Answer :
A little thought
will make it clear that the
answer must be fractional, and that in one case the numerator will be
greater and in the other case less than the denominator.
As a matter of
fact, the height of the larger cube must be ^{8}/_{7}
ft., and of the smaller ^{3}/_{7}
ft., if we are to have the answer in the smallest possible figures.
Here the lineal measurement is ^{11}/_{7}
ft.—that is, 1^{4}/_{7}
ft. What are the cubic contents of the two cubes?
First ^{8}/_{7} × ^{3}/_{7} × ^{8}/_{7} = ^{512}/_{343},
and secondly ^{3}/_{7} × ^{3}/_{7} × ^{3}/_{7} = ^{27}/_{343}.
Add these together and the result is ^{539}/_{343},
which reduces to ^{11}/_{7}
or
1^{4}/_{7}
ft.
We thus see that
the answers in cubic feet and lineal feet are precisely the same.
The germ of the
idea is to be found in the works
of Diophantus of Alexandria, who wrote about the beginning of the
fourth century. These fractional numbers appear in triads, and are
obtained from three generators, a,
b,
c,
where a
is the
largest and c
the smallest.
Then ab+c^{2}=denominator,
and a^{2}c^{2},
b^{2}c^{2},
and a^{2}b^{2}
will be the three numerators.
Thus, using the generators 3, 2, 1, we
get ^{8}/_{7},
^{3}/_{7},
^{5}/_{7}
and we
can pair the first and second, as in the above solution, or the first
and third for a second solution.
The denominator must always be a prime
number of the form 6n+1,
or composed of such primes.
Thus you can have 13, 19, etc., as denominators, but not 25, 55, 187,
etc.
When the principle
is understood there is no
difficulty in writing down the dimensions of as many sets of cubes as
the most exacting collector may require.
If the reader would like one,
for example, with plenty of nines, perhaps the following would satisfy
him: ^{99999999}/_{99990001}
and ^{19999}/_{99990001}.
