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find two different egyptian fractions that add up to one third

 Oct 4, 2019
 #2
avatar+106539 
+5

Egyptian Fractions have the form :  1 / positive integer 

 

1/3 =  1/a + 1/b

 

Let z  = 3

 

a,b  must be > 3

 

So let a  =  z + m

And let b =  z + n

 

So we have

 

1/ z  =    1/ [z + m]  +  1/[;z + n]

 

1/z  =  [ 2z + m + n] / [(z + m) (z + n)]      cross-multiply

 

(z + m) (z + n)  =  z (2z + m + n ]

 

z^2 + mz + nz + mn =  2z^2 + mz + nz

 

z^2 + mn = 2z^2

 

z^2  = mn       so

 

3*2 = mn

 

9 = mn

 

So the possibilites for m,n  are

 

m   n

1    9

3    3

 

 

So....the possible fractions are

 

1/ [3 + 1]   +  1/ [ 3 + 9]  =     1/4 + 1/12

1/[3 + 3] + 1/[3 + 3]  =  1/6 + 1/6

 

Only the first is what we need

 

So

 

1/4  + 1/12 

 

cool cool cool

 Oct 4, 2019
edited by CPhill  Oct 4, 2019
 #3
avatar
+4

Here is another way:
1 - Start with 1
2 - Divide 1 by 4 = 1/4
3 - Take one of the remaining 1/4
4 - Subdivide that second 1/4 into 3 parts, or:
5 - (1/4) / 3 = (1/4) x (1/3) = 1 / 12
6 - Add 1/4 in (2) above to 1 / 12 in (5) above, or:
7 - 1 / 4 + 1 / 12 = 1 / 3 - which is what you want. 

 Oct 4, 2019

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