| Squaring a circle, i.e. constructing
a square whose area equals that of a given circle, is one of the three famous
construction problems of antiquity. Its (negative) solution was eventually
obtained in 1882 from the following theorem by the German mathematician
Lindemann:
Theorem
The expression
(*) A1e^a1 + A2e^a2 + ... +
Ane^an
in which all coefficients A
and a are algebraic, A's are non-zero and a's are all different, cannot
vanish.
Remark
Note that, in the simplest
case of a single term ea, the theorem implies that the only point on the
graph of y = ex with both x and y rational is (0,1). Although the set
of rational points is dense in the plane, the graph of y = ex somehow
manages to cut the plane without passing but through just one of them.
The theorem is also associated
with the French mathematician Hermite(1822-1901) who, in 1873, proved
the special case in which the coefficients and exponents were rational
numbers. Hermite applied his theorem to prove that e, the base of natural
logarithms, is transcendental. (Indeed, if P(e) = 0 for a polynomial with
integer coefficients, then we'll get an expression (*) with a's being
integers 0,1,...,n.)
From the Lindemann's theorem
it follows that the number is
also transcendental. This follows from the wonderful Euler's identity
E^i + 1 = 0
Indeed, on the left we have
an expression in the form (*). Since it equals zero, the exponent i can't
be algebraic. Hence, is transcendental.
Now returning to the problem
of squaring a circle. The area of a circle with radius 1 is exactly .
To construct a square with this area we must be able to construct a segment
of length . Numbers for which it's possible to construct (with a compass
and a straightedge) a segment of that length are called constructible.
For now, please take my word for it that constructible numbers are algebraic.
Also, a product of two constructible numbers is also constructible. Thus
if were construcitble so would be . But then it also would be
algebraic in contradiction with the Euler's identity and the Lindemann's
theorem.
Thus it's impossible to square
a circle using a straightedge and a compass; but like the problem of angle
trisection, this one can be solved by other means. Have a look at the
diagram on the right. Assume a circle of unit radius is rolled half a
turn on a straight line. Then the distance between the points A and B
will be exactly . If we draw a semicircle on AC = AB+1 as a diameter,
and continue the vertical radius of the right circle to the intersection
with the semicircle at a point D, then AB*BC = BD2. Which, of course,
solves the famous problem because AB = and BC = 1.
Note, however, that the distance
measurement with the help of a rolling circle is routinely used on modern
cars as a part of the odometer mechanism.
 squaring a circle
References
R.Courant and H.Robbins, What
is Mathematics?, Oxford University Press, 1996
H.Dorrie, 100 Great Problems
Of Elementary Mathematics, Dover Publications, NY, 1965.
W.Dunham, Journey through Genius,
Penguin Books, 1991
M.Kac and S.M.Ulam, Mathematics
and Logic, Dover Publications, NY, 1968.
R.B.Nelsen, Proofs Without
Words, MAA, 1993
|