THE QUADRATURE OF THE PARABOLA 91
e area of the
exhausts the
gures ‘ in the
this purpose
with the same
added to the
Le number of
ie segment,
scribed ‘ in the
■ the point of
lei to Qq, and
q. H QV, Vq
, rm be drawn
jurve in R, r,
5 of the next
lized manner’,
acting FQ, Pq
RM=3PW.
= 2 RY.
’Qq■ (Prop. 21.)
next addition
As PRQ, Prq,
(Prop. 22.)
are is greater
7er before the
ent at P and
larallel to PV,
the triangle PQq is half of the parallelogram and therefore
more than half the segment. And so on (Prop. 20).
We now have to sum n terms of the above geometrical
series. Archimedes enunciates the problem in the form, Given
a series of areas A, B,C,D ... Z, of which A is the greatest, and
each is equal to four times the next in order, then (Prop. 23)
A + B + G+... + Z+±Z=±A.
The algebraical equivalent of this is of course
i+è+(i) a +...+(ir- 1 = l-.-l(è)'
i-i
To find the area of the segment, Archimedes, instead of
taking the limit, as we should, uses the method of reductio ad
absurdum.
Suppose K = f. A PQq.
(1) If possible, let the area of the segment be greater than K.
We then inscribe a figure ‘in the recognized manner’ such
that the segment exceeds it by an area less than the excess of
the segment over K. Therefore the inscribed figure must be
greater than K, which is impossible since
A + B + G+ ...+Z < §A,
where A = A PQq (Prop. 23).
(2) If possible, let the area of the segment be less than K.
If then A PQq = A, B = ^ A, G = \B, and so on, until we
arrive at an area X less than the excess of K over the area of
the segment, we have
A+B+C+ ... +X + $X = $ A=K.
Thus K exceeds A + B + G + ... + X by an area less than X,
and exceeds the segment by an area greater than X.
It follows that A + B + G + ... + X > (the segment); which
is impossible (Prop. 22).
Therefore the area of the segment, being neither greater nor
less than K, is equal to K or f A PQq.
On Floating Bodies, I, II.
In Book I of this treatise Archimedes lays down the funda
mental principles of the science of hydrostatics. These are
