Unit Fractions
Show that z/pq = 1/pr + 1/qr, where r = (p + q)/z.

For r=(p+q)/z, we want to show that 1/pr + 1/qr = z/pq.
zq  p
1
1
q
p
q p
q p
z
 





pr qr qpr qpr
 p  q pqq  p pqq  p pq
pq
 z 
z
--------------------------------------------------------------------------------------------------------------------Take z=2, p=1, q=7, and obtain the unit fraction decomposition of 2/7 as given in the Rhind
papyrus.
z
2
 p  q  8

and r 
 4
 z  2
pq 17
z
2
1
1
1 1



 
pq 17 14  74  4 28
-------------------------------------------------------------------------------------------------------------------- Represent 2/99 as the sum of two different unit fractions in 3 different ways.
z
2
36

and r 
 18
pq 333
2
2
1
1
1
1




99 318 3318 54 594
z
2
20

and r 
 10
pq 911
2
2
1
1
1
1




99 910 1110 90 110
z
2
100

and r 
 50
pq 199
2


2
1
1
1
1




99 150 9950 50 4950
--------------------------------------------------------------------------------------------------------------------By taking z=1, p=1, q=n in the relation of (a), obtain the more particular relation
1/n=1/(n+1) + 1/n(n+1), and show that when n is odd, this leads to a representation of
 2/n as a sum of two unit fractions.
z
1
n 1
z
1
1
1
1
1

and r 
 n  1. We have that

 , so 

pq 1n
1
pq pr qr
n n  1 nn  1
Now, let q=n such that n=k+1, i.e. n is odd.
z
1
z
1
1
1
1
1

and r  k  2. We have that

 , so


pq 1k  1
pq pr qr
k  1 k  2 k  2k  1


k  1
1
k  1  1
1
1




.
k  1k  2 k  2k  1 2k  1k  2 2k  1 2n 
---------------------------------------------------------------------------------------------------------------------

Show that if n is a multiple of 3, then 2/n can be broken into a sum of 2 unit fractions of
which one is 1/2n.
We want to show that
, for some integer x. If n is a multiple of 3, then we
can write
, where m is some integer. So, we now need to show that 2/(3m)
can be broken into a sum of two unit fractions, one of which is 1/2(3m), i.e.
. Now, we must solve for x:
.
Now, we have
. So, we should go back and check that when
x=2m, we get 2/3m:
. We do in fact get
what we were looking for. So, if n is a multiple of 3, then 2/n can be broken into a
sum of 2 unit fractions of which one is 1/2n and the other is 1/2m, for
.
--------------------------------------------------------------------------------------------------------------------Show that if n is a multiple of 5, then 2/n can be broken into a sum of 2 unit fractions of
which one is 1/3n.
We want to show that
, for some integer x. If n is a multiple of 5, then we
can write
, where m is some integer. So, we now need to show that 2/(5m)
can be broken into a sum of two unit fractions, one of which is 1/3(5m), i.e.
. Now, we must solve for x:
.
Now, we have
. So, we should go back and check that
when x=5m, we get 2/5m:
. We do in fact get what we were
looking for. So, if n is a multiple of 5, then 2/n can be broken into a sum of 2 unit
fractions of which one is 1/3n and the other is 1/3m, for
.