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Jim Hefferon
proofs
Commits
934bc5ce
Commit
934bc5ce
authored
Nov 03, 2016
by
Jim Hefferon
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bean_fcn.asy
asy/bean_fcn.asy
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1
bean_fcn.pdf
asy/bean_fcn.pdf
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bean_fcn_comp.asy
asy/bean_fcn_comp.asy
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bean_fcn_comp.pdf
asy/bean_fcn_comp.pdf
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ibl.tex
ibl.tex
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JIBLM.tex
jiblm/JIBLM.tex
+0
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hefferon.tex
jiblm/hefferon.tex
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iblmaxlength.pdf
output/iblmaxlength.pdf
+0
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preface.tex
preface.tex
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asy/bean_fcn.asy
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934bc5ce
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...
@@ 19,5 +19,5 @@ add(c_box,filltype=Fill(white));
path dtoc = bean4{dir(30)}..codomain_shift*bean1;
path dtoc_drawn=subpath(dtoc,0.3,0.7);
draw(dtoc_drawn,black,ArcArrow,margin=NoMargin);
draw(dtoc_drawn,black,ArcArrow
(TeXHead)
,margin=NoMargin);
label("$f$",point(dtoc,0.5),align=N);
asy/bean_fcn.pdf
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934bc5ce
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asy/bean_fcn_comp.asy
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934bc5ce
...
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@@ 29,13 +29,13 @@ add(b_box,filltype=Fill(white));
path dtoc = bean4{dir(30)}..codomain_shift*bean1;
path dtoc_drawn=subpath(dtoc,0.3,0.7);
draw(dtoc_drawn,black,ArcArrow,margin=NoMargin);
draw(dtoc_drawn,black,ArcArrow
(TeXHead)
,margin=NoMargin);
label("$f$",point(dtoc,0.5),align=N);
path ctob = codomain_shift*bean4{dir(30)}..cocodomain_shift*bean1;
path ctob_drawn=subpath(ctob,0.3,0.7);
draw(ctob_drawn,black,ArcArrow,margin=NoMargin);
draw(ctob_drawn,black,ArcArrow
(TeXHead)
,margin=NoMargin);
label("$g$",point(ctob,0.5),align=N);
path dtob = bean4{dir(25)}..cocodomain_shift*bean1;
path dtob_drawn=shift((0,1))*subpath(dtob,0.20,0.80);
draw(dtob_drawn,black,ArcArrow,margin=NoMargin);
draw(dtob_drawn,black,ArcArrow
(TeXHead)
,margin=NoMargin);
label("$g\circ f$",point(dtob_drawn,0.5),align=N);
asy/bean_fcn_comp.pdf
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934bc5ce
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ibl.tex
View file @
934bc5ce
...
...
@@ 29,7 +29,8 @@
We begin with results about the integers
$
\Z
=
\set
{
\ldots
,

2
,

1
,
0
,
1
,
2
,
\ldots
}$
.
In this chapter, ``number'' means integer.
Some statements refer to the natural numbers
$
\N
=
\set
{
0
,
1
,
2
,
\ldots
}$
.
Some statements refer to the natural numbers
$
\N
=
\set
{
0
,
1
,
2
,
\ldots
}$
or the positive integers
$
\Z
^
+=
\set
{
1
,
2
,
\ldots
}$
.
% .....................................
...
...
@@ 401,7 +402,7 @@ we will show that the following implication holds.
The
\definend
{
Principle of Mathematical Induction
}
is that
completing both steps proves
that the statement is true for all natural
numbers greater than or equal to the initial number.
numbers greater than or equal to the initial number
~
$
i
$
.
For the example statement about odd numbers and squares,
the intuition behind the principle is first that the base step
...
...
@@ 752,8 +753,8 @@ We prove this in three stages.
\begin{exes}
\begin{exercise}
Show that~
$
q
$
and~
$
r
$
are unique, assuming that they exist.
(O
ne way to proceed is to suppose that
$
a
=
bq
_
0
+
r
_
0
=
bq
_
1
+
r
_
1
$
with
$
0
\leq
r
_
0
,r
_
1
<b
$
and then show that
$
q
_
0
=
q
_
1
$
and~
$
r
_
0
=
r
_
1
$
.
)
\hint
o
ne way to proceed is to suppose that
$
a
=
bq
_
0
+
r
_
0
=
bq
_
1
+
r
_
1
$
with
$
0
\leq
r
_
0
,r
_
1
<b
$
and then show that
$
q
_
0
=
q
_
1
$
and~
$
r
_
0
=
r
_
1
$
.
\end{exercise}
\begin{answer}
Suppose that
$
q
_
0
,r
_
0
\in\Z
$
and
$
q
_
1
,r
_
1
\in\Z
$
are such that
...
...
@@ 965,11 +966,13 @@ the greatest integer less than or equal to~$x$.
$
p
_
0
+
\cdots
+
p
_{
n

1
}$
would be less than
$
na
$
, so at least
one must be at least average.
\end{answer}
\begin{exercise}
If you put
$
n
$
many papers into fewer than
$
n
$
many
pigeonholes then at least one hole gets at least two papers.
\begin{exercise}
If you have
$
n>
0
$
many pigeonholes and
more than
$
n
$
many papers
then at least one hole gets at least two papers.
\end{exercise}
\begin{answer}
Form the list where there are~
$
p
_
i
$
many papers in the
$
i
$
th pigeonhole.
Form the list
$
p
_
0
,
\ldots
,p
_{
n

1
}$
where there are~
$
p
_
i
$
many papers in the
$
i
$
th pigeonhole.
The average number of papers per hole is greater than~
$
1
$
,
so the maximum number is greater than~
$
1
$
.
Since each
$
p
_
i
$
is a natural number, the maximum must be at least~
$
2
$
.
...
...
@@ 1920,7 +1923,7 @@ we allow here that some exponents are zero).
Prove that
in the prime factorization of
$
\gcd
(
a,b
)
$
the exponent of
$
p
_
i
$
is
$
\min
(
\set
{
e
_
i,f
_
i
}
)
$
.
(
The proof is much the same
that in the prime factorization of
$
\lcm
(
a,b
)
$
(
Much the same proof shows
that in the prime factorization of
$
\lcm
(
a,b
)
$
the exponent of
$
p
_
i
$
is
$
\max
(
\set
{
e
_
i,f
_
i
}
)
$
.
Taken together the two show that
$
\gcd
(
a,b
)
\cdot\lcm
(
a,b
)=
ab
$
.)
\begin{answer}
...
...
@@ 2173,9 +2176,9 @@ The definition that we gave allows sets to contain anything.
This turns out to be naive.
For, if sets can contain anything then we naturally think of
the set that contains everything, all sets.
Note that it
ha
s itself as an element.
So, consider the set of all sets that don't contain themselves as
element
s~
$
D
=
\setbuilder
{
S
}{
S
\notin
S
}$
.
Note that it
contain
s itself as an element.
In this way we are led to the set of all sets that don't contain
themselve
s~
$
D
=
\setbuilder
{
S
}{
S
\notin
S
}$
.
\begin{exes}
\begin{exercise}
Show that assuming
$
D
$
is an element of itself leads to a contradiction.
...
...
@@ 2223,10 +2226,10 @@ we are still in number theory.
\begin{df}
Let
$
A
$
and
$
B
$
be sets.
Their
\definend
{
union
}
is the collection of elements
from either set
from either set
,
$
A
\union
B
=
\setbuilder
{
x
}{
\text
{$
x
\in
A
$
or
$
x
\in
B
$}}$
.
Their
\definend
{
intersection
}
is the collection of elements
from both sets
from both sets
,
$
A
\intersection
B
=
\setbuilder
{
x
}{
\text
{$
x
\in
A
$
and
$
x
\in
B
$}}$
.
\end{df}
...
...
@@ 3108,7 +3111,7 @@ $G$ just consists of arbitrary pairings.
\begin{problem}
[
\midlength
]
The
\
textit
{
hailstone function
}
$
\map
{
h
}{
\N
}{
\N
}$
is defined by cases,
The
\
definend
{
hailstone function
}
$
\map
{
h
}{
\N
}{
\N
}$
is defined by cases,
\begin{equation*}
h(n)=
\begin{cases}
n/2
&
\text
{
 if
$
n
$
is even
}
\\
...
...
@@ 3123,7 +3126,7 @@ using a different formula when the input is even than when it is odd.
How many steps does it take?
\item
How many steps does it take starting with
$
n
=
11
$
?
\end{items}
(The
\
textit
{
Collatz conjecture
}
is that for every natural number starting
(The
\
definend
{
Collatz conjecture
}
is that for every natural number starting
value, iteration will eventually reach~
$
1
$
.
No one knows if it is true.)
\begin{answer}
...
...
@@ 4024,7 +4027,7 @@ is an equivalence.
\label
{
PlaneLinesAsClasses
}
Let
$
\mathcal
{
L
}$
be the set of lines in the Euclidean plane and consider
the relation
$
R
=
\setbuilder
{
\sequence
{
\ell
_
0
,
\ell
_
1
}
\in
\mathcal
{
L
}^
2
}{
\text
{
the two are parallel
,
or are equal
}}$
.
$
R
=
\setbuilder
{
\sequence
{
\ell
_
0
,
\ell
_
1
}
\in
\mathcal
{
L
}^
2
}{
\text
{
the two are parallel or are equal
}}$
.
\begin{items}
\item
List five elements of~
$
R
$
.
\item
Where
$
\ell
$
is a vertical line,
...
...
@@ 4103,8 +4106,8 @@ symmetric or not, and transitive or not, so there are eight
possible combinations.
\begin{exes}
\begin{exercise}
There are four
combinations
that
are not reflexive (e.g., one is: not reflexive, not symmetric, and
Four of the
combinations
are not reflexive (e.g., one is: not reflexive, not symmetric, and
not transitive).
For each, give an example relation on
$
A
=
\set
{
0
,
1
,
2
}$
.
\end{exercise}
...
...
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preface.tex
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934bc5ce
...
...
@@ 28,7 +28,9 @@ Besides, it is a lot of fun.
\medskip
\noindent\textsc
{
Topics.
}
We start with elementary number theory, not logic and sets,
We cover sets, functions and relations, and elementary number theory.
We start with number theory instead of sets
for the same reason
that the baseball team's annual practice starts with tossing the ball and
not with reading the rulebook.
...
...
@@ 39,7 +41,7 @@ whereas weeks of preliminary material is less of a lure.
But the background is good stuff also and
students are on board once they see where it is going.
In the second and third chapters we do
sets, functions, and relations
, keeping the
the other material
, keeping the
intellectual habits that we established at the start.
...
...
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