History of Math

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Hey guys, first timer here, but i really need help with thie Math hw, for my college history of math coarse.


SOME HOMEWORK PROBLEMS ON MATHEMATICS IN ANCIENT EGYPT

I. Problems about Unit Fractions [25 points]
(a) Show that if p, q, z are whole numbers and if r = (p+q)/z, then z/pq = 1/pr + 1/qr . If r is also an integer, the terms in the sum will be unit fractions. [According to Howard Eves, “This method for finding possible decompositions of a fraction into two unit fractions is indicated on a papyrus written in Greek sometime between A.D. 500 and 800, and found at Akhmim, a city on the Nile River.”]
 
Add the two right hand fractions. The LCD pqr. The denominator pq cancels with the denominator of the left hand side and you are there.
 
I got the 2nd equation to be z = (q+p)/r

but how does that show that its a whole number?
 
(b) Use the result above to express 2/7 as the sum of unit fractions. If you match things suitably, your answer will match Ahmes’ unit fraction decomposition of 2/7.
(c) Derive two different unit fraction decompositions for 2/99. Show the work.
(d) Specialize the general fact in Part (a) to show that every unit fraction 1/n can be expressed as the sum of two unit fractions. Do this in such a way that if n is odd, then both of the denominators in the sum are even. Explain.


I am supposed to use the answer from there to get these answers as well... i just dont get it.. 22 questions, and this one set is the only one i cant get... grrr
 
The whole process was predicated with if p,q,r and z are integers, then this is true. If r isn't then you don't get unit fractions.
 
You want 2/7 so if p=1,q=7 and r=4 gives z=(1+7)/4 = 2, an integer so
1/pr+1/qr=
1/4+1/28 = 2/7

For 2/99 we may try p=1,q=99,r=50
z=(1+99)/50 = 2 an integer so
1/50+1/(99*50) = 1/50+1/4950 = 2/99
You can try other factors of 99 for other p & q that give z=2.
 
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