| Some more speculation. Have a look at https://en.wikipedia.org/wiki/Ancient_Egyptian_multiplicatio... Egyption multiplication was (implicitly) based on powers of two. In some sense weirdly similar to modern bit-twiddling. Note that 4/3 in binary is 1.0101 0101 0101 0101 0101 0101 0101 0101... So that suggests the following algorithm, expressed in modern day Python: def mul4_3(x):
table = []
while x > 0:
table.append(x)
x //= 4
return sum(table)
I deliberately used a 'table', because that's what a scribe would do.Repeat this function four times, and you will have multiplied by 256/81. I have no clue whether they would have done anything resembling this procedure; this is just to show that it's plausible given how their multiplication worked. I don't know how this relates to 'Egyption fractions'. P.S. Just for fun the same thing for 22/7: def f22_7(x):
y = (x << 1) + x
while x > 0:
x >>= 3
y += x
return y
Not actually harder to execute by hand, I'd say, but perhaps harder to come up with? |
This is consistent with the notion from Jarrosson, because that means that 4/3 can be represented as
which is a sum of fractions with unitary numerator - the representation ancient Egyptians are said to absolutely prefer.The approximated ratio of the circumference to its diameter can be represented by four iterations ( (4/3)^4 ) of infinite series of sums of fractions with unitary numerator.