|
|
Re: Ariadne's thread
Posted:
Dec 21, 2011 3:35 AM
|
|
One single equation containing three variables can't be
solved. Ahmes did not use algebra. He used algorithms
or step by step procedures that required a good knowledge
of conversions of 1 into a variety of unit fraction series
- the more the better. Beginners had to learn those
algorithms and conversions. If they excelled - when they
could perform the arithmetical operations 'in their sleep' -
they were taken to the higher levels of challenging problems
and hidden theorems enveloped in the same numbers. We can
reach the advanced levels and unveil the secrets promised
by Ahmes in the opening lines of the Rhind Mathematical
Papyrus (copy of a lost scroll of the Middle Kingdom from
around 1850 BC) playing with the numbers, recognizing to
our surprise that the same problems could have been used
for beginners and 'college' classes and 'graduates' alike.
Picture that we could use the same math textbook from
primary school to seminary ...
RMP 36 - a pair of granaries
3 '3 '5 times '4 '53 '106 '212 equals 1
Imagine a square whose diagonal measures 2 royal cubits.
Its area measures 2 square cubits.
Imagine a circle whose diagonal measures 3 royal cubits.
Using the value '135 of 424 for pi, the area of the
circle measures 7 '15 square cubits.
6/2 (plus 22/7) 28/9 ... 424/135
Build a granary on the square (inner diagonal 2 royal
cubits). Build a granary on the circle (inner diameter
3 royal cubits).
Fill the round granary to a height of 1 royal cubit.
You will need 53 quadruple-hekat or 212 hekat of barley.
Fill 212 hekat of grain in the square granary. The barley
will reach a height of 3 '3 '5 royal cubits.
Fill the square granary to a height of 1 royal cubit. You
will need 2 cubic cubits or 3 khar or 15 quadruple-hekat
or 60 hekat of barley. Fill 60 hekat of barley in the
round granary. It will reach a height of '4 '53 '106 '212
royal cubits.
|
|
