1.40 (a) Let and .
1.45 (a)a, ab, aab, aaabaabaaaab, aaba etc.
p(x): x is a string of a's and b's that starts with a and
that does not have substring 'bb'.
1.47(b) but they are not equal.
For let . Then there exist strings each
in and . Hence each is in and
. Hence .
Hence .
For let and , and .
Then . But .
They have for example aa in common.
Hence .
2. Textbook p. 78:
2.49 Basis Step: Let n = 1. Then LHS = 1 and RHS = 2/3. Hence LHS > RHS.
Induction: Assume true for n = k.
Then for n = k+1, LHS = .
Hence we need to prove that .
Taking common denominator for LHS and then subtracting from both sides, we
obtain to prove.
RHS = . Hence we need to show .
After taking the square of the both sides, and simplifying 3k-1 > 0 is obtained,
which is true since .