Transcript Nikolai Lobachevsky (1792

Nikolai Lobachevsky
(1792-1856)
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• Born: 1 Dec 1792 in Nizhny Novgorod, Russia
Died: 24 Feb 1856 in Kazan, Russia
• Went to school and studied in Kazan under Martin
Bartels (1769 - 1833), who was a friend of Gauss.
• 1811 Master's Degree in physics and mathematics.
• 1814 he was appointed to a lectureship
• 1816 he became an extraordinary professor.
• 1822 he was appointed as a full professor
• 1820 -1825 dean of the Math. and Physics Department
• 1827 he became rector of Kazan University
Nikolai Lobachevsky
(1792-1856)
Accepting the possibility of HAA
Lobachevsky redefines parallels in the
following way:
16. All straight lines which in a plane go out
from a point can, with reference to a given
straight line in the same plane, be divided into
two classes - into cutting and non-cutting. The
boundary lines of the one and the other class
of those lines will be called parallel to the
given line.
Nikolai Lobachevsky
(1792-1856)
Nikolai Lobachevsky
(1792-1856)
Depending on the angle HAD which is called
(p) there are two possibilities.
Either
1. (p)=π/2 and there is only one parallel to CB
through A,
or
2. (p)<π/2 there are exactly two Lobachevsky
parallels.
This means that there are infinitely many lines
through A which do not intersect CB.
Nikolai Lobachevsky
(1792-1856)
In the case that there are two parallels
there are also two sides of parallelism.
l’
The side of parallelism of l’
(p)
p
(p)
The side of parallelism of l’’
l’’
Nikolai Lobachevsky
(1792-1856)
Lobachevsky needs to prove the following.
His definition of parallelism uses the point A,
he has to show that in fact parallelism is a
symmetric property of two lines
a) 17. A straight line maintains the characteristic of
parallelism at all its points.
b) 18. Two lines are always mutually parallel.
Nikolai Lobachevsky
(1792-1856)
He shows that HOA is not possible with
the first four of Euclid’s axioms (which he
made more precise e.g. #3).
19. In a rectilineal triangle the sum of the
three angles can not be greater than two right
angles.
Nikolai Lobachevsky
(1792-1856)
He also shows that if one triangle has the
property that the sum of the interior angels
is 180°, then all triangles have this
property.
This is the dichotomy between Euclidean
geometry (HRA) and non-Euclidian
geometry (HAA).
20. If in any rectilineal triangle the sum of the
three angles is equal to two right angles, so is
this also the case for every other triangle.
Nikolai Lobachevsky
(1792-1856)
Finally he shows that the dichotomy extends to
parallels.
• In Euclidean geometry there is exactly one parallel line
to a given line through a given point not on that line.
• In non-Euclidean geometry there are exactly two parallel
lines, in Lobachevsky’s sense, which implies that there
are infinitely many lines through the point that do not
intersect the given line.
22. If two perpendiculars to the same straight line are parallel to
each other then the sum of the three angles in a rectilineal
triangle is equal to two right angels.
Lobachevsky’s Euclidean and nonEuclidean geometry
•
Euclidean or plane geometry is the geometry
in which the equivalent assumptions
1. For all lines and points p: (p) = π/2 and equivalently
2. The sum of the interior angles of any triangle = π
•
hold.
Non-Euclidean (imaginary) geometry is the
geometry in which the equivalent assumptions
1. For all lines and points p: (p)< π/2 and equivalently
2. The sum of the interior angles of any triangle < π
hold.
Nikolai Lobachevsky’s
Non-Euclidean geometry
Now assume that for all lines and points p:
(p)< π/2.
In this case parallels can have any angle
which is less than π/2 and moreover for
any given angle there is always a pair of
parallels whose angle is that angle
23. For every given angle  there is a line p
such that (p)= .
Nikolai Lobachevsky’s
Non-Euclidean geometry
Parallels cease to have the same distance
to each other at all points rather:
24. The farther parallel lines are prolonged on
the side of their parallelism, the more they
approach on another.
B
x
B’
s’
s
A
x
A’
Moreover in the above figure s’=se-x.