The Transcendentality of
By definition, the
number
is the ratio of the circumference to the diameter of a circle. This
ratio is the same for all circles.
is an irrational number. It cannot be represented as the ratio of two
integers, regardless of the choice of integers. Equivalently, it cannot
be represented as an unending, periodic decimal.
is also a transcendental number. It is not a root of any algebraic
equation of the form
a0 +
a1x + a2x2 +
+ anxn
= 0
where the ai
are all rational numbers and n is finite. For comparison,
is also an irrational number. But
is not transcendental since it is a root of the equation
x2 -
2 = 0.
Both
and
are irrational, but only
is transcendental. What makes the difference? One important argument
is that a line of length
can be constructed using classical techniques (i.e., using compass
and straight-edge in a finite number of steps). But, because of the
way
is defined, a line of length
cannot be so constructed. (A curve can. Mark off a unit
segment. Bisect the segment. Using the midpoint as center, scribe the
appropriate circle. The circle has length .)
To illustrate, let
us first consider .
To make a construction that produces a line of this length, we begin
with two unit-length segments placed end to end so that one segment
is at right angles with the other. We then, connect the free ends to
complete a right triangle. The new line has length
Now, consider .
A circle of unit diameter has its circumference = .
Draw a unit circle, and locate its center. From the center produce a
set of n radial lines each separated from its neighbor by an angle 2/n.
Approximation for
n = 9
Isoseles Triangle
Now, connect the
ends with straight line segments to form a set of isosceles triangles.
The sum of the lengths of these straight-line segments approaches the
circumference of the circle as n approaches infinity (see figure for
n = 9).
To construct a line
of length ,
we have but to produce the length AB n times along any line. In the
figure, n = 9, AB is a typical straight line segment completing an isosceles
triangle, and Circumference
9(AB) (see figure).
Now, let us use
some algebra to calculate the length AB in the general case. We begin
by redrawing part of our previous picture.
One isosceles triangle
(ABC
with one vertex at the center C and two more vertices on the circle
at A and B) is selected (see figure). The angle subtended at C is 2/n.
We wish to calculate the length AB, and then to estimate how good an
approximation the value n(AB) is to the actual circumference of the
circle.
The line CM bisects
the angle 2/n,
and meets the line AB at right angles. Thus, the triangle CMA is a right
triangle, and the length AM = ½(AB).
AM/(Radius) = AM/(½)
= sin (/n)
AM = ½ sin (/n)
& AB = sin (/n).
Finally, n(AB) =
n sin (/n).
Since the actual
circumference of the circle is ,
we now write
n(AB) = {n
sin (/n)}/
=
{(sin (/n))/(/n)}
=
{(sin ())/()}
where =
/n.
The value n(AB)
differs from
by the multiplicative factor (sin ())/(),
with
= /n.
Notice that, in the limit as n
,
0 and (sin ())/()
1. The value n(AB) does indeed approach
in the limit. But, while the limit exists, the
actual function f()
= (sin ())/()
does not exist at
= 0. It becomes the indeterminate form 0/0. Thus, although n(AB)
may approach the actual circumference to any arbitrary precision we
might desire, the actual value n(AB) =
can never be obtained from any construction of the type outlined above.