Skip to main content\(\newcommand{\im}{{\mathrm{im}}}
\newcommand{\given}{{\, :\,}}
\newcommand{\hanoi}{\mathrm{hanoi}}
\newcommand{\lcm}{{\mathrm{lcm}}}
\newcommand{\ID}{{\mathrm{ID}}}
\newcommand{\chr}{{\mathtt{chr}}}
\newcommand{\ord}{{\mathtt{ord}}}
\newcommand{\set}[1]{\left\lbrace #1 \right\rbrace}
\newcommand{\seq}[2]{( #1 )_{#2}}
\newcommand{\gr}[1]{\left( #1 \right)}
\newcommand{\abs}[1]{\left| #1 \right|}
\newcommand{\size}[1]{\# #1}
\newcommand{\id}[1]{\mathrm{id}_{#1}}
\newcommand{\ip}[1]{\left\langle #1 \right\rangle}
\newcommand{\floor}[1]{\left\lfloor #1\right\rfloor}
\newcommand{\ceiling}[1]{\left\lceil #1\right\rceil}
\newcommand{\indemph}[1]{{\textbf{\emph{#1}}\index{#1}}}
\renewcommand{\implies}{\to}
\renewcommand{\iff}{\leftrightarrow}
\newcommand{\lxor}{\oplus}
\newcommand{\cP}{\wp}
\newcommand{\pset}{\cP}
\newcommand{\ZZ}{\mathbb{Z}}
\newcommand{\QQ}{\mathbb{Q}}
\newcommand{\NN}{\mathbb{N}}
\newcommand{\CC}{\mathbb{C}}
\newcommand{\RR}{\mathbb{R}}
\newcommand{\barr}[1]{\overline{#1}}
\newcommand{\sage}{\textsc{Sage}}
\newcommand{\magma}{\textsc{Magma}}
\newcommand{\bob}{\text{Bob}}
\newcommand{\true}{\text{T}}
\newcommand{\false}{\text{F}}
\newcommand{\ahat}{\hat{a}}
\newcommand{\ah}{\ahat}
\newcommand{\bhat}{\hat{b}}
\newcommand{\abar}{\bar{a}}
\newcommand{\cA}{\mathcal{A}}
\newcommand{\cF}{\mathcal{F}}
\newcommand{\cW}{\mathcal{W}}
\newcommand{\todo}[1]{{\color{red} TODO: #1}}
\renewcommand{\div}{\operatorname{div}}
\newcommand{\gexp}[3]{#1^{#2 #3}}
\newcommand{\gexpp}[3]{\displaystyle\left(#1\right)^{#2 #3}}
\newcommand{\glog}[3]{\log_{#1}^{#3}#2}
\def\dbend{{!!!}}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\definecolor{fillinmathshade}{gray}{0.9}
\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
\)
Exercises 1.1.7 Exercises
1.
Give the definition for these terms. Be sure to set up any notation that is required.
-
-
negation of a proposition
-
conjunction of two propositions
-
disjunction of two propositions
-
exclusive disjunction of two propositions
-
conditional of two propositions
-
biconditional of two propositions
-
converse of a conditional
-
contrapositive of a conditional
-
2.
Complete the following truth table.
\begin{equation*}
\begin{array}{|c|c||c|c|c|c|c|}
\hline
p\amp q\amp p\lor q\amp p\lxor q \amp p\land q \amp p \implies q \amp p \iff q\\
\hline
\true\amp \true\amp \amp \amp \amp \amp \\\hline
\true\amp \false\amp \amp \amp \amp \amp \\\hline
\false\amp \true\amp \amp \amp \amp \amp \\\hline
\false\amp \false\amp \amp \amp \amp \amp \\\hline
\end{array}
\end{equation*}
3.
Identify each of the following as a proposition or not. For each proposition, give the truth value.
-
Greensboro is the capital of North Carolina.
-
Letβs meet at the dining hall at 7pm.
-
Squares have three sides.
-
-
4.
Identify each of the following as a proposition or not. For each proposition, give the truth value.
-
\(\displaystyle 1 + 1 = 2\)
-
-
If
\(2 + 3 = 7\) , then
\(11 - 4 = 7\text{.}\)
-
If
\(11 - 4 = 7\) , then
\(2 + 3 = 7\text{.}\)
-
\(3\) is odd or
\(253\) is odd.
-
If the sun came up today, then squares have four
5.
Identify each of the following as a proposition or not. For each proposition, give the truth value.
-
-
-
If cats are reptiles, then dogs are mammals.
-
If dogs are mammals, then cats are reptiles.
-
Cats are reptiles if and only if dogs are mammals.
6.
What is the negation of each of these propositions?
-
There are
\(60\) minutes in an hour.
-
\(169\) is a perfect square.
-
Alice is less than
\(5\) feet tall.
-
Bob has more than
\(20\) books.
-
\(\displaystyle 5 \cdot 6 = 30\)
7.
Let
\(r\) be the proposition βIt is raining.β and let
\(w\) be βThe ground is wet.β
Express each of the propositions as an English sentence.
-
\(\displaystyle \lnot r\)
-
\(\displaystyle r\land w\)
-
\(\displaystyle r \implies w\)
-
\(\displaystyle \lnot r \implies\lnot w\)
-
\(\displaystyle \lnot r\land\lnot w\)
8.
Let
\(p\) be the propositions βThere are
\(8\) pints in a gallon.β and let
\(q\) be βThere are
\(4\) quarts in a gallon.β.
Write each of these propositions using \(p\) and \(q\) and logical operators (including negations).
-
There are
\(8\) pints in a gallon, or there are
\(4\) quarts in a gallon.
-
If there are
\(8\) pints in a gallon, then there are not
\(4\) quarts in a gallon.
-
There are not
\(8\) pints in a gallon, or there are
\(8\) pints in a gallon and
\(4\) quarts in a gallon.
-
There are
\(8\) pints in a gallon if and only if are
\(4\) quarts in a gallon.
-
There are not
\(8\) pints in a gallon, but there are
\(4\) quarts in a gallon.
9.
Let
\(p\) be the proposition βYou have the flu.β and let
\(q\) be βYou miss the final exam.β and let
\(r\) be βYou pass the course.β
Express each proposition as an English sentence.
-
\(\displaystyle p \implies (q\lor r)\)
-
\(\displaystyle (\lnot q\land r)\lor r\)
-
\(\displaystyle p \implies (q\land r)\)
-
\(\displaystyle (\lnot p\lor\lnot q) \implies r\)
-
\(\displaystyle (p\land q)\lor (\lnot q\land r)\)
10.
Write each of these statements in the form βif \(p\) then \(q\text{.}\)β
-
It is necessary to study every day to pass MAT 253.
-
Alice gets caught whenever she cheats.
-
To pass MAT 253, it is sufficient to attend class every day.
-
The warranty is good only if you bought the computer less than
\(1\) year ago.
-
Bob will get a good job unless he does not learn discrete mathematics.
11.
State the converse, contrapositive, and inverse of each of these conditional statements.
-
If it is snowing tonight, I will stay home.
-
The home team wins whenever it rains.
-
Alice eats a desert only if she eats her vegetables.
-
When Bob stays up late, it is necessary for him to sleep until noon.
-
A positive integer is prime only if it has no positive divisors other than
\(1\) and itself.
12.
Construct a truth table for each of these compound propositions.
-
\(\displaystyle (p\land q) \implies r\)
-
\(\displaystyle (p\lor q) \implies\lnot r\)
-
\(\displaystyle p \implies (q\land r)\)
-
\(\displaystyle \lnot q \implies (r\lor\lnot p)\)
-
\(\displaystyle p\land (q\lor\lnot r)\)