```source("http://macosa.dima.unige.it/r.R")    # If I have not already loaded the library
---------- ---------- ---------- ---------- ---------- ---------- ---------- ----------
# For what positive value of x  x^2 = 2 ?
# I have to look for x such that  x^2 - 2 = 0
f = function(x) x^2-2
graphF(f, -3,3, "blue")

# Between 0 and 2 the graph of  f  cross the x axis.
# I define a procedure that bisects the interval and selects the subinterval in which
# F changes the sign.
Zero=function(f,a,b) {m=a+(b-a)/2;y1=sign(f(a));y2=sign(f(m)); if(y1==y2) a<<- m else b<<- m}
# In  a  and  b  the extremes of this subinterval are stored
a=0;b=2
Zero(f,a,b); c(a,b)
#  1  2
Zero(f,a,b); c(a,b)
#   1.0  1.5
Zero(f,a,b); c(a,b)
#  1.25  1.50
Zero(f,a,b); c(a,b)
#  1.375  1.500
# ...
Zero(f,a,b); c(a,b)
#  1.414213  1.414214
Zero(f,a,b); c(a,b)
#  1.414214  1.414214
#
# The procedure "solution" automates this process.
#
# NOTE. Why (to calculate the mean of a and b) we use a+(b-a)/2 instead of (a+b)/2 ?
# If I use a calculator with an 8-digit display (that works with 8 digits) for
# computing (5.0000001 + 5.0000003) / 2 I obtain 10.00000/2 and then 5; if I
# compute (5.0000003 + 5.0000004) / 2 I obtain 10.00001/2 and the 5.0000005.
# I get results that are out of [a,b].
# If I use a+(b-a)/2 I cannot get values outside [a,b].
```