Ancient egyptian math

Question

(a) Show that z/pq = 1/pr + 1/qr, where r = (p+q)/z. This method for finding possible decompositions of a fraction into 2 unit fractions.
(b) Take z = 2, p = 1, q = 7, and obtain the unit fraction decomposition of 2/7 as given above.
(c) Represent 2/99 as the sum of two different unit fractions in 3 different ways.

Answer 1

For part a, let's start by substituting in r = (p+q)/z to the right side of the equation. We get:

1/p(p +q)/z + 1/q(p +q)/z moving over z ->

z/p(p+q) + z/q(p + q) making a common denominator ->

zq/pq(p+q) + zp/pq(p + q) adding the fractions ->

(zq + zp)/pq(p+q) factoring out z ->

z(p+q)/pq(p+q) simplifying ->

z/pq

We got the left side of the equation! Part a is done.

Now for part b: let's plug in the values they gave us to that equation.

2/(1*7) = 1/1r + 1/7r simplify and substitute for r ->

2/7 = 1/1(p+q)/z + 1/7(p+q)/z plug in the values ->

2/7 = 1/1(1+7)/2 + 1/7(1+7)/2 simplify ->

2/7 = 2/8 + 2/56 simplify ->

2/7 = 1/4 + 1/28 Check with your calculator for yourself!

Finally, part c. We can follow the same logic as we did for part b, but first let's find possible values for z, p, and q. z will probably be 2, but there are multiple factors we can use to get 99.

z = 2

1) p = 99 and q = 1

2) p = 33 and q = 3

3) p = 11 and q = 9

1)2/99 = 1/99(99+1)/2 + 1/1(99+1)/2

2/99 = 2/9900 + 2/100

2/99 = 1/4950 + 1/50

2)2/99 = 1/33(33+3)/2 + 1/3(33+3)/2

2/99 = 2/1188 + 2/108

2/99 = 1/594 + 1/54

3)2/99 = 1/11(11+9)/2 + 1/9(11+9)/2

2/99 = 2/220 + 2/180

2/99 = 1/110 + 1/90