Solutions to Problem Set 7

Math 310-01/02

9-20-2017

1. (a) Prove that if n is even, then $n^2 + 3 n + 16$ is even.

(b) Give a specific counterexample to show that the converse is false.

(a) Let n be an even integer. Then $n = 2 m$ for some $m \in \integer$ . So

$$n^2 + 3 n + 16 = (2 m)^2 + 3(2 m) + 16 = 4 m^2 + 6 m + 16 = 2(2 m^2 + 3 m + 8).$$

Hence, $n^2 + 3 n + 16$ is even.

(b) The converse is: "If $n^2
   + 3 n + 16$ is even, then n is even." However, if $n = 1$ , then $n^2 + 3 n + 16 = 20$ , which is even, while n is odd. Thus, the converse is false.


2. Prove that if $x^2 + 5 x - 14
   \le 0$ , then $x \le 2$ .

I'll prove the contrapositive: If $x > 2$ , then $x^2 + 5 x - 14 > 0$ . Suppose that $x > 2$ . Then $x^2 > 4$ and $5 x > 10$ , so

$$x^2 + 5 x > 4 + 10 = 14.$$

Therefore, $x^2 + 5 x - 14 > 0$ .


3. (a) Show that if $\ln x + \ln
   (x - 7) = \ln 8$ , then $x = 8$ or $x = -1$ .

(b) Show that it's not true that if $x = 8$ or $x = -1$ , then $\ln x + \ln (x - 7) = \ln 8$ .

(a)

$$\eqalign{\ln x + \ln (x - 7) & = \ln 8 \cr \ln x(x - 7) & = \ln 8 \cr e^{\ln x(x - 7)} & = e^{\ln 8} \cr x(x - 7) & = 8 \cr x^2 - 7 x - 8 & = 0 \cr (x - 8)(x + 1) & = 0 \cr}$$

Therefore, $x = 8$ or $x =
   -1$ .

(b) Plugging $x = 8$ into $\ln x + \ln (x - 7) = \ln 8$ gives $\ln 8 +
   \ln 1 = \ln 8$ , which is true. The answer checks.

Plugging $x = -1$ into $\ln x + \ln (x - 7) = \ln 8$ gives $\ln (-1) +
   \ln (--8) = \ln 8$ , which doesn't make sense --- the log of a negative number is undefined. The answer doesn't check.


4. Premises: $\displaystyle
   \left\{\matrix{(A \lor C) \ifthen \lnot B \cr D \ifthen \lnot A
   \cr}\right.$ .

Prove: $A \ifthen \lnot(B \lor
   D)$ .

$$\matrix{ 1. \hfill & (A \lor C) \ifthen \lnot B \hfill & \hbox{Premise} \hfill \cr 2. \hfill & D \ifthen \lnot A \hfill & \hbox{Premise} \hfill \cr 3. \hfill & A \hfill & \hbox{Premise for conditional proof} \hfill \cr 4. \hfill & A \lor C \hfill & \hbox{Constructing a disjunction} \hfill \cr 5. \hfill & \lnot B \hfill & \hbox{Modus ponens (1,4)} \hfill \cr 6. \hfill & \lnot D \hfill & \hbox{Modus tollens (2,3)} \hfill \cr 7. \hfill & \lnot B \land \lnot D \hfill & \hbox{Constructing a conjunction (5,6)} \hfill \cr 8. \hfill & \lnot(B \lor D) \hfill & \hbox{DeMorgan (7)} \hfill \cr 9. \hfill & A \ifthen \lnot(B \lor D) \hfill & \hbox{Conditional proof (3,8)} \quad\halmos \hfill \cr}$$


We are what we think. All that we are arises with our thoughts. With our thoughts we make the world. - The Buddha


Contact information

Bruce Ikenaga's Home Page

Copyright 2017 by Bruce Ikenaga