English | Español

# Try our Free Online Math Solver! Online Math Solver

 Depdendent Variable

 Number of equations to solve: 23456789
 Equ. #1:
 Equ. #2:

 Equ. #3:

 Equ. #4:

 Equ. #5:

 Equ. #6:

 Equ. #7:

 Equ. #8:

 Equ. #9:

 Solve for:

 Dependent Variable

 Number of inequalities to solve: 23456789
 Ineq. #1:
 Ineq. #2:

 Ineq. #3:

 Ineq. #4:

 Ineq. #5:

 Ineq. #6:

 Ineq. #7:

 Ineq. #8:

 Ineq. #9:

 Solve for:

 Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:

# 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. (b) Decompose 1/4 into unit fractions. (c) Decompose 2/5 into unit fractions. 2. Explain how any unit fraction 1/n can be decomposed into other unit fractions. 3. (a) Write 1 as a sum of 4 distinct unit fractions. (b) Write 1 as a sum of 5 distinct unit fractions. (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

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 tk, and if t1 +t2 +. . .+tk is still less than a/b , then
let tk+1 be the largest unit fraction less than both tk and a/b .

Step 3. If t1 + . . . + tk+1 equals a/b , the decomposition is found. Otherwise, repeat step 2

Why does this method never result in an infinite sequence of ti?

Solution: Let is a fraction in simplest terms. Initially, this
algorithm will have etc. until This will eventually happen
by problem 5, since there exists a k such that At that point, there is
some n with such that In this case, Suppose that there exists nk such that for some k. Then we have and This shows that once we have found nk such that and we no longer have to worry about tk+1 being less than tk, since tk, and also while On the other hand, once we have found such an nk, the sequence {ak} must be decreasing
by problem 4. Since the ak 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 

A juicy number is an integer j > 1 for which there is a sequence a1 < a2 < . . . < ak of positive
integers such that ak = j and such that the sum of the reciprocals of all the ai 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. 3. Let p be a prime. Given a sequence of positive integers b1 through bn, 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 bn is the term divisible by p (i.e. bn = kp) since the order
of addition doesn’t matter. We can then write where b is not divisible by p (since none of the bi 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 bi 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.

 Prev Next