The Egyptians dealt with fractions in a different way than we do now. Fractions were viewed as a sum of *different* unary fractions. Unary factions are those with a 1 in the Numerator
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Re: Good solutions.spkr4thedead51September 21 2005, 03:12:12 UTC
7/3 = 2r1, therefore the first is 1/(2+1)
3/7 - 1/3 = 2/21
and haha, i realized that I made a typo (as I was doing this all in my head)
21/2 = 10r1, therefore, 2/21 = 1/11 + a/b
a/b = 2/21 - 1/11 = 1/(21*11) = 1/231
3/7 = 1/3 + 1/11 + 1/231. The other one doesn't work actually. oops :-(
There is an interesting algorithm (from the Akim Papyrus) that lets you see if the fraction can be written as the sum of just 2 unit fractions:
take a/b and find the factor-pairs of the denominator b (b = c*d, e*f, h*i, etc), then sum the pair and divide by the numerator (N1=(c+d)/a, N2=(e+f)/a, etc). if you get N to be an integer, then the fraction can be represented by 2 unit fractions where the two denominators are the first factor times the result of the check (a/b = 1/(c*N) + 1/(d*N).
for example:
6/35 has the factor pairs of 1*35, and 5*7
(1+35)/6 = 6, (5+7)/6 = 2
so 6/35 = both 1/(1*6) + 1/(35*6) and 1/(5*2) + 1/(7*2)
He is in fourth grade. I was trying to send him a home school assignment and my company email was refusing to send to aol. Putting it up on my LJ page was the easiest way I could think of...
Comments 13
2/3 = 1/2 + 1/6
3/5 = 1/2 + 1/10
3/8 = 1/3 + 1/24
2/7 = 1/4 + 1/28
3/7 = 1/3 + 2/21 = 1/3 + 1/12 + 3/21*12 = 1/3 + 1/12 + 1/84
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What algorithm gives 2/21 -> 1/12 +3/21*12?
Randy
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3/7 - 1/3 = 2/21
and haha, i realized that I made a typo (as I was doing this all in my head)
21/2 = 10r1, therefore, 2/21 = 1/11 + a/b
a/b = 2/21 - 1/11 = 1/(21*11) = 1/231
3/7 = 1/3 + 1/11 + 1/231. The other one doesn't work actually. oops :-(
There is an interesting algorithm (from the Akim Papyrus) that lets you see if the fraction can be written as the sum of just 2 unit fractions:
take a/b and find the factor-pairs of the denominator b (b = c*d, e*f, h*i, etc), then sum the pair and divide by the numerator (N1=(c+d)/a, N2=(e+f)/a, etc). if you get N to be an integer, then the fraction can be represented by 2 unit fractions where the two denominators are the first factor times the result of the check (a/b = 1/(c*N) + 1/(d*N).
for example:
6/35 has the factor pairs of 1*35, and 5*7
(1+35)/6 = 6, (5+7)/6 = 2
so 6/35 = both
1/(1*6) + 1/(35*6) and
1/(5*2) + 1/(7*2)
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I did NOT like fractions ... couldn't really figure them out till 9th grade algebra.
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2^(-1)+4^(-1)=1.3^(-1)
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. Regards
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