Download Euclid`s 5th Axiom (on the plane): That, if a straight line falling on two

Euclid’s 5th Axiom (on the plane):
That, if a straight line falling on two
straight lines makes the interior
angles on the same side less than two
right angles, the two straight lines, if
produced indefinitely, meet on that
side on which are the angles less than
the two right angles.
1
2
1   2  180o
 the lines will cut when extended .
Playfair,
an 18th century
Scottish scientist,
formulated the
axiom (on the
plane):
Given a line and a point not on the
line, it is possible to draw exactly one
line through the given point parallel to
the line.
We’ll show that this is in fact
equivalent to Euclid’s Fifth
Postulate.
First, we remark that, under the first
four axioms of Euclid, the following
statements hold.
(1) Opposite angles are the same.
1
 2  1
(2) If two triangles have both sides
equal to each other and the angles
between the sides are also equal, then
they are congruent.
2
1

(3) In any triangle, if one of the sides
is produced, the exterior angle is
greater than either of the opposite
angles.
1   ,
&
2   .
Proof.
(Space for you to down the proof.)
Find l 2 so that 1   2 .
To show that
1   2
 l1 // l2 .

2
P
1
l2
l1
Direct Proof/Argument.
Indirect Proof/Argument.
Suppose l1 and l2 meet in the left.
Then we have the hypethetical diagram
where PQR is a triangle.
.
Q
2
1
1
R
 P2
l2
l1
We redraw portion of the previous
diagram. Observe that 1
is a exterior angles of the
triangle PQR.
Q
1 R
1   2
2
P
( exterior   opposite )
a contraditon (1   2 is given ).
Hence the lines cannot meet in the left.
Likewise , they cannot meet in the right.
To show “5th axiom” = “// axiom”
(I) “5th axiom”

“// axiom”
We know from above that
1   2
 l1 // l2 .
P
2
l2
1
l1
We are required to show that
l2 is the only line passing
through P that is parallel to l
1
.
Suppose not. Then there is another
line l3 passing through P that is
parallel to
Let
l3
l1
.
lie as in the diagram.
l3
P

1
2
l2
l1
From the previous diagram we have
  180o   2  1    1  180o   2
 1    180
o
(1   2 ).
P


1
l3
l1
5th axiom  l1 and l3 have to meet!
Similarly if
l3
lie as in the diagram.
Convince yourself that this is
mirror image of the previous case.
P
l3
1
l2
l1
(I) “// axiom”

“5th axiom”
Re fer to the following diagram,
we are required to show that
1   3  180o  l1 meets l3 in the right.
P
3
l3
1
Left
l1
Right
Find l 2 so that 1   2 .
We know that ( from lecture)
1   2  l1 // l2 .

2
P
3
1
l2
l3
l1
// axiom  l1 and l3 cannot be parallel.
Contradiction  no such l3 exist.
Thus to finish the proof ,
we just need to show that
l1 and l3 cannot meet in the left.
Suppose they do meet in the left.
Then we have the hypethetical diagram
where PQR is a triangle.
.
Q
3
 P
1
1
R
3
l3
l1
We redraw portion of the previous
diagram. Observe that 1 and 3
are the exterior angles of the
triangle PQR.
Q
1 R180 o  1
1  180  3
o
180o  3 3
( exterior   opposite )
P
3  180o  1
 1   3  360o  (1  3 )
 2(1   3 )  360o
 (1  3 )  180o
a contraditon. Hence the lines cannot meet in the left.
That is, they have to meet in the right!
Together with the 5th axiom of Euclid,
or equivalently, the Playfair postulate,
we have the following.
3
2
Parallel lines
1
1   2   3 .
The fifth axiom implies that the sum
of the interior angles of any triangle is
equal to two right angles, that is,
180 degrees.
Indeed, Euclid’s fifth axiom,
the Playfair axiom,
the Pythagoras’ theorem,
and the statement that
the sum of the interior angles of a
triangle is equal to 2 right angles,
are all equivalent.
That is, we won’t change the
Euclidean geometry if we replace
the fifth axiom by anyone of the
other statements.
The details can be found in
Reference: http://www.cut-theknot.org/triangle/pythpar/PTimpliesPP
.shtml
Because the parallel axiom, or
the Playfair axiom, appears to be
so natural and intuitive, many had
tried, unsuccessfully, to derive it
from the first four axioms.
Despite of this, for two thousand years
nobody really doubted that the parallel
axiom can be changed or replaced.
It was held as an absolute truth.
One of the most famous stories about Gauss
depicts him measuring the angles of the great
triangle formed by the mountain peaks of
Hohenhagen, Inselberg, and Brocken for evidence
that the geometry of space is non-Euclidean.
Gauss was apparently the first to arrive at the
conclusion that no contradiction may be
obtained this way. In a private letter of 1824
Gauss wrote:
“The assumption that (in a triangle)
the sum of the three angles is less than
180o leads to a curious geometry, quite
different from ours, but thoroughly
consistent, which I have developed to
my entire satisfaction.”
Lobachevsky and Bolyai built their geometries
on the assumption:
Through a point not on the line there exist
more than one line parallel to the line.
This is equivalent to Gauss' assumption that
the sum of angles in a triangle is less than
180 degree.
Rightfully, the new geometry created is called
Non-Euclidean geometry.
The related result is used by
Einstein in his General Theory of
Relativity – space-time is
non-Euclidean!
Black Hole.
In a 1919 test of the general theory of relativity, stars
(marked with short horizontal lines) were
photographed during a solar eclipse.
The rays of starlight were bent by the Sun's gravity on
their way to the earth. This is interpreted as evidence in
favor of Einstein's prediction that gravity would cause
deviations from Euclidean geometry.
Terence Tao
Got his PhD at the age of 20 from
Princeton University.
Became full professor in UCLA at 25.
Awarded the Fields Medal at the age of
31.
Contribution includes Kakeye’s problem.
•
Grigori Perelman
In 1994, Perelman sequestered
himself away to tackle the problem,
and for the following 8 years gave
no signs of life.
In May 2003, he announced that he
had solved the Poincare Conjecture
and the Thurston Geometrization
Conjecture.