This article offers an indirect proof of an implication of Euclid’s
fifth postulate: that *through any given point off a line, there
exists one and only one other line parallel to the first. *We will
show this by assuming the contrary and exposing the contradiction in
such an assumption.

In
“The
Universal Validity of Euclidean Geometry,”
I had irrefutably demonstrated that two parallel lines do not by
definition intersect. As in that demonstration,
let us assume that L_{1} and L_{2}
have slope m. At the y-axis, L_{1} has coordinates (0,y_{1}),
and L_{2} has coordinates (0,y_{2}). The vertical
distance between L_{1} and L_{2} along the y-axis is
(y_{2}-y_{1}).

Let us
assume, for the purposes of this proof, that L_{1} has a *
second *line parallel to it through the point (0,y_{2}). We
will call the line L_{3}. If L_{3} is to be a unique
line, it cannot be equal to L_{2 }and must thus differ from L_{2},
which is y=mx+y_{2}. Because we defined L_{3} as
passing through (0,y_{2}), it cannot differ from L_{2}
in that respect. It can only differ in its slope, which would have to
be unequal to m. Yet m is also the slope of L_{1}, and
mutually parallel lines are defined to have the same slope. Thus, if L_{3}
is y=nx+y_{2}, where n≠m, it cannot by definition be parallel
to L_{1}. The second line we assumed to be parallel to L_{1}
is shown not to be parallel to it by the implications of that very
assumption.

Also, L_{3
}will eventually intersect L_{1}. L_{1 }is y=mx+y_{1}
and L_{3} is y=nx+y_{2}, where n≠m. To demonstrate my
contention, we can try setting the two equal to each other and find
the x-value at which they intersect:

mx+y_{1}=nx+y_{2}

mx-nx=y_{2-}
y_{1}

(m-n)x=y_{2-}
y_{1}

**x = (y**_{2-}y_{1})/(m-n)

Because,
by our conditions, n≠m and y_{2}≠y_{1}, both the
numerator and denominator of the above expression have nonzero real
values. Dividing them produces a real value for x, which hence *must
occur *somewhere along the progression of L_{1} and
L_{3}. Of course, if L_{3
}intersects
L_{1, }the two lines cannot be
parallel. L_{1 }has only one parallel line to it at a
distance of separation of (y_{2}-y_{1}). That parallel
line is L_{2}, which will never intersect L_{1}.

Thus, we
have proved that *through any given
point off a line, there exists one and only one other line parallel to
the first, *by showing any contrary instance of it to be false and
contradictory.

**G. Stolyarov II is a
science fiction novelist, independent filosofical essayist, poet,
amateur mathematician, composer, contributor to
Enter Stage Right,
The Autonomist,
Le Quebecois Libre, and the
Ludwig von Mises Institute, Senior Writer
for
The Liberal Institute, and Editor-in-Chief
of
The Rational Argumentator, a magazine
championing the principles of reason, rights, and progress. His newest
science fiction novel is ***
Eden against the Colossus*. His latest
non-fiction treatise is *
A Rational Cosmology*. Mr. Stolyarov can
be contacted at
gennadystolyarovii@yahoo.com.

**Read****
Mr. Stolyarov's** **new comprehensive treatise,** **
A*** Rational Cosmology, *explicating such
terms as the universe, matter, space, time, sound, light, life,
consciousness, and volition, at
http://www.geocities.com/rational_argumentator/rc.html.