since and
, if we view l as a transversal for m and n
we have alternate interior angles congruent, and so
.
To prove the converse, let l’ be a line and P a point not on l’. We have
demonstrated how we can create a line through P parallel to l’ by using
perpendiculars, so we will let l be such a line, created by dropping a
perpendicular from P to l’ with foot Q, then forming line l
perpendicular to
at point P. We now prove there can be no other
lines through P parallel to l’. Suppose m is a line through P parallel to
l’. Drop a perpendicular from Q to m with foot F. Now
, so by statement 3 either
Since
Now since
and
, or
and
.
clearly share point Q, we must have
.
can only intersect m once, F = P. But this contradicts
the Protractor Postulate, since l and m are both perpendicular to
P.
at
To prove statement 4 from EPP, suppose that l is parallel to m and m is
parallel to n , and that l and n are not equal. We prove
. Suppose
for contradiction that l and n intersect. Then since l is parallel to m, by
statement 1, n and m intersect, a contradiction. Thus l and n cannot
intersect and are therefore parallel.
To prove the converse, let l’ be a line and P a point not on l’. We have
demonstrated how we can create a line through P parallel to l’ by using
perpendiculars, so we will let l be such a line, and suppose m is a
second line through P parallel to l’. But then l is parallel to l’ and l’ is
parallel to m so by statement 4, l is parallel to m, which cannot be since
they share the point P. Thus no such line m can exist. €
Whew!
Our next Tour de Force is to prove that the EPP is equivalent to the
claim that the angle sum for every triangle is 180. One direction of the
equivalence is easy; the other is messier and requires some properties
of real numbers.
Theorem: The statement that the angle sum for any triangle is 180 is
equivalent to the EPP.
~ We’ll do the easy direction first; let
through point C construct a line parallel to
be any triangle, and
.
Now,
(I’ll leave you
to fill in the betweenness details necessary to make all that work) and
by the converse to the Alternate Interior Angles Theorem (equivalent
to EPP),
and
.
Substitution gives
.
OK, now for the harder direction. We prove it by contradiction, so we
will actually show that if there is a line and external point through
which many parallels exist, then there is a triangle whose angle sum is
less than 180.
We need a lemma first, one vaguely reminiscent one part of the proof
of the Saccheri-Legendre Theorem.
Lemma: Let
be a segment and let points QN be a point such that
pPQQN is a right angle. Then for every g > 0 there exists a point T on
such that
.
~ Choose a point PNon the same side of
as QNsuch that pQPPN is
a right angle (i.e.
). On ray
find the point T1 with
QT1 = PQ. Next, find the point T2 with T1T2 = PT1. Continue in this
way, choosing Tn so that Q*Tn-1*Tn and PTn-1 = TnTn-1. This creates a
series of isosceles triangles
with base angles so that
. Moreover, each of the Tn’s are
interior to angle pQPPN. This gives us:
,
But
. Thus the sum of the
measures of the angles we are forming with vertex V must be less than
90. Now suppose for contradiction that
for every
. By the Archimedean Property of Real Numbers there is an n
for which
, so that
,
which contradicts our previous conclusion about this sum. Thus there
must be some for which
and so some for
which
.€
Now the other direction of the main theorem:
Let l be a line and P a point not on l through which there is more than
one line parallel to l. Build a line m in the usual way by dropping a
perpendicular with foot Q from P to l and creating a line m
perpendicular to
through P. Now pick another line m’ through P
parallel to l. The angle m’ makes with
is not of measure 90 so on
one side or the other of
makes an angle less than 90 with m’. On
that side pick a point S on m’ so that S is on the same side of m as Q.
Choose a point R on m such that R is on the same side of
as S.
Finally, using our lemma above, choose a point T on l such that T is on
the same side of
as S, and
. We will
show that the angle sum for
is less than 180.
First note that T was constructed to be on the S side of
it were not on the Q side of
then
and that if
would have to intersect m’ =
by Plane Separation. This cannot happen since m’ is parallel to l.
So T is interior to pQPS, and we have that
by additivity of angles. Moreover, we
chose S so that S is interior to pQPR, so we have
. This gives us:
or
or
€