# Unit Fractions

In an earlier post about Egyptian Fractions I discussed whether it was possible to present *any *(proper) fraction as a sum of unit fractions and posed the following questions :

(1) Can every proper fraction *a/b* (where *a,b *are positive integers) be expressed as a sum of unit fractions?

(2) For any given *a/b*, how many ways are there of representing it as a sum of unit fractions, supposing it to be possible ?

(3) Is there a way to reduce the number of terms in the series of unit fractions to a minimum, supposing *a/b *can be so represented ?

These are my investigations (undertaken *without* looking up the relevant article in Wikipedia which I recommend you do.)

Take two positive whole numbers **a **and **b **and assume that **a/b **is a ‘proper’ fraction, i.e. **a < b **as in **3/8** or **4/15
**The challenge is to find a method to transform

**a/b**into a sum of

*unit fractions*, i.e. fractions which all have

**1**as numerator such as

**1/6**,

**1/47**and so on.

In the trivial case where

**a = 1**we already have a ‘series’ of unit fractions, so we shall assume

**a ≠ 1**. Also, we will assume that

**a/b**is in its lowest terms, i.e. that

**a**and

**b**do not have a common factor greater than unity

**(note 1**).

Now

**a = 1 + (na**

**–**

**b)**

**b n nb**

** **(since if you multiply out you obtain **(b + na – b) = na = a **)

**nb nb b**

For example, **7 = 1 + (5.7 ‒ 24) = 1 + 11
**

**24 5 5.24 5 120**

Note that if ** na < b **the numerator ** (na ‒ b)** **< 0 ** (is negative) and, since **nb > 0 **(i.e. is positive) this makes the whole second fraction negative and we end up with a subtraction, not an addition.

For example, if **n = 2 **we have **(na ‒ b) = 2.7 ‒ 24 = ‒10 **and

** 7 **** = 1 ‒ 10 = 1 ‒ 5
**

**24 2 2.24 2 24**

With the values we have chosen for **a** and **b**, it is clear that we need to have **n **at least **4**, **n ≥ 4
**Can we choose

*any*value for

**n**so long as it is greater or equal to

**4**? Seemingly, yes. For example, if

**n = 187**

** 7 = 1 + (187.7 ‒ 24) = 1 + 1285
**

**24 187 187.24 187 4488**

However, we have still got the second fraction to reduce to a series of unit fractions and there may be more or less economical ways of doing this.

If we start with the lowest values of **n **that make **an ‒ b **positive, we find that, for **a = 7, b = 24 ** we need to have **n **at least **4**. In such a case

** 7 **** = 1 + (4.7 ‒ 24) = 1 + 4 = 1 + 1
**

**24 4 4.24 4 4.24 4 24**

So we have transformed **7/24 ** into a sum of just two unit fractions.

However, if we try out different values of **a** and **b **we will find that this was just a lucky fluke since, in general, the second fraction does not cancel down so readily. We can obtain the condition for this to happen, namely that **(na ****–**** b) = n ** or ** b = (a ‒ 1)n **

This gives the useful rule :

If ** ** **b = (a ‒ 1)n ** then ** a = 1 + 1 …………..(i)
**

**b n b**

We can check this by making up **a/b **in such a way that the condition is met. For instance, **65 = 5.13 **and, according to **(i)**,

**6 = 1 + 1 **which is the case.

**65 13 65 **

Similarly, since **51 = 3.17 ** we find that

**4 = 1 + 1
**

**51 17 51**

This situation is, however, highly unusual.

In general, given an arbitrary* *proper fraction **a/b**, and arbitrary **n**, we will have a unit fraction and a second fraction which is *not* unitary, i.e.

**a **** = 1 + c **where **c ≠****1
**

**b n d**

For example, with **a = 7**, **b = 24**, **n = 5 **we obtain

** 7 **** = 1 + (35 ****–**** 24)**** = 1 + 11
**

**24 5 5.24 5 120**

We can now express **11/120 **as the sum of **1/n **and **c/d **but this only pushes the problem further down the line. With **n = 17 **

**11 = 1 + (17.9 ‒ 120) = 1 + 33
**

**120 17 17.120 17 2040**

The second fraction is sufficiently small to be neglected in rough calculations but it is still there. And if we now express **33/240 **as the sum of **1/n **and **e/f**, we shall, unless we are very lucky, have a further non-unit fraction to deal with. Clearly, what we require is for the second fraction to disappear altogether, which will happen when **(en — f) = 0**, or **f **is an exact multiple of **e**, or when **(en — f) = n **in which case it will cancel with the **n **in the denominator (the case we have already considered). Can we, by a judicious choice of **n**, guarantee that this string of unit fractions will terminate either with zero or with a unit numerator?

*Answer*: It is always possible to express a proper fraction **a/b **as a terminating series of unit fractions. To do this, we only have to look for the *smallest* possible choice of **n **where **an **–** b > 0 **(positive). In the original case with **a = 7**, **b = 24**, it so happens that **4 **is the smallest choice since **3 ×7 = 21** falls short** **of **24 **whereas **5 × 7 = 35 **is too far above.

In the case of **7/24 **we achieved our goal immediately which will not usually be the case, so let us look at another case to see how this checks out.

**7 **** = 1 + (3.7 ****–**** 16)**** = 1 + 5
**

**16 3 3.16 3 48**

Here, **3 **is the smallest choice giving a positive second fraction. We continue :

** 5 **** = 1 + (10.5 ****–**** 48)**** = 1 + 2
**

**48 10 10.48 10 480**

** **nce again, we choose the smallest possible **n**. The second fraction reduces to **1/240 **and if we hadn’t spotted that straight off we would obtain

** 2 **** = 1 + (2.240 ‒ 480) = 1 + 0
**

**480 40 240.480 240**

We have thus expressed **7/16 **as the sum of unit fractions

** **** 7 = 1 + 1 + 1
**

**16 3 10**

**240**

Is something like this going to occur if we systematically choose the smallest possible number for **n **compatible with **(an ‒ b) > 0 **?

** **In fact, yes. It all depends on the successive numerators of the non-unit fractions getting smaller each time. For example, **5 < 7 **in

** 7 = 1 + 5 **and **2 < 5 **in ** 5 = 1 + 2
**

**48 10 480 48 10 480**

If the numerators always decrease, and are always positive integers, we will eventually arrive at a numerator which is either zero or unity. Can we show that this is necessarily the case ? Yes.

If **a/b** is a proper fraction, i.e. **b > a**, and **b **is not an exact multiple of **a**, then **b/a **will lie between **(n ‒ 1) **and **n **because we must have **(an ‒ b) **positive. Thus, **b/a **will be squeezed between successive integers or

** (n ‒ 1) < b/a < ** **n
**

**The distance between**

**b/a**and the

*smallest*integer

**n**such that

**n > (b/a)**will thus always be less than unity, for example the distance between

**16/7**and

**3**is less than unity since

**16/7**lies between

**2**and

**3**. Now, since

**(n ‒ b) < 1 **means **(na ‒ b) < a
**

**a**

However, **(na ‒ b) **is the numerator of the second fraction in

**a **** = 1 + (na ****–**** b) b
b**

**n nb**

** **This shows that the numerator of the second fraction in the expansion is always less than the numerator we started with. Expanding the second fraction, we have the same situation

**(na ****–**** b)**** = 1 + (m(na ****–**** b) ‒ nb) **

** nb m mnb **

** **Here, the numerator of the second fraction is less than the numerator we started with while the denominator is greater, since it has been multiplied by **m**. Thus, the last fraction in an expansion gets steadily smaller and, since the numerator is always a positive integer (or zero), we end up either with zero or a unit fraction.

More formally, is we define **[b/a]** as the *smallest *integer equal to or greater than **b/a**, we have the relation

**a **** = 1 + ([b/a]a ****–**** b)**

**b [b/a]** ** nb **

and, if we have **([b/a]a ****–**** b) > 0**, then **([b/a] ‒ b/a) < 1 **so

** ([b/a]a ‒ b) < a **

Supposing **([b/a]a ‒ b) ≠ 1 **we set

**([b/a]a**whence

*‒ b) = c*** c **** = 1 + [nb/c] ‒ nb
**

**nb [nb/c]**

**[nb/c]nb**

and we have ** ([nb/c] ‒ nb) < c < a **and so on.

** ** Also, note that we have assumed **a < b** ** ** which means that

**na < nb ** so that **(na –**** b) < (nb – b) **or **b(n – 1) ** ** **

For example, in the case of **7 **we took **n = 4 **

**a **** = 1 + (na ****–**** b)**

**b n nb **

** ****(na ****–**** b) **** = 1 + m(na ****–**** b) ****–**** nb**

** nb m mnb **

** **Let **(na ****–**** b) = c **and let **m **be the least multiple of **c < nb
**Now,

**mc < mb < a**so the numerator is always diminishing and since

**0**

**£**

**numerator**we must eventually get back to

**1**or

**0**.

Try another example : **a = 8, b = 57, n = 8 **

** ****8 **** = 1 + (64 ****–**** 57)**** = 1 + 7 c = 7 **and **7 < 8
**

**57 8 8.57 8 456**

** **** 7 **** = 1 + 462 ****–**** 456 **** = 1 + 6 d = 6 < 7 < 8
**

**456 66 66.456 66 30096**

** **** 6 **** = 1 + 30096 ****–**** 30096 **** = 1 . + 0
**

**30096 5016 5016.30096 5016**

** **Thus

**8 **** = 1 + 7
**

**57 8 456**

** **** = 1 + 1 + 6 .
**

**8 66 30096**

** **** = 1 + 1 + 1 .
**

**8 66 5016**

** **** **We conclude that it will always be possible to express a fraction as a sum of unit fractions. ** ***S.H. *

** **