G
OLDBACH
(1690–1764) conjectured that every even
number greater than 2 is the sum of two
PRIME
num-
bers. For example, 4 = 2 + 2, 6 = 3 + 3,…, 20 = 3 +
17,…, 50 = 13 + 37,…, 100 = 3 + 97, and so forth. No
one to this day has been able to establish whether this
claim holds true for each and every even number. The
problem has since become known as Goldbach’s con-
jecture, and it remains one of the most famous
unsolved mathematical problems of today. Between
March 20, 2000, and March 20, 2002, a $1-million
prize was offered to anyone who could solve Gold-
bach’s conjecture. The prize went unclaimed.
Goldbach’s conjecture is equivalent to the chal-
lenge of proving that every integer greater than 5 is the
sum of three primes. (If the number is odd, subtract 3
and write the resultant number as a sum of two primes.
If it is even, subtract 2.)
The source of the difficulty with this problem is
that primes are defined in terms of multiplication,
while the problem involves addition. It is often very
difficult to establish connections between these two
separate operations on the integers.
In 1931, Russian mathematician L. Schnirelmann
(1905–38) proved that every positive integer is the sum
of at most 300,000 primes. Although this result seems
ludicrous in comparison to the original problem, it is a
significant first step to solving the conjecture: it shows
at least that it is possible to put a bound on the number
of primes representing the integers.
In 1937 another Russian mathematician, Ivan
Vinogradoff (1891–1983), proved that every suffi-
ciently large number can be written as a sum of four
primes. What constitutes “sufficiently large” is not
known (Vinogradoff’s work only shows that there can-
not be infinitely many integers that require more than
four prime summands)—but at least a small bound has
been placed on “most” integers. In 1973 Jing-Run
Chen proved that every sufficiently large even number
is the sum of a prime and a number that is either prime
or has two prime factors.
Work on solving the Goldbach conjecture contin-
ues. As of the year 2003, it has been confirmed by com-
puter that the conjecture holds true for all numbers up
to 6 ×1016.
See also
BRUTE FORCE
.
golden ratio (divine proportion, extreme and mean
ratio, golden mean, golden section) A line segment
connecting two points Aand Bis said to be divided by
a third point Pin the golden ratio if the ratio of the
whole length AB to the length AP is the same as the
RATIO
of the length AP to the length PB:
=
The value of this ratio is denoted ϕ(the Greek letter
phi) and can be computed as follows:
Set the length PB to be one unit. Then, since
= ϕ, the length AP is ϕunits, and the length
of the entire segment AB is ϕ+ 1. We have:
=
yielding the
QUADRATIC
equation ϕ2= ϕ+ 1.
This has two solutions. Selecting the solution
that is larger than 1 yields:
ϕ= = 1.618033988…
(The second solution to this quadratic equation is
= 1 – ϕ= –0.618033988….)
The golden ratio was studied and made famous by
the Pythagoreans, the followers of the Greek mathe-
matician P
YTHAGORAS
(ca. 500
B
.
C
.
E
.). They discov-
ered it in their study of the
PENTAGRAM
, the figure that
appears when one draws in the diagonals of a regular
pentagon. The sides of a pentagram divide each other
in the golden ratio.
The following method shows how to construct a
point Pthat divides a given line segment AB into the
golden ratio:
Given a line segment AB, draw a perpendicular
line through A. Find the
MIDPOINT
Mof AB
and locate on the perpendicular line a point C
whose distance from Ais the same as the
length AM. (Use a circle with center Aand
radius AM as an aid.) Now draw a circle with
center Cand radius BC to locate a second
point Don the perpendicular line such that
length BD equals length BC. Let Pbe the point
on the line segment AB such that length AP
equals length AD. (Use a circle with center A
1 + √
–
5
––––
2
1 + √
–
5
––––
2
ϕ
––
1
ϕ+ 1
––––
ϕ
AP
––
PB
AP
––
PB
AB
––
AP
230 golden ratio