CS 390 Solutions to Homework 2


1. Textbook p. 38:

1.40 (a) Let tex2html_wrap_inline139 and tex2html_wrap_inline141 .

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) tex2html_wrap_inline205 but they are not equal.
For let tex2html_wrap_inline207 . Then there exist strings tex2html_wrap_inline209 each in tex2html_wrap_inline211 and tex2html_wrap_inline183 . Hence each tex2html_wrap_inline187 is in tex2html_wrap_inline217 and tex2html_wrap_inline219 . Hence tex2html_wrap_inline221 .
Hence tex2html_wrap_inline205.

tex2html_wrap_inline223
For let tex2html_wrap_inline139 and tex2html_wrap_inline141 , and tex2html_wrap_inline229 .
Then tex2html_wrap_inline231 . But tex2html_wrap_inline233 . They have for example aa in common.
Hence tex2html_wrap_inline223.

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 = tex2html_wrap_inline263 .
Hence we need to prove that tex2html_wrap_inline265 .
Taking common denominator for LHS and then subtracting tex2html_wrap_inline267 from both sides, we obtain tex2html_wrap_inline269 to prove.
RHS = tex2html_wrap_inline271 . Hence we need to show tex2html_wrap_inline273 .
After taking the square of the both sides, and simplifying 3k-1 > 0 is obtained, which is true since tex2html_wrap_inline277 .