1) Rewrite this expression so that it does not use the ! operator (use 
   DeMorgan's Law):

   !(pt.get_x() > -1 && pt.get_x() < 1 && 
     pt.get_y() > -1 && pt.get_y() < 1)

Solution:

    First pass:
    (!(pt.get_x() > -1) || !(pt.get_x() < 1) ||
    !(pt.get_y() > -1) || !(pt.get_y() < 1))

    Final pass:
    (pt.get_x() <= -1 || pt.get_x() >= 1 ||
     pt.get_y() <= -1 || pt.get_y() >= 1)



2) Are the following 2 expressions equivalent?

Lec 001: a >= b && a != c
         !(a < b || a == c)

Solution: Yes

Lec 002: a >= b && a != c
         (a < b || a == c)

Solution: No



3) Write a for loop that outputs the integers from n to m given that n < m.
   Do not change the values of n and m.

Solution:

for (int i = n; i <= m; i++)
{
   cout << i << " ";
}