Skip to main content ☰ Contents Index You! < 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{\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\,}$}}}
\)
Section 6.7 Proof by counterexample
Sometimes we want to prove that \(P \not\lgcimplies Q\text{;}\) i.e. that \(P \lgccond Q\) is not a tautology.
Recall.
The equivalence
\begin{equation*}
P \lgccond Q \lgcequiv (P \lgcand C_1 \lgccond Q) \lgcand \dotsb \lgcand (P \lgcand C_m \lgccond Q)
\end{equation*}
holds for any set of cases
\(C_1, C_2, \dotsc, C_m\) such that
\(C_1 \lgcor \dotsb \lgcor C_m\) is a tautology. (See
Section 6.4 .)
So if
\(P \lgcand C_i \lgccond Q\) is
not a tautology for at least one
\(i\text{,}\) then
\(P \lgccond Q\) also cannot be a tautology. Again, this also works in the universal case since
\(\forall\) distributes over
\(\lgcand\) (
Proposition 4.2.6 ).
counterexample
relative to the logical implication \(P \lgcimplies Q\text{,}\) a statement \(C\) such that \(P \lgcand C \lgccond Q\) is false
Worked Example 6.7.1 .
In
Exercise 6.12.8 , you are asked to prove the following statement by proving the contrapositive.
If \(2^n - 1\) prime, then \(n\) is prime.
Prove that the converse of this statement is false .
Solution .
The converse statement is “If \(n\) is prime, then \(2^n - 1\) is prime.” But the case \(n=11\) is a counterexample :
\begin{equation*}
2^{11} - 1 = 2047 = 23 \cdot 89
\end{equation*}
is not prime even though \(n = 11\) is prime.
Check your understanding.
