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Unit Fractions
A unit fraction is a fraction of the form 1/n, where n is
a positive integer. In this problem, you
will find out how rational numbers can be expressed as sums of these unit
fractions. Even if you
do not solve a problem , you may apply its result to later problems.
We say we decompose a rational number q into unit
fractions if we write q as a sum of 2 or
more distinct unit fractions. In particular, if we write q as a sum of k
distinct unit fractions, we
say we have decomposed q into k fractions. As an example, we can decompose 2/3
into 3 fractions:
1. (a) Decompose 1 into unit fractions.
Answer:
(b) Decompose 1/4 into unit fractions.
Answer:
(c) Decompose 2/5 into unit fractions.
Answer:
2. Explain how any unit fraction 1/n can be decomposed into other unit fractions.
Answer:
3. (a) Write 1 as a sum of 4 distinct unit fractions.
Answer:
(b) Write 1 as a sum of 5 distinct unit fractions.
Answer:
(c) Show that, for any integer k > 3, 1 can be decomposed
into k unit fractions.
Solution : If we can do it for k fractions, simply replace the last one
(say 1/n) with
Then we can do it for k +1 fractions. So,
since we can do it for k = 3, we
can do it for any k > 3.
4. Say that a/b is a positive rational number in simplest
form , with a ≠ 1. Further, say that n is
an integer such that:
Show that when is written in simplest form, its numerator is smaller than a.
Solution: Therefore,
when we write it in simplest form, its numerator
will be at most a(n + 1) − b. We claim that a(n + 1) − b < a. Indeed, this is
the same as
an − b < 0 <> an < b <> b/a > n, which is given.
5. An aside: the sum of all the unit fractions
It is possible to show that, given any real M , there exists a positive integer k
large enough
that:
Note that this statement means that the infinite harmonic series,
grows without
bound, or diverges. For the specific example M = 5, find a value of k , not
necessarily the
smallest, such that the inequality holds . Justify your answer.
Solution: Note that
Therefore,
if we apply this
to n = 1, 2, 4, 8, 16, 32, 64, 128, we get
so, adding in 1/1 , we get
so k = 256 will suffice.
6. Now, using information from problems 4 and 5, prove that the following
method to decompose
any positive rational number will always terminate:
Step 1. Start with the fraction a/b . Let t_{1} be the largest unit fraction 1/n
which is less than or
equal to a/b .
Step 2. If we have already chosen t_{1} through t_{k},
and if t_{1} +t_{2} +. . .+t_{k} is still less than a/b
, then
let t_{k+1} be the largest unit fraction less than both t_{k}
and a/b .
Step 3. If t_{1} + . . . + t_{k+1} equals a/b , the decomposition is found. Otherwise, repeat step 2
Why does this method never result in an infinite sequence of t_{i}?
Solution: Let
is a fraction in simplest terms. Initially, this
algorithm will haveetc. until
This will eventually happen
by problem 5, since there exists a k such thatAt
that point, there is
some n with such that
In this case,
Suppose that there exists n_{k} such that
for some k. Then we have
and This shows that once we have found n_{k}
such that and
we no longer have to worry about t_{k+1}
being less than t_{k}, since
t_{k}, and also while
On the other hand, once we have found such an n_{k},
the sequence {a_{k}} must be decreasing
by problem 4. Since the a_{k} are all integers, we eventually have to
get to 0 (as there is no
infinite decreasing sequence of positive integers). Therefore, after some finite
number of steps
the algorithm terminates with
which is what we wanted.
Juicy Numbers [100]
A juicy number is an integer j > 1 for which there is a
sequence a_{1} < a_{2} < . . . < a_{k} of positive
integers such that a_{k} = j and such that the sum of the reciprocals of
all the a_{i} is 1. For example,
6 is a juicy number because , but 2 is not
juicy.
In this part, you will investigate some of the properties
of juicy numbers. Remember that if
you do not solve a question, you can still use its result on later questions.
1. Explain why 4 is not a juicy number.
Solution: If 4 were juicy, then we would have
The . . . can only possible contain
1/2 and 1/3 , but it is clear that are all
not equal to 1.
2. It turns out that 6 is the smallest juicy integer. Find
the next two smallest juicy numbers,
and show a decomposition of 1 into unit fractions for each of these numbers. You
do not need
to prove that no smaller numbers are juicy.
Answer:
3. Let p be a prime. Given a sequence of positive integers
b_{1} through b_{n}, exactly one of which
is divisible by p , show that when
is written as a fraction in lowest terms , then its
denominator is divisible by p. Use this fact
to explain why no prime p is ever juicy.
Solution: We can assume that b_{n} is the
term divisible by p (i.e. b_{n} = kp) since the order
of addition doesn’t matter. We can then write
where b is not divisible by p (since none of the b_{i}
are). But then Since b
is not divisible by p, kpa + b is not divisible by p, so we cannot remove the
factor of p from
the denominator. In particular, p cannot be juicy as 1 can be written as 1/1 ,
which has a
denominator not divisible by p, whereas being juicy means we have a sum
where and so in particular none of the b_{i}
with i < n are divisible by
p.
4. Show that if j is a juicy integer, then 2j is juicy as
well.
Solution: Just replace
5. Prove that the product of two juicy numbers (not
necessarily distinct) is always a juicy
number. Hint: if j_{1} and j_{2} are the two numbers, how can
you change the decompositions of
1 ending in to make them end in
Solution: Let Then
and so j_{1}j_{2} is juicy.
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