18.2: Basics and Examples
- Page ID
- 83495
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\dsum}{\displaystyle\sum\limits} \)
\( \newcommand{\dint}{\displaystyle\int\limits} \)
\( \newcommand{\dlim}{\displaystyle\lim\limits} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\(\newcommand{\longvect}{\overrightarrow}\)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\ket}[1]{\left| #1 \right>}\)
\(\newcommand{\bra}[1]{\left< #1 \right|}\)
\(\newcommand{\braket}[2]{\left< #1 \vphantom{#2} \right| \left. #2 \vphantom{#1} \right>}\)
\(\newcommand{\braopket}[3]{\left< #1 \vphantom{#2}\vphantom{#3} \right| #2 \vphantom{#1}\vphantom{#3} \left| #3 \vphantom{#1}\vphantom{#2} \right>}\)
\(\newcommand{\qmvec}[1]{\mathbf{\vec{#1}}}\)
\(\newcommand{\op}[1]{\hat{\mathbf{#1}}}\)
\(\newcommand{\expect}[1]{\langle #1 \rangle}\)
\(\newcommand{\dfn}[1]{\emph{\textbf{#1}}}\)
What properties should a relation on a set have to be useful as a notion of “equivalence”?
- Each object in the set should be equivalent to itself. So the relation should be reflexive.
- Equivalence should be bidirectional. That is, a pair of equivalent objects should be equivalent to each other. So the relation should be symmetric.
- We should be able to infer equivalence from chains of equivalence. So the relation should be transitive.
a relation on a set that is reflexive, symmetric, and transitive
symbol for an abstract equivalence relation (instead of the letter \(R\) that we've been using for abstract relations up until now)
Let \(\mathscr{L}\) be the set of all possible logical statements built out of the statement variables \(p_1, p_2, p_3, \ldots\text{.}\) Show that logical equivalence of statements is an equivalence relation on \(\mathscr{L}\text{.}\)
Solution
Reflexive. We have \(A \Leftrightarrow A\) for every statement \(A\text{,}\) since \(A\) has the same truth table as itself.
Symmetric. If \(A \Leftrightarrow B\text{,}\) then \(A,B\) have the same truth table, so \(B \Leftrightarrow A\text{.}\)
Transitive. If \(A \Leftrightarrow B\) and \(B \Leftrightarrow C\text{,}\) then \(A\) has the same truth table as \(B\text{,}\) which has the same truth table as \(C\text{.}\) So \(A\) has the same truth table as \(C\text{,}\) i.e. \(A \Leftrightarrow C\text{.}\)
an equivalence of integers, where two integers are equivalent if they have the same remainder when divided by \(n\)
integers \(m_1,m_2\) are equivalent modulo \(n\)
Verify that equivalence modulo \(n\) is an equivalence relation.


