CS 381 Solutions to Homework 8

pp. 199 - 201

8. Basis Step: Let $n = 1$. Then $LHS = 1^{3} = 1$ and $RHS = [1*(1+1)/2]^{2} =1^{2} =1$.
Hence $LHS = RHS$.
Inductive Step: Assume that $1^{3} + 2^{3} + ... + n^{3} = [n(n+1)/2]^{2}$.
Then $1^{3} + 2^{3} + ... + n^{3} + (n+1)^{3} = (1^{3} + 2^{3} + ... + n^{3}) + (n+1)^{3}$
= $[n(n+1)/2]^{2} + (n+1)^{3}$
= $(n+1)^{2}[n^{2}/4 + (n+1)]$
= $(n+1)^{2}[n^{2} + 4n + 4]/4$
= $(n+1)^{2}(n+2)^{2}/4$
= $[(n+1)(n+2)/2]^{2}$, which is the $RHS$ of the equality for $n+1$.

14. Basis Step: Let $n = 2$. Then $LHS = 2! = 2$ and $RHS = 2^{2} = 4$.
Hence $n! < n^{n}$ when $n = 2$.
Inductive Step: Assume that $n! < n^{n}$.
Then $(n+1)! = n! (n+1) < n^{n}(n+1)$. Since $n^{n} < (n+1)^{n}$, $n^{n}(n+1) < (n+1)^{(n+1)}$.
Hence $(n+1)! < (n+1)^{(n+1)}$.

24. Basis Step: The smallest $n$ is $4$.
When $n=4$, $n^{2} - 7n + 12 = 16 -28 + 12 = 0$. Hence $n^{2} - 7n + 12$ is nonnegative.
Inductive Step: Assume that $n^{2} - 7n + 12 \geq 0$. Then $(n+1)^{2} -7(n+1) + 12 = n^{2} + 2n + 1 -7n - 7 + 12$
= $(n^{2} -7n + 12) + 2n - 6$
$n^{2} - 7n + 12 \geq 0$ by thge induction hypothesis and $2n - 6 \geq 0$ since $n > 3$. Hence $(n+1)^{2} -7(n+1) + 12 \geq 0$ if $n > 3$.

31 a, b: See the back of the textbook pp. S-23   S-24.

42 b) Basis Step: Let $n = 2$. (For $n = 1$ it is simpler.)
Then since $A_{1} \subseteq B_{1}$ and $A_{2} \subseteq B_{2}$, by the property 7 of set operation $(A_{1} \cap A_{2}) \subseteq (B_{1} \cap B_{2})$.
That is $\cap_{k=1}^{2} A_{k} \subseteq \cap_{k=1}^{2} B_{k}$.
Inductive Step: Assume that $A_{k} \subseteq B_{k}$ for $k = 1, ..., n$ and $\cap_{k=1}^{n} A_{k} \subseteq \cap_{k=1}^{n} B_{k}$ .
Then $\cap_{k=1}^{n+1} A_{k} = (\cap_{k=1}^{n} A_{k}) \cap A_{n+1}$ and $\cap_{k=1}^{n+1} B_{k} = (\cap_{k=1}^{n} B_{k}) \cap B_{n+1}$.
Hence again by the property 7 of set operation,
$(\cap_{k=1}^{n} A_{k}) \cap A_{n+1} \subseteq (\cap_{k=1}^{n} B_{k}) \cap B_{n+1}$.
Hence $\cap_{k=1}^{n+1} A_{k} \subseteq \cap_{k=1}^{n+1} B_{k}$.

Textbook pp. 209 - 210

4 b) $f(2) = f(3) =f(4) = f(5) = 1$
d) $f(2) = f(3) =f(4) = f(5) = 1$