Download History of the Three Greek Problems

The Three Greek Problems
1.) Doubling the Cube
2.) Trisecting the Angle
3.) Squaring the Circle
History of the Three Greek
Problems
These three problems have an interesting
history, especially “doubling the cube”.
According to legend, when ancient Athens
was faced with a plague, they went to the
oracle of Apollo at Delos for advice. The
answer they received was to double the
cubical alter to Apollo. However, instead of
doubling the alter they doubled the length of
each edge, making it eight times the original
size. This has been called the “Delian
problem” ever since.
Constructible Lengths
The rules of constructible lengths are:
1.) Given 2 points, we may draw a line through
them, extending it indefinitely in each direction.
2.) Given 2 points, we may draw the line segment
connecting them.
3.) Given a point and a line segment, we may draw a
circle with center at the point and radius equal to
the length of the line segment.
Lemma 1
Given segments of lengths 1, a, and b, it is possible
to construct segments of lengths a+b, a-b (when a >
b), ab, and a/b (when b ≠ 0).
a
ab
1
b
A Field, F, is a set of numbers that is
closed under addition, multiplication,
and division.
For example, rational numbers and real
numbers are in a field. When two
rational numbers are added, multiplied,
or divided the answer is a rational
number.
The same is not true for integers.
If F is a field, a quadratic extension of F
is the set
ab k
such that a, b, and
k are elements of F.
EX. 2  3 2 is an element of a
quadratic extension of the rational
numbers. This is because 2 and 3 are
rational numbers.
WHEN IS A NUMBER
CONSTRUCTIBLE ?
A number, a, is constructible if and
only if a  Fⁿ where
QF
 F²  …  Fⁿ
F¹ is a quadratic extension of Q,
like Fi+1 is a quadratic extension of
Fi.
Consider:
Volume = L
=1
=1
1
1
1
Volume = L
=2
Side =
3
2
LEMMA 2
If 3 2 is in a quadratic extension of a
field, then 3 2 must lie in the field itself.
If this is true,
a b k .
3
2 looks like a  b k or
Where a, b, and k are in the field.
If 3 2
this:
is in a quadratic extension of a field F, then it will look like
3
2  ab k
 2   a  b k 
3
3
3

 

2  a  3ab k  3a b  b a k
2
 2   a  b k 
3
3
or
3
2
3
=0
3

 

2  a  3ab k  3a b  b a k
3
2
2
3
=0
3
2  a b k,a b k
By the previous equations, 3 2 can not be constructed because
there is only one real root of 3 2.
Since there is only one real root of 3 2 , then b=0. When
b=0, 3 2 must equal a, which would mean that since a is in
the field 3 2 must also be in the field. That means that 3 2 is
a rational number, which is not true. This shows the
contradiction.
Work Cited

Hadlock, Robert. Field Theory and
Its Classical Problems. The
Mathematical Association of America.
1978.