We sketch the proof of the easy half of the theorem.
This proves one half of (2); the other half is a matter of direct computation.
the left half of the interval
the upper half of the unit disc
Thus F is greater by a half.
F is half the sum of the positive roots. [Or: half of the sum]
Half of the sets of R miss i and half the remaining miss j.
Then J contains an interval of half its length in which f is positive.
On average, about half the list will be tested.
Now F is half as long as G. [Or: as G is]
We divide N in half.
The length of F is thus reduced by half.