NAME:

Instructor: Nataša Jonoska

MGF 1106-008, Finite Math

Fall 2002

TEST # 1, September 25, 2002

This test has 110 points, but will be graded out of 100. Every problem is graded out of 10 points, unless otherwise stated.

There are 8 problems (4 pages) on the test, so be sure you have them all. Read the entire test before starting. If you don't understand something, please ask.

Write out your solutions in a clear and precise manner. If explanation is needed, write it using complete sentences! Points will be taken off for not complete or precise statements.

Partial credit will be given for partial answers. Please show your work entirely. If you use the back of the last page, please so indicate.

1. (2 pts each) Indicate whether the following statements are true (T) or false (F): (a) 5,7,8,0,2=0,5,2,7,8

Solution: True
(b) $\{\emptyset\}= \emptyset$

Solution: False
(c) $\sim(p\lor q)\,\Leftrightarrow \,\,\sim p \,\lor \sim q$

Solution: False
(d) The contrapositive of a conditional statement $r\rightarrow s$ is equivalent to $\sim s\rightarrow \sim r$ .

Solution: True
(e) $A\cup A=A$

Solution: True

2. (3 pts each) In the following statements fill in the missing values such that the statements become true statements. (a) $\{2,3,7\}\cap \{2,7\}=$

Solution: 2,7
(b) The converse of $p\rightarrow q$ is

Solution: $q\rightarrow p$
(c) The negation of $r\rightarrow (\sim s)$ is

Solution: $r\land s$
(d) By DeMorgan's laws, the set $(A\cup B)'$ is equal to the set

Solution: $A'\cap B'$
(e) By DeMorgan's laws, $(\sim s)\lor (\sim r)$ is equivalent to

Solution: $\sim (s\land r)$
(f) Let A be a set. Then $\emptyset \cup A=$

Solution: A
(g) Let be a set. Then $\emptyset \cap A=$

Solution: $\emptyset$
(h) The subsets of $\{0,*,a\}$ are

Solution: $\emptyset$ , 0, *, a, 0,*, 0,a, a,*, 0,a,*
(i) The set $\{0,*,a,\triangle\}$ has $\dots\dots\dots$ (number of) subsets.

Solution: 16
(j) Let be the universal set and $B\subseteq U$ . Then

Solution: B
(k) If ``You have a calculator'' and ``You pass the test'' then the statement ``You cannot pass the test if you have a calculator'' written in symbolic form is

Solution: $p\rightarrow \sim q$

3. Consider the following statements:

(1)``If you own a motorcycle then you have no trouble parking.''

(2)``Everybody eats strowberies for breakfast.''

(a) Write the first statement in symbolic form.

Solution: We set p:``You own a motorcycle.'' and q:``You have trouble parking.'' Then statement (1) in symbolic form is: $p\rightarrow \sim q$ .

(b) Write the contrapositive and the negation of statement (1)in symbolic form.

Solution: Contrapositive: $q\rightarrow \sim p$

Negation: $p\land q$ (c) Express the contrapositive and the negation with words.

Solution: Contrapositive: If you have trouble parking, then you don't own a motorcycle.

Negation: You own a motorcycle and you have trouble parking. (d) Write the negation of the staement (2).

Solution: Some people don't eat strawberries for breakfast.

(e) Write a negation of ``Some people don't eat strowberies for breakfast.''

Solution: Everybody eats strawberries fro breakfast.

In the diagram (a) below, the regions are numbered 1 through 8. What numbers belong to the regions represented with sets:

$\begin{displaymath} (i)\quad (A\cap B)\cup C\qquad \qquad\qquad (ii) \quad (A'\cap C) \qquad\qquad\qquad (iii)\quad (A\cup B)\cap C' \end{displaymath}$

Solution: (i) 2,5,6,7,8

(ii) 7,8

(ii) 1,2,3

$\begin{figure}\begin{center} \mbox{ \epsfxsize =5.5in \epsfbox{q2diagram.eps}} \end{center}\end{figure}$

In the diagram (b), (i) write

in the regions that belong to the set $B-(A\cup C)$ .

Solution: See the figure above!
(ii) write $\triangle$ in the regions that belong to the set $A'\cap C' \cap B$ .

Solution: See the figure above!
(iii) Does the equality $B-(A\cup C)=A'\cap C' \cap B$ hold?

Solution: Yes!

5. (10 pts) Determine whether the following statements are equivalent and show your work:

`` My doctor lied, or I am hearing things.''

`` If I am hearing things then my doctor lied.''

Solution: Set p: `` My doctor lied.'' and q: ``I am hearing things.''

Then the first statement written in symbolic form is: $p\lor q$ and the second is $q\rightarrow p$ . We know that the conditional $q\rightarrow p$ is equivalent to $\sim q \lor p$ i.e. to $p\lor \sim q$ . Then $p\lor q$ and $p\lor \sim q$ are not equivalent since for and the statement $p\lor q$ is true but the statement $p\lor \sim q$ is false.

This could be observed if the truth table for both of these statements is constructed.

6. (15 pts) Determine whether the following argument is valid and explain your answer: If you rent videocassettes, you do not go to the movie theater. If you go to the movie theater, you pay attention to the movie. Therefore, you do not pay attention to the movie if you rent videocassettes.

Solution: Set: p: ``You rent videocassettes.'', q: ``You go to the movie theater.'' and r: ``You pay attention to the movie.''

Then the argument looks like:

$p\rightarrow \sim q$

$q\rightarrow r$

$p\rightarrow \sim r$

The logical statement that corresponds to the argument is: $(p\rightarrow \sim q)\land (q\rightarrow r)\rightarrow (p\rightarrow \sim r)$ . The argument is valid if this statement is true regardless of the values of p,q,r. One can check the validity of the argument by a truth table:

$\begin{figure}\begin{center} \mbox{ \epsfxsize =5.5in \epsfbox{truth-table.eps}} \end{center}\end{figure}$

Since the argument statement gets value false when p:=T, q:=F, and r:=T, the argument is not valid.

7. Using Euler diagrams determine the validity of the following argument:

``All mathematics teachers have publications. Some PhD's have publications. Therefore, some PhD's are mathematics teachers.''

Solution: Let M:= the set of all mathematics teachers.

P:= the set of all people that have publications.

D:= the set of all people with PhD.

The first statement syas that the set of mathematics teachers is a subset of the set of all people with publications $M\subseteq P$ . The second statement says that the set of people with PhD has a non-empty intersection with the set of people with publications $P\cap D\not = \emptyset$ . An Euler diagram showing this relationship is:

$\begin{figure}\begin{center} \mbox{ \epsfxsize =4in \epsfbox{euler.eps}} \end{center}\end{figure}$

Since the set of Mathematics teachers does not have to intesect the set of people with PhD, the argument is invalid.

8. (15 pts) Freshmen students at USF have been polled about what mathematics classes have they signed up during their first semester. 220 students said that they have signed up for Finite Math, 170 have signed up only Finite Math, 330 have signed up for Chaos & Fractals, and 180 sugned up only for Chaos & Fractals, 20 have signed up for Finite Math and Chaos & Fractals, 700 have signed up for College Algebra, no student has signed up for all three classes, 700 students have not signed up for any math class. (a) Draw a Venn Diagram that represents the polling and identify the number of students that would belong to each region.

Solution: FM=students that take Finite Math.

CF=students that take Chaos and Fractals.

CA= Students that take College Algebra.

$\begin{figure}\begin{center} \mbox{ \epsfxsize =5.5in \epsfbox{t-diagram.eps}} \end{center}\end{figure}$

(b) How many students were polled?

Solution: The sum of the number of students that correspond to each of the regions is 1770.

Solution: The number of students who have signed up for more than one math class is sum of the numbers in the regions that belong to intersections of at least two sets. They are: 130+20+30=180. This is approximately 10% of the total number of poled students.