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:

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 t₁ be the largest unit fraction 1/n which is less than or
equal to a/b .

Step 2. If we have already chosen t₁ through t_k, and if t₁ +t₂ +. . .+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 t₁ + . . . + 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 a₁ < a₂ < . . . < 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 b₁ 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 j₁ and j₂ are the two numbers, how can you change the decompositions of
1 ending in to make them end in

Solution: Let Then

and so j₁j₂ is juicy.