Download DECOMPOSITION OF RATIONAL NUMBERS INTO ODD UNIT

DECOMPOSITION OF RATIONAL NUMBERS INTO ODD UNIT FRACTIONS
DANA CLAHAN AND PHILIP PESCA
In what follows, we will use the symbol “∀” to denote the phrase “for all,” and we will denote the phrase
“there exists” by “∃.” In this note, we make rigorous a problem that can be found, for example, in the book,
Old and New Unsolved Problems in Plane Geometry and Number Theory’, by Victor Klee and Stan Wagon.
We have made the statement of the problem there rigorous, or, more precisely algorithmic.
Theorem (Fibonacci, thirteenth century). Suppose that p, q ∈ Z, where p, q 6= 0. Then ∃k ∈ N and
n1 , n2 , . . . , nk ∈ N such that
k
X
1
p
= ,
n
q
m=1 m
and, ∀j ∈ {1, 2, . . . , k − 1} , nj is the smallest natural number such that
j
X
1
p
< .
n
q
m
m=1
Example. The above theorem guarantees that ∃k ∈ N and n1 , n2 , . . . , nk ∈ N such that
k
X
1
3
= ,
n
4
m
m=1
and, ∀j ∈ {1, 2, 3, . . . , k − 1}, nj is the smallest natural number such that
j
X
1
3
< .
n
4
m
m=1
Definition (Fibonacci decomposition of a rational number). Suppose that p, q ∈ Z and are non-zero.
Suppose that ∃k ∈ N and n1 , n2 , . . . , nk ∈ N such that
k
X
1
p
= ,
n
q
m
m=1
and, ∀j ∈ {1, 2, . . . , k − 1} , nj is the smallest natural number such that
j
X
1
p
< .
n
q
m=1 m
Then we call the series
k
X
1
n
m=1 m
a Fibonacci decomposition of p/q. Since the nm ’s are uniquely determined, there is only one such decomposition for a given p/q and thus we can call this decomposition the Fibonacci decomposition of p/q.
The authors’ work on this note was partially supported by ENGAGE in STEM, which is administered by the Fullerton
College Office of Special Programs.
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DANA CLAHAN AND PHILIP PESCA
Example. Let’s find a Fibonacci decomposition of p/q = 3/4 in the above example. First, we find the
smallest n1 ∈ N such that n11 < 34 . n1 = 1 does not work because 11 = 1 ≮ 34 . But n1 = 2 does work since
3
1
2
3
1
1
2 = 4 < 4 . Next, we find the smallest n2 ∈ N such that 2 + n2 ≤ 4 , with a preference for equality if we
1
3
1
can get it. Notice that n2 = 4 yields equality, 2 + 4 = 4 , so just as Fibonacci promised, ∃k ∈ N (k = 2) and
n1 = 2, n2 = 4 such that n11 < 34 but n11 + n12 = 34 .
Exercise. Find the Fibonacci decomposition for 56 .
All of the fractions in a Fibonacci decomposition for a given rational p/q have 1 as their numerator and
are thus called unit fractions; more precisely, we have the following definition:
Definition. (Unit fraction) We call q > 0 ∈ Q (the set of all rational numbers) a unit fraction if and only
if the numerator of q’s expression as a quotient of positive integers in lowest terms, is 1.
Example. Since the numerator of 12 , which is in lowest terms, is 1, it follows that
1
2
is a unit fraction.
Example. Let’s consider the reciprocal of the rational from the previous example 2 = 21 . We need to show
that 2 is NOT a unit fraction. That is, we need to show that there is no positive integer b such that
2
1
= ,
1
b
which is equivalent to the equation 2b = 1, which has no integer solution. Thus there is no positive integer
b such that 2 = 1/b, and we are done. That is, 2 is not a unit fraction, as we claimed.
Exercise. Determine whether the following numbers are unit fractions.
2
(a) − .
4
1
(b) .
3
3
(c) − .
1
1
.
π
π
(e) − .
1
(d)
Prove your answer in each case.
Definition. (Odd and even unit fractions) We say that a unit fraction 1/u, where u ∈ N, is an odd unit
fraction if and only if u is odd. Otherwise, 1/u is called an even unit fraction.
Example.
1
2
is an even unit fraction since 2 is clearly even, and 1/3 is an odd unit fraction, since 3 is odd.
Exercise. Determine which unit fractions are even or odd in the previous exercise.
We are now prepared to state an open question which is the focus of this note:
Open Question: Suppose that p and q are positive integers and q is odd. Are there k ∈ N and odd
n1 , n2 , . . . , nk ∈ N such that
k
X
1
p
=
n
q
m=1 m
and, ∀j ∈ {1, 2, . . . , k − 1}, nj is the smallest positive integer such that
j
X
1
p
<
?
n
q
m=1 m
Example. Let’s test 52 to see if the answer to the open question in the case of this fraction, which is obviously
in lowest terms, is “yes” or “no.” The smallest odd integer n1 such that n11 < 25 is 3, since 1/1 and 1/2 are
5
6
6
5
1
both larger than 2/5. Now 31 < 25 =⇒ 15
< 15
. =⇒ 25 − 13 = 15
− 15
= 15
. =⇒ ∃k ∈ N and odd integers
1
1
2
n1 = 3, n2 = 15 such that n1 + n2 = 5 and n1 = 3 is the smallest positive integer such that 1/n1 = 13 < 52 .
Thus the answer to the open question is “yes” when p/q in lowest terms is 2/5, for example.
DECOMPOSITION OF RATIONAL NUMBERS INTO ODD UNIT FRACTIONS
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Remarks:
(1) The condition that q needs to be odd is necessary in the statement of the above open question, since
it is impossible to find a set of odd unit fractions such that the sum of the odd unit fractions has an
even denominator when expressed in lowest terms. This fact follows immediately from the fact that
the least common multiple of a finite collection of odd numbers is necessarily odd, since none of the
numbers in the collection have 2 as a factor and hence their least common multiple can only contain
primes which are not 2.
(2) We caution all readers that the open question asked does not simply ask for n1 , n2 , etc. such that
the sum of the reciprocals of these is p/q, because the answer to that question is easily yes: Indeed,
simply choose k = q and then let each of the ni ’s be q. For example, 2/5 can be written as 1/5 + 1/5.
The open question asks for more than this - it asks that these numbers ni be constructed in a suitable
way and not merely that their reciprocals add up to p/q!
E-mail address: [email protected]
Mathematics and Computer Science Division, Fullerton College, 321 E. Chapman, Fullerton, CA 92832-2095