Download Math 2800 Math Majors Seminar

Math 2800
.
Math Majors Seminar
Book 2
that
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to
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ete
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nez
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:
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solution
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nti
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of
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prove
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with
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all
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called
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:
easy
every
n
disks
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R
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then
step
#
in
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move
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and
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or
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R
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solution
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n
disks
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from
for
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.
1
and
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stack
2
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to
to
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modification
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stick
of
hyped
Hanoi
2
2
.
.
puzzle
stack
disk to
n
slight
0
1
top
there
disks )
nil
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a
if
that
inductive
the
given
,
,
disk
the
largest
roles of stacks
the
switch
more
finally
the
We
,
here
•
(
.
a
.
inductive
disks
Next
•
Hanoi puzzle has
about the
minimum
worry
of
Towers
starting
initial
case
bad
his
case
2
the
the
prove
to
however
for
called
is
that the
disks
of
)
that
want
stack
solution
This
.
P,
is
to
prove
a
To
Later
This
.
0
Next
moves
the
prone
)
possible number
We
.
by induction
every
statement
the
is
.
show
will
is
d
required
.
( Again
of stacks
this
uses
0 and 1.)
hypothesis K to interchange the roles
to
disks in state 2 from smallest (
the
largest (
top )
bottom )
'
on
.
on
A
statement
stronger
Theory
moves
following
of
Hanoi
problem
:
with
n
disks
has
a
solution
using
zny
.
( Initial
proof
(
The Towers
.
the
is
formula
2
'
1=1
-
.
•
First
move
Then
use
:
El
is
observe
'
,
from
stack
that
to
0
there
is
a
stack 2)
the
top
1
move
move
the
the
221
in
n
to
n
minimum
moves
disks
as
solution
using
a
predicted by
move
our
disks
from
number
of
that
the
n
.
disk
an
-
,
.
stack
from
hypothesis
disk puzzle
(n+ , )
given
moves
221
to stack 1
in
slack
0
largest disk from slack 0 to stack
then
,
the
more
inductive
the
by
Assuming
.
solved
be
Finally
.
Remark
can
n=
.
sleep )
( Inductive
puzzle
For
.
single desk
the
more
)
case
moves
I to
stack
required
.
2
in
although
221
moves
I
2.
.
didn't show this
.
Example
odd
Is
n
theorem
Proof
Now
if
odd
also
If
is
n
Remark
then
even
0=20
If
.
A
=
is
n
this
for
by
n
then
a
is
odd
two
cases
as
follows
k
then
,
so
,
,
must show
we
nh
=
that
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,
nt
:
=2k+i
nti
-
zfk )
is
even
.
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form
to
?
.
.
K )
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.
dots
chords
proof
.
odd
is
.
ktl
where
zckti )
is
integer
an
a
on
,
we
&
circle
have
then
.
.
.
.
a
.
.
&
.
and
divided the
.
.
&
.
•
.
.
.
.
.
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.
.
connecting
circle into
.
•
.
.
•
odd
is
.
'
.
.
DH
Ztk
n=
i
.
•
.
.
.
.
for positive integers
which is either even
works
above
positive
is
n
n=2k+1
-
.
•
n
The
.
•
.
!
odd
or
negative integer then
a
•
:
.
integer k
even
.
-
.
D
•
.
Conjecture
odd
or
or
some
.
either
is
If
.
2k
induction
•
False
even
even
integer
;
.
either
In
some
k
integer k
some
.
even
is
1=20+1
zkti
n=
Every integer
:
form
prone
for
n=2k
say
,
follows
result
The
Also
odd
or
is
ntl
so
either
We
.
integer
some
n=2kfor
if
even
is
integer is
it has the
odd
or
a is odd ,
If
for
positive integer
a
even
nez
I
positive
since
supposing
even
either
2kt
n=
Every
is
A
of
pay
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.
.
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in
regions
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.
.
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.
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.
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.
.
.
i. i.
.
.
.
pairs
•
theorem
If
subsets
of
F#
n=2
n=
1
:
is
S=
{0
S={ }
:
The
theorem
is
( which
is
Remark
{a
:
either
usually
}
a
,
Directproof
{
=
set
If
empty
{ }
-
and
valid
the
in
S
,
S=
1
{a
not
it's
or
,
,
as
a }
,
.
This
gives
Alternatively
Proof
subset
-
a
Gz
.
total
of
,
we
argue
in
can
by
prone
of
,
{1 }
{
,
0
I
,
}
.
.
denoted
5=0
This has
.
the theorem works
so
induction
by
or
only
that
in
case
directly
talking
abort
amttirset
}
an
,
then
every
)
(
value
every
;
subset
is
A E 5
.
-
-
.
2×2×2
x.
the
induction
agreement with
}
0
20=1
.
...
.
We
0
.
number
the
then
)
,
,
-
)
.
or
.
n
}
0
proved either
not
( we're
.
{
,
{
,
usually
,
itself
be
can
{ }
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a.
ag
choices
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?
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,
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.
{ }
are
set
preferable )
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S
subsets
two
the
is
subset
one
subsets of
;
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o
ie
.
}
I
,
(
elements
n
"
2
S={ }
:
n=o
S
has
S
set
a
the
fault
.
.
×
2
=
n=
ISI
subsets
"
induction
by
on
z
.
formula 20=1
For
.
the
A
E S
I
.
.
initial
case
151=0
,
5-
0
has
1
.
Now
to have
assumed
is
Bhas
We
It
subsets
'
li )
If
be A
All
Reformulated
Proof by
follows
as
S
S
{b }
on
any
b
So
B
E
.
A
.
.
have
n
B
nti
We
.
'
{ b } UB
=
POWER
OUTAGE
.
where |B' f-
n
.
.
.
handout
see
.
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of
horses
of
size
have
only
15171
,
then
any
horse
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S
in
.
For
n=i
we
set
of
a
,
+1
suppose
horses
+4
one
horse
In
.
conclusion holds for
this
the
case
result
horses
same
the
any
set
of
n
horses
.
5#f€'qa
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IBK
n
must show that
.
inductive step
the
size
set of size
.
in S
.
set of
every
color
set
151
choose
where
'
same
horse
n=
A
U
any
a
thus
ni
:
handout
the
is
any
induction
be
cases
value of
fixed
a
be
B
can
we
,
two
=
Let
.
have the
immediately
for
let
If
color
same
A
fan
,
horses
:
113171
in
then
example
:
subsets
"
since
.
subsets AEB
Theory
and
2
count
Another
the
conclusion holds for
the
suppose
www.saaeksakah
color
.
all
have
.
these
.
II
The
Write
We
flaw
in
R for
"
the
the
proof
"
above
"
statement
.
horses
n
proof
theorem For every
proof
For
Pz
inductive
horses
.
:
#
same
P,
→
Pz
flawed
is
1+2+3
E
R,
Ps
Then
as
+
(
+
...
color !
n+D=#z±→
predicted by
the Theorem
:
n > I
The formulae
the
2
of horses of the
.
proof of
Pi
flawed
to
consists
D_DDID÷
Another
a
Ps
→
etc
our
of
from
III. #
Pg
however
the inductive step
set
every
P
correctly conclude
is
1+2+3
,
+
works for
step
,
It
.
.
.
+
n
since
n=i
suppose the
2+3
+
.
.
.
+
n=
"
=
1
h¥
=
.
14¥
Formula works
n(n¥
.
.
for
n
;
then
D
.
We neglected
true
This
.
For
correct
n=l
,
the
+
by
.
.
.
the
+
0
1
I
I
2
z
g
4
5
by
I
2
24
to
( n+D
Formula
n !
,
+
2
q
g
16
32
works
case
H2t3t
n
The result follows
n
if
it
trying
to
stated it
merely
we
make when
first
as
were
.
initial
the
so
formula
the
predicted
.
:
1+2+3
as
hypothesis P ) ;
students
mistake
the
common
proofs
1=14*2
Assuming
(given
,
most
inductive
proof
Pn+
prod
is
write
A
to
:
=
n
+
.
MEI
for
"
=
nay
.
+
( n+i)
Proof
n
!
I
lxzx
=
!
.
'
.
Now
xn
1
'
!
2
a
1×2=2
=
3 !
,
theorem
a
.
n
Itzel
=
.
induction
some
=
>/4
a > 4
n
n
! > 2
!=
!
,
,
=
( n+D×n
1×01
.
I
>
.
4 !
"
>
for
24
.
some
,
=
(n+Dn
!
z5n
>
@+ I )
!
,
We have checked
assuming
( ntp !
1×2×3=6
For
!
as
by
required
.
induction
The
.
!
"
>
z×[
5×2
=
It
result follows
D
'
theorem
Eg
all
For
nzo
's
It
,
+
tat
+
...
In
2
<
.
1<2
.
It
tz
It
tz
1++2
<
+
+
z
¥
<
2
tat 's
<
2
by induction
Proof
Assuming
It
Since
.
Itty
+
1<2
tgt
.
.
+
.
,
In
inequality holds
the
<
for
2
for
n3o
some
n=o
.
,
'ftXy¥fyy¥
that's
conclusion
"
"
cz
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tzt tzttgt
lttzttattgt
is
+
.
.
.
.
z÷±
+
I
<
,
¥
1+1
<
,
2
=
D
required
as
O
.
.
even
:
0=2.0
The empty set I 3=0
The case
subsets
.
.
has
n=o
Not
1
odd
subset
gives
zkti
O=
:
0
,
2
:
'
has
no
solution
k€21
Recall :
itself
If
lskn
subset (the initial case of
.
then
.
an
S has
induction )
.
2
"
To
R⇒R+
prove
"
Recall
:
P
F
#
T
F
F
There
This
.
For
not
is
In the
Proof
Now
result
The
more
subjects ,
learning
nzl
have
we
,
great example
it
pore
initial
•
•
P,
n
n >
! 2,3
:
,
.
to conclude
)
prove
-
npzr(P
in
R
)
case
⇒
mathematical induction
:
)
Pn+ ,
;
I
have
we
1
induction
initial
sequential process
a
.
but
for
to
show
the
sake of
.
:
the
is
.
illustrate
induction
positive integer
the
fact that n
follows
by induction
1
5
-
>
-
5
.
satisfies
n
a
"
,
,
.
mathematics
to
by
case
general form of
simply
true
is
.
)
0
T
assuming
follows From
This
3>0
statement
(
of times
T
will
argument
then
"
number
Q
all
a
implication ( conditional
any
F
most
than
the hypothesis
use
,
1=2
It
P ⇒
F
More
can
If
Q
T
A
the
gives
Pn+ ,
,
is
n
-5
want
we
,
positive
a
>
integer
ntl
also
is
.
D
.
In
order to
prone
R, R
.
B. Pg
"
,
.
all
at
once
,
one
>
.
5
.
Note
If
:
one
initial
P,
( the
R
follows
P
p
.
.
.
}
P
,
.
-
.
.
,
P
,
and
R
)
.
get
you
,
i
and P⇒Pz
P,
P and
.
.
*
case
from
statements
above
satisfy the
can
;
(R^PD⇒
and
and
B
and
Pz
;
( P npzn
B)
⇒
,
Pq
;
AND
P.P.BR/P5
Example
(
print
a
other
than
eg
To
say
.
.
.
,
pk
Every integer
is
integer
an
1
and
p )
4=2×2
7=7
that
,
n
k
is
7 1
a
.
p>
2
can
he Factored
which
is
as
not divisible
product of
a
by
any
primes
.
positive integers
.
6=2×3
.
n=2
12=2×2×3
product of
primes
means
n
=p Rpj pn
:
,
for
Some
primes
p ,p
,
.
,
For
initial
the
Now suppose
primes
case
We
If
i)
c
Case cii )
If
p p.
,
ntl
.
.
;
pine
nti
we
,
,
q,
ab
,q
,
integers 2. 3,4, ;
can
also be factored
.
done
are
ie
are
qe
single factor consisting of
a
(
.
prime
one
a
,
primes
a
factor )
ntl
.
.
product
=
and b
:p pi pn
Then
qqi
.
as
-
qe
of
a
of primes
.
where
ab
=
prime 2
product
,b£{2,3 ; ;n}
a
where
k
,
l
,
.
,
.
pzpjipnqqqiiqe
factorization
prime
of nti
result follows by induction
is
=
as
the
factored
be
Can
n
.
composite ) then
.
hypothesis
...
,
is
Prime (
not
inductive
pp
nti
which
is
the
,
,
is
ntl
By
and
show that
must
.
Case
each of the
and
72
n
which
n=2
p
=
,
a
.
The
some
;
we'll
60
60
One
uses
refer
textbooks
induction
order
II
.
can
of
=
=
to
6×10
=
=
other techniques
type
of
result
.
induction
induction
(2×2)×(3×5)
]
(2×3)×(2×5)
stronger
prove
the
prime factors
a
it
call
just
4×15
this
that Prime
This
is
induction
complete
or
.
same
except for
order
the
different
prime factors
of the
factorization
( not just induction )
strong
as
a
integers
of
are
.
essentially unique
Fundamental Theorem of Arithmetic
to
prove
.
.
ie
.
up
to
changing
But this
of
One
p
the
steps
then
,
We
a
won't
is
we
say
following
The
this today
prove
k.net/L
If
Euclid 's Lerma
divisible by
is
are
k divides
•
that
•
k
is
equivalent
k
is
eg
.
|
1
by
"
=
nez
(
some
6
:
"
7×1
ab
such that
z
by p
if
dk
n=
for
DEZ
some
}
n
n
integer d
kln
Note
k
316
is
to
satisfies
⇒
p.
In
.
and
this
with
is
this
:
a
relation between
net
operation
,
kfn
-5140
.
5/0
.
-5/10
.
an
n
write
Don't
51.10
sine
pine
a
the symbolic
way
since
.
is
by
.
54T
d.
divisible
is
:
6T3
18
integer
Every
divisible
every
for
divides
3
of
factor of
a
n=dk
.
divisor
be
divisible
is
didn
k
n
a
a.
.
isisadiuismttdskgthk
:!
•
b
or
p
If
:
since
since
010
-10=-2×5
10
045
=
f- 2) xf5)
5=()#
0=0 )x5
0=471×0
Given
Eg
a
bek
,
Find
.
the
,
greatest
god ( 18,30 )
of
Divisors
of
Divisors
18
30
:
-18
-9 -6
,
,
The
gcd
-30
Y5Y0Te
greatest
( 100
gcd
(
64 )
,
of
Divisors
of
Divisors
Common
0
)
63821
:
Wolfram
However
to
-200
divisor
=
.
.
.
integer
's
,
-2
will
give
,
-1,1
,
63821
has
Algorithm
-1
...
,
.
and
0
Arithmetic )
of
-1
-2
,
,
,
→
it
and
18
2,3
-1,1
, 6,9 ,
42
,
denoted
is
gcd ( a ,b )
.
,
18
3,5 6,10 , 15,30
,
3,6
1,2 ,
,
,
of
63821g
-63821
of
b
is
30
6
:
gcd ( 18
,
30
)
=
6
.
;
64=21212×2×2×2
100=2×215×5
,
divisors
and
a
63821
.
Alpha
,
-3
,
3.
'
,
using
:
Euclid
using
-6
-
Every positive
Theorem
4
=
,
:
-2
,
-5
,
common
0
63821
-3
,
:
30
of
.
Common divisors
of 18 and
divisor
common
a
.
-
the
integer factorization
digits ; 300 digits
1
...
,
,
z
,
.
,
:
.
.
)
unique factorization
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appropriate
answer
is
is
m
,
n
)
as
is
software
quickly
very
.
product of primes ( the Fundamental
practically computable for large integers
a
such
as
Testing
only practically feasible
infeasible
.
Maple
for
for
or
primality
numbers
Mathematical
also
is
up
to
easy
or
.
about
Real numbers
R
:
{
=
It
But
careful
be
real
define
to
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numbers
real
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numbers
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100
1.36363636
=
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algorithm
an
.
.
99x=35⇒
YF
13¥
.
.
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ex
contain
not
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any
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larger number systems
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1,5T
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.
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algorithm
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:
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numbers
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.
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136.36363636
so
define
we
:
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how do
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got
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infinite
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infinitesimal
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In
.
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a
number
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;
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use
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like
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statement
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simply
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shorthand
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ie
for
all
For
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.
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there
M
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real
no
number
a
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by
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10,1
we
.
a
=
given
refer to
IRK
Calculus
The
contexts
{
=
by
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a
:
:
Z
a
1,2
1
,
3,4
2
,
Q
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and
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arises
5
,
.
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3
.
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a
C
4
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in
may
that
fact
contexts
mathematics
in
often
we
of
things
Part
.
different
mean
contexts
R
,
set
-
theory
are
a
has
no
then
21
5
6
sets
infinite
as
,
.
x
.
In
.
concept of
.
IN
different
in
,
In
If
"
whenever
.
The concept of infinity
due to
is
the
confusion
the
e×xM
that
such
N
exist
1211=10,1
,
symbols
relevance
there
E
¥
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itit
is
F
8
7
which
8
<
are
talking
in
is
and often
a
IRI
used
10
114
=
•
,
.
in
about
one-to-one
9
write
very
size
different
of
a
correspondence
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.
.
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set
.
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#
1
0.33333
=
.
.
.
.
.
F.
7
=
umber
These
These numbers
.
.
.
one
based
is
i
decimal
numbers
(
a
on
expansion
each
.
had
decimal
expansion
.
.
.
.
a€R
can
he
.
3.141592653589793298
have
=
unique
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=
...
.
2.0000000
real
have
}
'
'
1.9999999
=
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.
'
.
1.0000000
0.9999999
=
=
Note
.
0.99999
can
3.1415926535
=
2
argument
real
a
=
¥
fallacious
number
only
the
to
infinitely
many
expressed
as
decimal
a
.
.
i.
e.
finite
.
digits
but
.
these numbers
.
.
eg
they
are
are
.
not infinite
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's
that
0.333333
=
+3=0.333333
¥9999999
proof
atbz
.
let
:
.
.
=/
.
:
Reward
.
.
3 has
:
"
.
1
unique decimal
which has two
.
9=0.999999
=
and
...
.
.
.
=
6=1
Find the
.
a
Z
acts
TE
1,9999991=0.799999
19
zfIIIIa#
then
aa¥< b
means
a=b
8
79
÷
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9
±
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.
.
expansion
decimal
expansions
.
#
1
a
unlike
"
'
0.333333
÷=
Third
0.999999
fact that
at¥=a
.
.
average
of
a
and b
:
sets
For any
# s
,
does
what
the
cardinality
151
mean
?
start
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.
the
size
same
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IAI
:
H
|{
-45,11 }/=
.
1131
=
0
(
of S
if there
a
is
Is denoted
S )
what it
by saying
3,10 }|
,
of
size
i.e.
IN
{ 1,2 3,4 :}
,
:
1
,
For finite
§
Ii{!t
sets
lie
same
size
the
have
set
F- S
S
itself
12/1=1 Nl
is
.
}
.
These two sets
:
6
5
'
'
countably infinite
the
have
be listed
size
same
S
:
with
as
a
iff
finite
ie
.
{s
-
,
I
's
in
.sn
and
say
I 1
{ 12,3
in
element
)
;
,
.
:
IN
0
}
...
.
S
is
sequence ;
}
.
only finitely
proper
the
only
set
.
subset
may
The
is
converse
finite set cannot
a
also
true
-1
I
Z
-2
I
I
2
I
I
I
2
3
4
5
-3
3
I
6
:
satisfying 171=151
TES
.
21
B
bare
...
,
,
1st
infinite
an
s
to
its elements
if
,
A ,B
.
.
sets
.
'
1*1=11141
.
sets
.
§
Iif
sometimes
or
one-to-one correspondence between A
"
5
4
Lets be a set We
IN
4
3
2
.
,
,
No :
A
No={0,433,4
.
=
for too
means
can
IN
|S|
I
7
4
-
4
I
I
8
9
.
.
-
.
.
.
is
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0
's
,
.
's
it
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.
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,
Q={o
,
tg
's
'
-
,
i÷
3g
,
'
,
-
,
,
:} ? ! !
Theorem
-
(
,
.
.
I
E. ¥
-2g
Georg
Intact
,
¥
It
.
E
,
-
,
;
,
¥
,
'
,
¥
,
7
-6
,
÷
I
-7
,
-
,
F. as
,
E
'
,
8
,
,
-
,
-8
,
¥
,
.
Es
.
.
I
,
...
.
,
...
,
.
"
...
...
,
t.t.tt
! ! !
Cantor
,
...
,
3
is
countable
.
...
R
is
(o ,D={xtR
not
:
ii.
,
...
÷,
-
,
,
,
¥
,
Is
,
6
-5
,
he
,
I
,
5
,
÷
,
#
,
%
,
Iz
Is
,
I
,
-4
,
,
iii.
?
-
,
¥
,
I
4
,
Is
.
E. I. ¥
,
÷
,
E. E. Is
-3g
,
tg
,
,
⇒ if
,
-3
3
E
,
infinite )
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is
E. I. ¥
-
,
he
,
Q
.
,
-3
,
÷
,
'
I
÷
,
-2
,
ie
-2
,
E. E.
,
÷
2
,
I
,
't
,
¥
tz
its
,
¥
in
-1
1
,
(
:
anmtably infinite
uncountable
ocxci
} is
.
It
,
...
Suppose ( 0 D= { as
each number in this
Proof
,
for
G,
0.9
=
0,9
=
92
93
=
94
=
0
.
9,9
,
list
,
924925
}
0
.
95, 952
-
.
}
Choosing
.
,
decimal
a
,=
953954955 95g
9jE{ 0,42
.
.
.
9⇒
an
qsqaq
,
.
expanga
,
aj
"
.
93,9329394935936 Gp
09,9%9%4
95
,
94,95
9,69g
9,394 9,5
an
,,
9,93
,
'
'
'
is
the
digit
.
'
}
9
.
jtl decimal
ith
number ;
IN
ije
,
.
,
of the
...
957
.
,
'
!
.
choose
Now
chosen
that
so
which
in
b=
an
but
am
bit
be
case
for
the
ne
nth
!
0
an
7
some
,
Think
about
this
b=
.
.
This
IN
For
bzb
:b
bqbsbab
example
gives
But
a
when b.
...
,
,
choose
might
I
number be 6,1 )
btq
since
nth
the
.
decimal
What
,
is
digit
the
uncountable
of
an
.
This
is
a
so
;
{ 0,42
'
unless
=3
by
-
,
is
}
a
is
;i⇒
assumption
of b is
our
,
decimal digit
Ho1) It
,µ
distinguishing
contradiction
practical relevance of
countable ?
vs
.
b
e
.
So
,
.