Commit a0b9fe55 authored by Jim Hefferon's avatar Jim Hefferon

Adjustments suggested by JIBL referee

parent 27b53632
size(4cm,3cm);
import settings;
settings.tex="xelatex";
settings.outformat="pdf";
unitsize(0.8cm);
// size(4cm,3cm);
// Colors from "Still not getting any" by sp613g, https://kuler.adobe.com/#themeID/181115
import ibl;
......@@ -11,8 +16,8 @@ filldraw(codomain_shift*bean,fillpen=color_light2);
real r=.1cm;
real label_offset_x=0.12*r, label_offset_y=0.23*r;
picture d_box=new picture;
label(d_box,"$D$",bean1+(label_offset_x,label_offset_y));
add(d_box,filltype=Fill(white));
label(d_box,"$D$");
add(d_box,bean1+(label_offset_x,label_offset_y),filltype=Fill(white));
picture c_box=new picture;
label(c_box,"$C$",codomain_shift*bean1+(label_offset_x,label_offset_y));
add(c_box,filltype=Fill(white));
......
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size(4cm,3cm);
import settings;
settings.tex="xelatex";
settings.outformat="pdf";
unitsize(0.8cm);
// size(4cm,3cm);
// Colors from "Still not getting any" by sp613g, https://kuler.adobe.com/#themeID/181115
import ibl;
......
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% ibl_asy.sty
% Some LaTeX stuff to put in ibl drawings
\usepackage{jh}
\usepackage{../jh}
\usepackage{amsmath}
\usepackage{fourier}
\usepackage[T1]{fontenc}
......
size(3cm,2cm);
// Colors from "Still not getting any" by sp613g, https://kuler.adobe.com/#themeID/181115
pen color_light1=rgb(219,213,242);
pen color_dark1=rgb(22,64,89);
pen color_base=rgb(201,235,242);
pen color_dark2=rgb(69,115,110);
pen color_light2=rgb(242,226,196);
// pen color_light1=rgb(219,213,242);
// pen color_dark1=rgb(22,64,89);
// pen color_base=rgb(201,235,242);
// pen color_dark2=rgb(69,115,110);
// pen color_light2=rgb(242,226,196);
import ibl;
pair z0=(0,0);
pair z1=(0,0); // center of circle1
......
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size(3cm,2cm);
// Colors from "Still not getting any" by sp613g, https://kuler.adobe.com/#themeID/181115
pen color_light1=rgb(219,213,242);
pen color_dark1=rgb(22,64,89);
pen color_base=rgb(201,235,242);
pen color_dark2=rgb(69,115,110);
pen color_light2=rgb(242,226,196);
// pen color_light1=rgb(219,213,242);
// pen color_dark1=rgb(22,64,89);
// pen color_base=rgb(201,235,242);
// pen color_dark2=rgb(69,115,110);
// pen color_light2=rgb(242,226,196);
import ibl;
pair z0=(0,0);
pair z1=(0,0); // center of circle1
......
No preview for this file type
size(3cm,0);
// Colors from "Still not getting any" by sp613g, https://kuler.adobe.com/#themeID/181115
pen color_light1=rgb(219,213,242);
pen color_dark1=rgb(22,64,89);
pen color_base=rgb(201,235,242);
pen color_dark2=rgb(69,115,110);
pen color_light2=rgb(242,226,196);
// pen color_light1=rgb(219,213,242);
// pen color_dark1=rgb(22,64,89);
// pen color_base=rgb(201,235,242);
// pen color_dark2=rgb(69,115,110);
// pen color_light2=rgb(242,226,196);
import ibl;
pair z0=(0,0);
pair z1=(-1,0); // center of circle1
......
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......@@ -2,11 +2,12 @@
// Jim Hefferon 2014-Mar-18 PD
size(3cm,0);
// Colors from "Still not getting any" by sp613g, https://kuler.adobe.com/#themeID/181115
pen color_light1=rgb(219,213,242);
pen color_dark1=rgb(22,64,89);
pen color_base=rgb(201,235,242);
pen color_dark2=rgb(69,115,110);
pen color_light2=rgb(242,226,196);
// pen color_light1=rgb(219,213,242);
// pen color_dark1=rgb(22,64,89);
// pen color_base=rgb(201,235,242);
// pen color_dark2=rgb(69,115,110);
// pen color_light2=rgb(242,226,196);
import ibl;
pair z0=(0,0);
pair z1=(-1,0); // center of circle1
......
No preview for this file type
usepackage("../jh");
size(3cm,2cm);
// Colors from "Still not getting any" by sp613g, https://kuler.adobe.com/#themeID/181115
pen color_light1=rgb(219,213,242);
pen color_dark1=rgb(22,64,89);
pen color_base=rgb(201,235,242);
pen color_dark2=rgb(69,115,110);
pen color_light2=rgb(242,226,196);
// pen color_light1=rgb(219,213,242);
// pen color_dark1=rgb(22,64,89);
// pen color_base=rgb(201,235,242);
// pen color_dark2=rgb(69,115,110);
// pen color_light2=rgb(242,226,196);
import ibl;
pair z0=(0,0);
pair z1=(0,0); // center of circle1
......
size(3cm,2cm);
// Colors from "Still not getting any" by sp613g, https://kuler.adobe.com/#themeID/181115
pen color_light1=rgb(219,213,242);
pen color_dark1=rgb(22,64,89);
pen color_base=rgb(201,235,242);
pen color_dark2=rgb(69,115,110);
pen color_light2=rgb(242,226,196);
// pen color_light1=rgb(219,213,242);
// pen color_dark1=rgb(22,64,89);
// pen color_base=rgb(201,235,242);
// pen color_dark2=rgb(69,115,110);
// pen color_light2=rgb(242,226,196);
import ibl;
pair z0=(0,0);
pair z1=(0,0); // center of circle1
......
No preview for this file type
size(3cm,0);
// Colors from "Still not getting any" by sp613g, https://kuler.adobe.com/#themeID/181115
pen color_light1=rgb(219,213,242);
pen color_dark1=rgb(22,64,89);
pen color_base=rgb(201,235,242);
pen color_dark2=rgb(69,115,110);
pen color_light2=rgb(242,226,196);
// pen color_light1=rgb(219,213,242);
// pen color_dark1=rgb(22,64,89);
// pen color_base=rgb(201,235,242);
// pen color_dark2=rgb(69,115,110);
// pen color_light2=rgb(242,226,196);
import ibl;
pair z0=(0,0);
pair z1=(-1,0); // center of circle1
......
No preview for this file type
No preview for this file type
No preview for this file type
......@@ -2,11 +2,12 @@
// Jim Hefferon 2014-Mar-18 PD
size(3cm,0);
// Colors from "Still not getting any" by sp613g, https://kuler.adobe.com/#themeID/181115
pen color_light1=rgb(219,213,242);
pen color_dark1=rgb(22,64,89);
pen color_base=rgb(201,235,242);
pen color_dark2=rgb(69,115,110);
pen color_light2=rgb(242,226,196);
// pen color_light1=rgb(219,213,242);
// pen color_dark1=rgb(22,64,89);
// pen color_base=rgb(201,235,242);
// pen color_dark2=rgb(69,115,110);
// pen color_light2=rgb(242,226,196);
import ibl;
pair z0=(0,0);
pair z1=(-1,0); // center of circle1
......
No preview for this file type
usepackage("../jh");
size(3cm,2cm);
// Colors from "Still not getting any" by sp613g, https://kuler.adobe.com/#themeID/181115
pen color_light1=rgb(219,213,242);
pen color_dark1=rgb(22,64,89);
pen color_base=rgb(201,235,242);
pen color_dark2=rgb(69,115,110);
pen color_light2=rgb(242,226,196);
// pen color_light1=rgb(219,213,242);
// pen color_dark1=rgb(22,64,89);
// pen color_base=rgb(201,235,242);
// pen color_dark2=rgb(69,115,110);
// pen color_light2=rgb(242,226,196);
import ibl;
pair z0=(0,0);
pair z1=(0,0); // center of circle1
......
......@@ -3,7 +3,7 @@
% Jim Hefferon jhefferon (at) smcvt.edu
% --- Class structure: identification part
% ---
\ProvidesClass{test}[2013/04/13 version 0.99 of Inquiry-based Learning book class (Jim Hefferon)]
\ProvidesClass{test}[2013/04/13 version 1.1 of Inquiry-based Learning book class (Jim Hefferon)]
\NeedsTeXFormat{LaTeX2e}
% ========================================
......@@ -292,8 +292,9 @@ qed=\qedsymbol
\begin{flushright}
\large\color{darki}
\begin{tabular}{@{}l@{}}
Jim Hef{}feron \\
version \version
Jim Hef{}feron \\
Saint Michael's College \\
Version \version
\end{tabular}
\end{flushright}
}
......@@ -343,6 +344,6 @@ qed=\qedsymbol
% ====================================================
% VERSION number
\newcommand{\version}{1.0}
\newcommand{\version}{1.1}
\endinput
\ No newline at end of file
\endinput
......@@ -142,16 +142,16 @@ but ``Does $2/5$?''\spacefactor=1000\ is not sensible.
\end{answer}
\begin{exercise}
Generalize the first item to arbitrary divisors.
Generalize the first item to be a statement about sums of multiples of some $d\in\Z$.
Prove it.
\end{exercise}
\begin{answer}
A reasonable generalization is that for any $d\in\Z$
the sum and difference of multiples of~$d$ is again a multiple of~$d$.
the sum of multiples of~$d$ is again a multiple of~$d$.
For the proof,
let $a,b\in\Z$ be multiples of~$d$ so that $a=dm$ and~$b=dn$
for some $m,n\in\Z$.
Then $a+b=dm+dn=d(m+n)$ is a multiple of~$d$
and $a-b=dm-dn=d(m-n)$ is multiple of~$d$.
Then $a+b=dm+dn=d(m+n)$ is a multiple of~$d$.
\end{answer}
\end{exes}
\end{problem}
......@@ -161,7 +161,7 @@ Prove the first, and prove or disprove the second.
\begin{exes}
\begin{exercise}
The product of two evens is even.
Generalize to any divisor.
Formulate and prove the analogous statement for any $d\in\Z$.
\end{exercise}
\begin{answer}
Let $a,b\in\Z$ be even
......@@ -741,7 +741,7 @@ then~$A$ cannot contain~$k+1$ either, since it would be least.
% .....................................
\section{Division}
\begin{problem} \notetext{Division Theorem}
\begin{problem} \notetext{Division Theorem, or Division Algorithm}
For any integers $a,b$ with $b>0$ there are unique integers $q,r$ such that
$a=bq+r$ and~$0\leq r<b$.
Here, $a$ is called the \definend{dividend} and $b$ is the \definend{divisor},
......@@ -2063,7 +2063,6 @@ The set without any elements is the \definend{empty set} $\emptyset$.
\begin{problem} Decide each, with justification.
\begin{items}
\item $\set{1,3,5}\subseteq\set{1,3, 5, 7, 9}$
\item $\set{1, 3, 5}\in\set{1, 3, 5, 7, 9}$
\item $\set{1,3,5}\subseteq\setbuilder{n\in\N}{\text{$n$ is prime}}$
\item $\emptyset\subseteq\set{1, 2, 3, 4}$
......@@ -2165,11 +2164,11 @@ There, if we say that we are considering
the set of things less than~$100$ then
we are considering the set of integers less than~$100$.
\begin{df}
The \definend{characteristic function} of a set~$A$ is a map
$\charfcn{A}$ (or $\mathbb{1}_A$), whose domain is the universal set, such that
$\charfcn{A}(x)=1$ if~$x\in A$, and $\charfcn{A}(x)=0$ if~$x\notin A$.
\end{df}
%% \begin{df}
%% The \definend{characteristic function} of a set~$A$ is a map
%% $\charfcn{A}$ (or $\mathbb{1}_A$), whose domain is the universal set, such that
%% $\charfcn{A}(x)=1$ if~$x\in A$, and $\charfcn{A}(x)=0$ if~$x\notin A$.
%% \end{df}
\begin{problem}[\maxlength]
\notetext{Russell's Paradox}
......@@ -2255,17 +2254,19 @@ and on the right the dark region, the complement, is all but the set~$A$.
Another tool illustrating set relationships is this table describing
two sets.
\begin{center} \small
\begin{tabular}{cc|c}
$x\in A$ &$x\in B$ &row number \\ \hline
$0$ &$0$ &$0$ \\
$0$ &$1$ &$1$ \\
$1$ &$0$ &$2$ \\
$1$ &$1$ &$3$
\begin{tabular}{r|cc}
\multicolumn{1}{r}{\textit{row number}} &$x\in A$ &$x\in B$ \\ \hline
$0$ &$0$ &$0$ \\
$1$ &$0$ &$1$ \\
$2$ &$1$ &$0$ \\
$3$ &$1$ &$1$
\end{tabular}
\end{center}
We use $0$ and~$1$ instead of $F$ and~$T$ so that each
It uses $0$ and~$1$ instead of $F$ and~$T$ so that the right side of each
row is the binary representation of its row number.
That table gives $A=\set{2,3}$, $B=\set{1,3}$,
The top row says that $0\notin A$ and $0\notin B$,
while the second row says that $1\notin A$ but $1\in B$.
In total, $A=\set{2,3}$, $B=\set{1,3}$,
and~$\universalset=\set{0,1,2,3}$.
For each of these simple results about set operations,
apply the statement to the sets $A$ and~$B$.
......@@ -2546,7 +2547,7 @@ where $X$ is the universal set $\universalset=X$.
\begin{problem}
\notetext{De Morgan's Laws} Prove, for all sets $A$, $B$, and~$C$.
\begin{exes}
\begin{exercise}
\begin{exercise} \notetext{De Morgan's Laws}
$\setcomp{(A\intersection B)}=\setcomp{A}\union\setcomp{B}$
and
$\setcomp{(A\union B)}=\setcomp{A}\intersection\setcomp{B}$
......@@ -3178,6 +3179,12 @@ Thus the number of possible function is $|C|^{|D|}$.
\end{answer}
\end{problem}
\begin{df}
The \definend{characteristic function} of a set~$A$ is a map
$\charfcn{A}$ (or $\mathbb{1}_A$), whose domain is the universal set, such that
$\charfcn{A}(x)=1$ if~$x\in A$, and $\charfcn{A}(x)=0$ if~$x\notin A$.
\end{df}
A function may have multiple arguments; one example is the function
$\map{f}{\R^2}{\R}$ whose action is
$\sequence{x,y}\mapsto x^2-2y^2$.
......@@ -3309,7 +3316,8 @@ and $d\mapsunder{}\alpha$.
\item Compute $\composed{g}{f}$ on all arguments or show that the composition
is not defined.
\item Compute $\composed{f}{g}$ on all arguments or show that it is not defined.
\item Find the range of $f$, $g$, and any defined compositions
\item Find the range of $f$, $g$, as well as $\composed{f}{g}$
and~$\composed{g}{f}$ if they are defined.
\end{items}
\begin{answer}
\begin{items}
......@@ -3596,7 +3604,8 @@ Prove.
Thus the supposition is false and $f$ is one-to-one.
\end{answer}
\begin{exercise}
Do the other two cases hold?
If $\composed{g}{f}$ is onto, is $f$ onto?
If it is one-to-one, is $g$ one-to-one?
\end{exercise}
\begin{answer}
The other two cases do not hold.
......@@ -3745,7 +3754,7 @@ Then:
In Definition~\ref{def:InvImage}, we defined
$f^{-1}(c)$ to be the set $\setbuilder{d\in D}{f(d)=c}$.
While this earlier notation is standard, it conflicts
While this earlier notation is standard, it conflicts with
what we just saw in Definition~\ref{def:InverseFcn}.
For instance,
if $\map{g}{\R}{\R}$ is $g(x)=2x$ then $g^{-1}(8)$ could mean
......@@ -4000,7 +4009,7 @@ A relation that satisfies all three conditions is an
% \end{problem}
\begin{problem} \label{ex:EquivModMIsEquivalenceRelation}
Fix a divisor $m\in\Z^+$.
Fix $m\in\Z^+$.
Show that the relation
$\setbuilder{\sequence{a,b}\in\Z^2}{a\equiv b\pmod m}$
is an equivalence.
......@@ -4282,8 +4291,8 @@ $\eqclass{1}=\eqclass{11}=\eqclass{21}=\cdots\,$.
Exhibit the equivalence classes.
\begin{exes}
\begin{exercise}
$x_0\equiv x_1 \pmod R$ if they have the same parity
(are both even or both odd), with $X=\N$
Two natural numbers $x_0,x_1\in\N$ are related if they have the same parity
(are both even or both odd).
\end{exercise}
\begin{answer}
To verify that it is an equivalence we will check that it is reflexive,
......@@ -4307,20 +4316,19 @@ $\eqclass{1}=\eqclass{11}=\eqclass{21}=\cdots\,$.
and the odd numbers
$\set{1,3,5,\ldots}=\eqclass{1}=\eqclass{3}=\eqclass{5}=\cdots$.
\end{answer}
\begin{exercise}
$i\equiv n \pmod R$ if
$i\equiv n\pmod 3$ (they leave the same remainder on division by $3$),
with $X=\Z$
\begin{exercise}
Two numbers $m,n\in\Z$ are related if they leave the same remainder
on division by~$3$, that is, if $m\equiv n\pmod 3$.
\end{exercise}
\begin{answer}
The relation is reflexive because for every integer,
on division by~$3$ it leaves the same remainder as itself.
The relation is symmetric because if, on division by~$3$,
$i$ leaves the same remainder as~$n$
then $n$ leaves the same one as~$i$.
Transitivity is as easy: if $i_0$ leaves the same remainder as~$i_1$,
which in turn leaves the same remainder as~$i_2$ then all three leave
the same remainder and in particular $i_0$ and~$i_2$ do.
$m$ leaves the same remainder as~$n$
then $n$ leaves the same one as~$m$.
Transitivity is as easy: if $m_0$ leaves the same remainder as~$m_1$,
which in turn leaves the same remainder as~$m_2$ then all three leave
the same remainder.
There are three equivalence classes.
One is
......@@ -4329,8 +4337,9 @@ $\eqclass{1}=\eqclass{11}=\eqclass{21}=\cdots\,$.
$\set{\ldots,-5,-2,1,4,7,\ldots}=\eqclass{1}=\eqclass{-2}=\eqclass{4}=\cdots$.
The last is $\eqclass{2}$.
\end{answer}
\begin{exercise}
$x_0\equiv x_1\pmod R$ if $x_0-x_1\in \Z$, with $X=\R$
\begin{exercise}
Two real numbers $r_0,r_1\in\R$ are related if
$r_0-r_1\in \Z$.
\end{exercise}
\begin{answer}
Reflexivitity $x\equiv x\pmod R$ holds because $x-x=0\in\Z$ for any $x\in\R$.
......@@ -4392,7 +4401,8 @@ Therefore~(iii) implies~(i).
\begin{df}
A \definend{partition}~$\partition{P}$ of a set $X$ is a
collection of nonempty subsets of~$P\subseteq X$ such that every
collection of nonempty subsets of~$P\subseteq X$, called \definend{parts},
such that every
element $x\in X$ is in exactly one of the $P$'s.
That is, $\partition{P}$ partitions~$X$ if and only if each
$P\in\partition{P}$ is nonempty,
......@@ -4590,8 +4600,8 @@ Suppose $\map{f}{D}{C}$.
\begin{exercise}
Consider the map
$\map{\hat{f}}{\partition{P}}{\range(f)}$ whose action is:
$\hat{f}(P)$ is defined to be~$f(d)$ where $d\in P$.
Show that~$\hat{f}$ is a function and that it is one-to-one.
$\hat{f}(P)$ is defined to be~$f(d)$ for any $d\in P$.
Show that~$\hat{f}$ is well-defined and that it is one-to-one.
\remark every function can be modified to be onto by
changing its codomain to equal its range.
This exercise gets one-to-one-ness by modifying the function's domain.
......
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......@@ -7,7 +7,7 @@ it can be used with first year students.
\medskip
\medskip
\noindent\textsc{Approach.}
This course is inquiry-based (sometimes called Moore method
or discovery method).
......
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