Skip to main content
Contents Index
Search Book
Search Results:
No results.
Readability settings Prev Up Next
\(\require{cancel}
\newcommand{\nth}[1][n]{{#1}^{\mathrm{th}}}
\newcommand{\bbrac}[1]{\bigl(#1\bigr)}
\newcommand{\Bbrac}[1]{\Bigl(#1\Bigr)}
\newcommand{\correct}{\boldsymbol{\checkmark}}
\newcommand{\incorrect}{\boldsymbol{\times}}
\newcommand{\inv}[2][1]{{#2}^{-{#1}}}
\newcommand{\leftsub}[3][1]{\mathord{{}_{#2\mkern-#1mu}#3}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\I}{\mathbb{I}}
\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
\DeclareMathOperator{\sqrtop}{sqrt}
\newcommand{\lgcnot}{\neg}
\newcommand{\lgcand}{\wedge}
\newcommand{\lgcor}{\vee}
\newcommand{\lgccond}{\rightarrow}
\newcommand{\lgcbicond}{\leftrightarrow}
\newcommand{\lgcimplies}{\Rightarrow}
\newcommand{\lgcequiv}{\Leftrightarrow}
\newcommand{\lgctrue}{\mathrm{T}}
\newcommand{\lgcfalse}{\mathrm{F}}
\newcommand{\boolnot}[1]{{#1}'}
\newcommand{\boolzero}{\mathbf{0}}
\newcommand{\boolone}{\mathbf{1}}
\newcommand{\setdef}[2]{\left\{\mathrel{}#1\mathrel{}\middle|\mathrel{}#2\mathrel{}\right\}}
\newcommand{\inlinesetdef}[2]{\{\mathrel{}#1\mathrel{}\mid\mathrel{}#2\mathrel{}\}}
\let\emptyword\emptyset
\renewcommand{\emptyset}{\varnothing}
\newcommand{\relcmplmnt}{\smallsetminus}
\newcommand{\union}{\cup}
\newcommand{\intersection}{\cap}
\newcommand{\cmplmnt}[1]{{#1}^{\mathrm{c}}}
\newcommand{\disjunion}{\sqcup}
\newcommand{\cartprod}{\times}
\newcommand{\words}[1]{{#1}^\ast}
\newcommand{\length}[1]{\abs{#1}}
\newcommand{\powsetbare}{\mathcal{P}}
\newcommand{\powset}[1]{\powsetbare(#1)}
\newcommand{\funcdef}[4][\to]{#2\colon #3 #1 #4}
\newcommand{\ifuncto}{\hookrightarrow}
\newcommand{\ifuncdef}[3]{\funcdef[\ifuncto]{#1}{#2}{#3}}
\newcommand{\sfuncto}{\twoheadrightarrow}
\newcommand{\sfuncdef}[3]{\funcdef[\sfuncto]{#1}{#2}{#3}}
\newcommand{\funcgraphbare}{\Delta}
\newcommand{\funcgraph}[1]{\funcgraphbare(#1)}
\newcommand{\nmathrel}[1]{\mathrel{\not #1}}
\newcommand{\relset}[3]{#1_{{} #2 #3}}
\newcommand{\gtset}[2]{\relset{#1}{\gt}{#2}}
\newcommand{\posset}[1]{\gtset{#1}{0}}
\newcommand{\geset}[2]{\relset{#1}{\ge}{#2}}
\newcommand{\nnegset}[1]{\geset{#1}{0}}
\newcommand{\neqset}[2]{\relset{#1}{\neq}{#2}}
\newcommand{\nzeroset}[1]{\neqset{#1}{0}}
\newcommand{\ltset}[2]{\relset{#1}{\lt}{#2}}
\newcommand{\leset}[2]{\relset{#1}{\le}{#2}}
\newcommand{\natnumlt}[1]{\ltset{\N}{#1}}
\DeclareMathOperator{\id}{id}
\newcommand{\inclfunc}[2]{\iota_{#1}^{#2}}
\newcommand{\projfunc}[1]{\rho_{#1}}
\DeclareMathOperator{\proj}{proj}
\newcommand{\funcres}[2]{\left.{#1}\right\rvert_{#2}}
\newcommand{\altfuncres}[2]{\left.{#1}\right\rvert{#2}}
\DeclareMathOperator{\res}{res}
\DeclareMathOperator{\flr}{flr}
\newcommand{\floor}[1]{\lfloor {#1} \rfloor}
\newcommand{\funccomp}{\circ}
\newcommand{\funcinvimg}[2]{\inv{#1}\left({#2}\right)}
\newcommand{\card}[1]{\left\lvert #1 \right\rvert}
\DeclareMathOperator{\cardop}{card}
\DeclareMathOperator{\ncardop}{\#}
\newcommand{\EngAlphabet}{\{ \mathrm{a}, \, \mathrm{b}, \, \mathrm{c}, \, \dotsc, \, \mathrm{y}, \, \mathrm{z} \}}
\newcommand{\ShortEngAlphabet}{\{ \mathrm{a}, \, \mathrm{b}, \, \dotsc, \, \mathrm{z} \}}
\newcommand{\eqclass}[1]{\left[#1\right]}
\newcommand{\partorder}{\preceq}
\newcommand{\partorderstrict}{\prec}
\newcommand{\npartorder}{\npreceq}
\newcommand{\subgraph}{\preceq}
\newcommand{\subgraphset}[1]{\mathcal{S}(#1)}
\newcommand{\connectedsubgraphset}[1]{\mathcal{C}(#1)}
\newcommand{\permcomb}[3]{{#1}(#2,#3)}
\newcommand{\permcombalt}[3]{{#1}^{#2}_{#3}}
\newcommand{\permcombaltalt}[3]{{\leftsub{#2}{#1}}_{#3}}
\newcommand{\permutation}[2]{\permcomb{P}{#1}{#2}}
\newcommand{\permutationalt}[2]{\permcombalt{P}{#1}{#2}}
\newcommand{\permutationaltalt}[2]{{\permcombaltalt{P}{#1}{#2}}}
\newcommand{\combination}[2]{\permcomb{C}{#1}{#2}}
\newcommand{\combinationalt}[2]{\permcombalt{C}{#1}{#2}}
\newcommand{\combinationaltalt}[2]{{\permcombaltalt{C}{#1}{#2}}}
\newcommand{\choosefuncformula}[3]{\frac{#1 !}{#2 ! \, #3 !}}
\DeclareMathOperator{\matrixring}{M}
\newcommand{\uvec}[1]{\mathbf{#1}}
\newcommand{\zerovec}{\uvec{0}}
\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\,}$}}}
\)
Activities 17.5 Activities
Activity 17.1 .
In each of the following, describe the requested combination of relations in words. (That is, in the form โa is related to b if โฆโ.) Try to โsimplifyโ your description, if possible.
In
Taskย h and
Taskย i , the symbol
\(\mathord{\equiv}_k \) represents a relation on
\(\Z \text{,}\) where
\(m \equiv_k n \) means that
\(m \) and
\(n \) have the same remainder when divided by
\(k \text{.}\) (It may help to know that this is equivalent to
\(k \) dividing the difference
\(m - n \text{.}\) )
(a)
\(\mathord{\lt} \cup \mathord{\gt} \) on
\(\R \text{.}\)
(b)
Union of โlonger thanโ and โshorter thanโ on
\(\words{\Sigma} \) for some alphabet
\(\Sigma \text{.}\)
(c)
Union of โlonger thanโ, โshorter thanโ, and โsame length asโ on
\(\words{\Sigma} \) for some alphabet
\(\Sigma \text{.}\)
(d)
Intersection of โlonger thanโ and โshorter thanโ on
\(\words{\Sigma} \) for some alphabet
\(\Sigma \text{.}\)
(e)
The complement of
\(\mathord{\le} \) on
\(\R \text{.}\)
(f)
The inverse of
\(\mathord{\le} \) on
\(\R \text{.}\)
(g)
The inverse of โ
\(x \mathrel{R} y \) if
\(2 x + 3 y = 0 \) โ on
\(\R \text{.}\)
(h)
The intersection of
\(\mathord{\equiv_5} \) and
\(\mathord{\equiv_7} \) on
\(\Z \text{.}\)
(i)
The intersection of
\(\mathord{\equiv_2} \) and
\(\mathord{\equiv_4} \) on
\(\Z \text{.}\)
Activity 17.2 .
(a)
\(A = \R \text{,}\) \(R \) is
\(\mathord{\lt} \text{.}\)
(b)
\(A \) is the set of all straight lines in the plane,
\(R \) means โis parallel to.โ
(c)
\(A \) is the set of all straight lines in the plane,
\(R \) means โis perpendicular to.โ
(d)
\(A = \words{\Sigma} \) for some alphabet
\(\Sigma \text{,}\) \(R \) means โis the same length as.โ
(e)
\(A = \words{\Sigma} \) for some alphabet
\(\Sigma \text{,}\) \(R \) means โis shorter than.โ
(f)
\(A = \words{\Sigma} \) for some alphabet
\(\Sigma \text{,}\) \(x \) is some fixed choice of letter in
\(\Sigma \text{,}\) \(R \) means โcontains the same number of occurrences of
\(x \) as.โ
(g)
(h)
Activity 17.3 .
(a)
Suppose
\(R \) is a relation on a set
\(A \text{.}\) Convince yourself that
\(R \union \inv{R} \) is symmetric. (See the
Symmetric Relation Test .)
(b)
Recall that
\(\mathord{\mid} \) represents the relation โdividesโ on sets of integers. Draw the directed graph for
\(\mathord{\mid} \) on the set
\(A = \{2,4,6,8,10,12,14,16\} \text{.}\) Then describe how to obtain the graph for the symmetric relation
\(\mathord{\mid} \cup \inv{\mathord{\mid}} \) as an undirected graph from the graph of
\(R \) using only an eraser.
Activity 17.4 .
Assume that \(R \) and \(S \) are nonempty relations on a set \(A \) that both have the property. For each of \(\cmplmnt{R} \text{,}\) \(R \union S \text{,}\) \(R \intersection S \text{,}\) and \(\inv{R} \text{,}\) determine whether the new relation
must also have that property;
might have that property, but might not; or
cannot have that property.
Any time you answer
Statementย i or
Statementย iii , outline a proof. Any time you answer
Statementย ii , provide two examples: one where the new relation has the property, and one where the new relation does not. (You may use graphs to describe your examples.)