# Neutral Geometry

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• July 9th 2010, 10:51 AM
MATNTRNG
Neutral Geometry
Prove the following Theorem. The Euclidean Parallel Postulate is equivalent to the following statement: If a line is perpendicular to one of two parallel lines, then it is perpendicular to the other.

I have already been able to prove that if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other by assuming that the Euclidean Parallel Postulate is true and arriving at a contradiction (indirect proof).

I am having trouble proving the Euclidean Parallel Postulate by assuning that if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other is true.

Thanks in advance
• July 9th 2010, 11:54 AM
wonderboy1953
Since you're saying that the lines are parallel, then you're using circular reasoning in proving the Parallel Postulate (on your third paragraph).
• July 9th 2010, 03:44 PM
MATNTRNG
Any thoughts on what I should do?
• July 11th 2010, 04:00 AM
HallsofIvy
Quote:

Originally Posted by wonderboy1953
Since you're saying that the lines are parallel, then you're using circular reasoning in proving the Parallel Postulate (on your third paragraph).

No, he is trying to prove that the parallel postulate is equivalent to the "If a line is perpendicular to one of two parallel lines, then it is parallel to the other". That is he must:
1) Assume the parallel postulate and show that the "perpendicular" theorem follows.
2) Assume the "perpendicular" property and show that the parallel postulate follows.

What, exactly, is the phrasing of the "Euclidean parallel postulate" you are using? (There are several equivalent forms.) Since one of those is "the sum of angles on one side of a line crossing two parallel lines is one straight angle", that should follow from the "perpendicular" theorem fairly easily.
• December 19th 2012, 05:24 PM
cathectio
Re: Neutral Geometry
Imagine one circle above another equal to it in 3D. Where they meet on an intersecting piece of paper, they will be perpendicular between 2 pairs of points, though they are parallel to each other in space. BUT A POINT IS NOT A LINE, SO EVEN THOUGH THE CIRCLES ARE COMPRISED OF LINES, THEY DEFEAT EUCLID'S PARALLEL POSTULATE.
• December 19th 2012, 05:28 PM
cathectio
Re: Neutral Geometry
In other words, they are everywhere parallel and everywhere perpendicular, but not in a single plane, and thus their perpendicularity does not exist as between 'lines' per se, again, even though the two circles are indeed comprised of two 'lines.'
• December 19th 2012, 05:30 PM
cathectio
Re: Neutral Geometry
As for what this has to do with the principle of "neutrality", would you please explain that to me?
• January 1st 2013, 02:21 PM
blmalikov
Re: Neutral Geometry
Assume your "perpendicular postulate" and assume the negation of the parallel postulate to derive a contradiction.

Assume you have two lines l_1 and l_2 that are traversed by another line p. Let p be perpendicular to l_1. Either l_1 and l_2 are parallel or not. If they are, then by your postulate p is perpendicular to both, i.e. the sums of the interior angles add up to 180 on both sides, and by the negation of the parallel postulate l_1 and l_2 intersect on both sides. Contradicting the notion of parallel lines. We have only one more option. If l_1 and l_2 are not parallel, then there is a side of p such that the interior angles add up to less than 180 degrees. By the negation of the parallel postulate, if l_1 and l_2 are extended indefinitely they will never intersect. The same reasoning applies to the other side of p. And we are left with that l_1 and l_2 never intersect - the definition of parallel. Thus we reach a contradiction.
• January 1st 2013, 03:36 PM
Plato
Re: Neutral Geometry
Quote:

Originally Posted by cathectio
As for what this has to do with the principle of "neutrality", would you please explain that to me?

In that study of axiomatic geometry the term neutral geometry
has very specific meaning. Namely the axiom set is that of David Hilbert (as modified by RL Moore) minus the parallel axiom.

In neutral geometry there is no axiom that says; For every line $\ell$ and every point $P\notin\ell$ then there is at most one line $j$ such that $P\in j$ and $j\| \ell$.

In other words, there is no parallel axiom.
• January 1st 2013, 04:43 PM
cathectio
Re: Neutral Geometry
David Hilbert (as modified by Mr. Moore) is definitely adequate authority. I, however, mean something different by "Neutral Operations" and thus "Neutral Geometry":
If two line segments a and b are added in a contiguous straight line, that is a "distance", commonly thought of.
And if they are multiplied by each other, then that is an "area". commonly thought of.
So what happens when
a+b = a*b
a+b-b = ab-b
a = b(a-1)
a/(a-1) = b AND b HAS BEEN ISOLATED AND DEFINED IN TERMS OF a AND 1, SUCH THAT IF WE SUBSTITUTE A VALUE FOR a, b IS DETERMINED:
Let a = 5
a/(a-1) = b = 5/4
5 + 5/4 = 5 * 5/4 = 25/4; CHECK √
Is Nature now allowed to decide whether the 25/4 will be a distance or an area? Is she "neutral"? THAT IS WHAT I MEAN BY "NEUTRAL GEOMETRY".

ALSO:
ab = a^b
ab/a = a^b /a
b = a^(b-1)
b^(1/(b-1)) = a
Let b = 63
b^(1/(b-1)) = a
63^(1/62) = a = 1.06910810446532

a*b = a^b?
1.06910810446532 * 63 = 67.3538105813149
1.06910810446532^63 = 67.3538105813307 PRETTY DAMN CLOSE ANYWAYS. Probably an XL calculation error ... theory's just fine.

CG