Skip to main content
Contents Index Dark Mode Prev Up Next \(\newcommand{\im}{{\mathrm{im}}}
\newcommand{\given}{{\, :\,}}
\newcommand{\hanoi}{\mathrm{hanoi}}
\newcommand{\lcm}{{\mathrm{lcm}}}
\newcommand{\ID}{{\mathrm{ID}}}
\newcommand{\chr}{{\mathtt{chr}}}
\newcommand{\ord}{{\mathtt{ord}}}
\newcommand{\set}[1]{\left\lbrace #1 \right\rbrace}
\newcommand{\seq}[2]{( #1 )_{#2}}
\newcommand{\gr}[1]{\left( #1 \right)}
\newcommand{\abs}[1]{\left| #1 \right|}
\newcommand{\size}[1]{\# #1}
\newcommand{\id}[1]{\mathrm{id}_{#1}}
\newcommand{\ip}[1]{\left\langle #1 \right\rangle}
\newcommand{\floor}[1]{\left\lfloor #1\right\rfloor}
\newcommand{\ceiling}[1]{\left\lceil #1\right\rceil}
\newcommand{\indemph}[1]{{\textbf{\emph{#1}}\index{#1}}}
\renewcommand{\implies}{\to}
\renewcommand{\iff}{\leftrightarrow}
\newcommand{\lxor}{\oplus}
\newcommand{\cP}{\wp}
\newcommand{\pset}{\cP}
\newcommand{\ZZ}{\mathbb{Z}}
\newcommand{\QQ}{\mathbb{Q}}
\newcommand{\NN}{\mathbb{N}}
\newcommand{\CC}{\mathbb{C}}
\newcommand{\RR}{\mathbb{R}}
\newcommand{\barr}[1]{\overline{#1}}
\newcommand{\sage}{\textsc{Sage}}
\newcommand{\magma}{\textsc{Magma}}
\newcommand{\bob}{\text{Bob}}
\newcommand{\true}{\text{T}}
\newcommand{\false}{\text{F}}
\newcommand{\ahat}{\hat{a}}
\newcommand{\ah}{\ahat}
\newcommand{\bhat}{\hat{b}}
\newcommand{\abar}{\bar{a}}
\newcommand{\cA}{\mathcal{A}}
\newcommand{\cF}{\mathcal{F}}
\newcommand{\cW}{\mathcal{W}}
\newcommand{\todo}[1]{{\color{red} TODO: #1}}
\renewcommand{\div}{\operatorname{div}}
\newcommand{\gexp}[3]{#1^{#2 #3}}
\newcommand{\gexpp}[3]{\displaystyle\left(#1\right)^{#2 #3}}
\newcommand{\glog}[3]{\log_{#1}^{#3}#2}
\def\dbend{{!!!}}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\definecolor{fillinmathshade}{gray}{0.9}
\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
\)
Exercises 2.3.8 Exercises
1.
Give the definition for these terms. Be sure to set up any notation that is required.
image or range of a function
image of a set under a function
2.
Let \(f(x) = x^2 + 1\) and \(g(x) = x + 2\) be functions from \(\RR\) to \(\RR\text{.}\)
Find the composition
\(h = f \circ g\text{.}\) What is the image of
\(h\text{?}\)
Find the composition
\(i = g \circ f\text{.}\) What is the image of
\(i\text{?}\)
3.
Construct a function that is
injective but not surjective.
surjective but not injective.
neither surjective nor injective.
both injective and surjective.
4.
Let \(f \colon \RR \to \RR\) be defined by \(f(x) = 3x+ 1\) Prove that
5.
Let \(\cW\) be the set of words in the Oxford English Dictionary, and let \(\cA\) be the English alphabet. Let \(f \colon \cW \to \cA\) be the function defined by
\begin{equation*}
f(x) = \text{first letter of }x.
\end{equation*}
What is the image of
\(f\text{?}\)
Is
\(f\) is surjective? Justify.
Is
\(f\) is injective? Justify.
6.
Give an explicit example of a function from \(\ZZ\) to \(\ZZ\) that is
injective, but not surjective.
surjective, but not injective.
injective and surjective.
neither injective nor surjective.
Justify your answers.
7.
Let \(S = \set{-2, -1, 0, 3, 4, 5}\) Find \(f(S)\) for each of the following. Be sure to express the answer as a set.
\(\displaystyle f(x) = 1\)
\(\displaystyle f(x) = 2x + 1\)
\(\displaystyle f(x) = x^2\)
\(\displaystyle f(x) = \ceiling{\frac{x}{2}}\)
\(\displaystyle f(x) = \floor{\frac{x^2 + 1}{3}}\)
8. Let
\(A\) and
\(B\) be sets, and let
\(f \colon A \to B\) be a function. Let
\(S\) and
\(T\) be subsets of
\(A\text{.}\) Prove
\(f(S \cup T) = f(S) \cup f(T)\text{.}\)
9.
Let \(A\) and \(B\) be sets, and let \(f \colon A \to B\) be a function. Let \(S\) and \(T\) be subsets of \(A\text{.}\)
Prove
\(f(S \cap T) \subseteq f(S) \cap f(T)\text{.}\)
Give an explicit example that shows the inclusion above can be proper.
10. Let
\(A\) and
\(B\) be sets, and let
\(f \colon A \to B\) be a function. Let
\(S\) and
\(T\) be subsets of
\(A\text{.}\) Prove that if
\(f\) is injective, then
\(f(S \cap T) = f(S) \cap f(T)\text{.}\)
