Midterm Review:
Chapters:
Definition 3.2 / Divisible)
byblbla) provided
divisible
a
that bL=a
.
Definition
lleven)
3.
c. such
integer
integer
by
Prove : The
•
by two
provided
✗ is
,
then
even
xty
.
are even
is
divisible by 2. Likewise, yiseven
3.5
if
ty=2at2b=2latb)
that Xty=2C
a
/ prime )
-
Anintegerpiscalkdprime provided
3×1=325
11=10
only positive divisors ofpareteandp
.
Therefore
Proving
→
is
,
×
✗
83×2--6>5,503×5
'
Algebra
add up 1-090
are not
congruent :
:
land) ,V→( Or) ,
Ay
_
True
( not)
>
"
,
"
✗ →B
,
is
prime
"
-
Trueittx
-
_
hypothesis
B- conclusion
✗
y=Tittx=y
Biconditiona/ theorems:
"
T
"
when both A $ Bare
-
"
-
when Ais not
y
F
✗
→
T
F
y
-
-
otherwise
False
✗→
and x=F otherwise
y=y→x , Fifty
y=F
X←y=T
V
n
T
satisfied
TF
T
FF
X
TF
T
T
T
T
F
TF
F
F
F
>
×
F
T
logically equivalent
>
×
F
T
T
F
T
T
satisfied
are
TF
→
Truth table :
satisfied
when atleast one is
yandtix)Vy
T
Itandonlyit
"
Both
False
y=Fiftx=True$y
←
_
/ implies)
ifandonlyiflitt )X=True$y=True
✗
T
statement false
/ False) : counterexample
→
✗
a
'y=5o / True)
✗
→
/ namely ,
:<to
A>
5,502115
conditional statements : If A. then B
✗
c
angles
TX
Ex:
integer
'
Boolean
:
"
-
an
9=80 / True)
'
11=45-4--45
prime
>
22's that
therefore 5-
"
there is
-
XVY-trueiffx-trueory-true.lv
not A
,
/ 2)3=-8 /False) :( counterexample
n?
4×1--4<584×2--8 > 5,504×5
"
Since
'
.
2×2--4<5 and 2×3=6
AWB
.
satisfied but not B
Ais
counterexample Ex:
Two complementary
,
check all factors between 185
"
21g
that
.
find where
→
integer its odd
an
Ex : show that b-
A and B
such
'
→
that psi and
"
Suppose xandyare
n=3→n3= 33=27 / true)
n= -2
23--11
=
.
n=l→n3=P=l / true)
odd
46=211
A-
integer
Foreveryinlegern ,n3is positive :
47=2×+1
Definition
,
→
finding
2×+1
Ex : show that 47
Observe that ✗
Disproof :
there is
✗ such that a--2×1-1
odd : prove
.
atb , such
integer aisodd
-
integers
even
is
2lx.weknowanintegerasucntnatx-2a.li/Lewise,21y,weknowaninl-egerbisuchthat
y=2b
5?
integers
even
definition 3.1
integers by
even
itis divisible
of two
sum
Itxandyare
1=4
3.4 / odd)
Definition
a
a=2b
provided
-
20 is divisible
5120=(5×5-20)
an
→
there is
where
bla=lbC=a)
Even : prove
Ex : it
aand bare integers
-
→
✗✓
y
T
F
F
F
T
T
T
T
F
F
T
T
T
pnavethl
value
game
all possible
situation
Ex :-(
←
T
T
T
F
F
T
F
TF
F
FT
xnylandtixlvtiy)
lxny)
fix)vty1
×
y
T
T
F
F
F
F
T
F
F
T
F
T
T
F
T
F
T
T
F
T
F
F
F
T
T
T
T
✗
Ay
-
>
×
ay
TF
T
T
T
←
F
Chapter :
List :an ordered
of
sequence
Ex:( 1.2.35> alistot
Set Builder Notation :
objects
form :{ dummy variable
integers
Ex:{× / PIX) }
1) → an empty list
→ property
_
c-
ambient set : conditions}
is
Theorem 8.6 :
Variabkhguintnat true
EX :Ñ=§X:XEZ,X≥o}={
E-
{✗
theorem :
multiplication
A-
of elements
length
possible
C- 27 :2X
{ ✗ C-
elements
nk-nx-nxa.fi/- replacement
is
}
✗ C- I :X ≥o
is infinite / All even numbers
27%2×+4=6 } is finite / it's
equation
divisible
that
,
-81 }
=
allowed
an
are
}
only
by 2)
Subset : Aoindbaresets , Aisasubsetof B
kfactors
In)k=nxln- ( replacement
not allowed)
provided every
element of Aisalsoan element
OTB
K factors)
↓
Mlk D) =/n-ktdt-xiletA-81.si ,z } }
ntothelallingk
.
"
'
"
-
1. IEA
Factorial :
Double factorial :
otanoddpositiveinkgernisthe product of
all odd numbers from
7- ton
Ex :9 ! !
5
7-
9! !
__
3
-
Exiifnisodd ,aren ! ! 81h ! ) !
no,3 ! !
3. (3-12)=3
-
-
,
while /3 ! ) !
/ / In
-
→
j=O
4.
{ { 1,2771A
5.
∅≤
A-
→
always count the empty set
as
well
set :
Ex : list
all the elements
01-2%33?
2£ ?- {0,817,827,83%81,2%8433,82/37,842,37}
- I
""
expression
)=(n -0)(n-1) .tn
i
.
.
-
K-1
-
←
{ 1,2} .CA
A
power set of A :2A={B:B≤ }
-132.111--61=720
→ upper level
-
3.
Power
equal ?
Product notation :
#
{HEA
I
(9-27.17-27.15-2) (3-2)=945
9.
2.
)=(n)k
↓
empty
t¥t+at
repeats
Ex:{ 1,17
set
lower level
all elements
tneamiuntofnum
you can put
together
together
↳ index
14¥
> how
many numbers
cardinality:
youth
(6-5)=(6-0)/6 1) (6-2)/6-31=6×5114×3--360
Ex :/
-
object and Aisaset:XtA
87 means set / I"means list They
"
"
.
.
not equal the
anemptysetbyitself
something
reatnumbers
do
sets names :3
-4N OENI
,
Cardinality otapowerset 124=21^1 counting howmanyelementsisintheset
>
.
lnathlra/ numbers)
/ integers),
)
(rational numbers ,☒ / real numbers)
but -1
elements inaset
18181,2711--181,2771=2
✗ is an
famous
>
-
/ At
180 } / =/ → asetofanemptysetistillhasanelementwhichistheemptyset
Sets:
"
-
81,4371=3-33
101=0
if
number of elements inaset written as
,
etc
-
.
1ER
2$ # {0,817,827,83%81,2%8438,82/37,84437}=8
→
""
e↓mptyTÉ¥¥
set
Ex:{ lit
tneamaintotnuma" elements
together
you can put
together
)
has one answer
Union: theunionofttandpsistheset
:{ × :
AUB
↳ includes
intersection :
XEAORXEB }
flips the quarters then passes
everything ineitheaorts
Symbols : Let
set difference :
A- B-
↳ A- Bis
A
A-
{ 42,3
}(AnB=0don't
have
numbers
for
everything
}
2
thats not in B
.
every
]-
symmetric
istheset A
symmetric difference
B= / A- B) v43
↳
-
of A and B
A)
IT / (X))
:
integer, there is
ytzgfx
There
difference : the
beafor.MU/a-aboUtaUariableX
an
integer ysuch that
ty=o→X=3y= -3 (31-1-37)=0
✗
within
/
( x>
IX))-=V- ✗
t.tl/EZ,FyEZ,Xty--O / true)
>
the set difference of Aandpsistheset
( FX ,
multiple quantities
13=845,671 Theythe same
D- and B
{ × :xtAandX&B
_
→ →
the intersection of Aandrsistheset
everything within
down
,
AnB={ × :XEAandXEB }
↳
Negation:
×
is
c-
some
23×+9=0 /false)
integers
such
that torah
integers
,Xty=0→x= -39=1 (1-3)+11=10
Multiple negation:
TN-XEZT-y-CZYEXK-T-xc-Z-rfd-ytzyz.tl
everthingwithinnsnotin
FXEZV-y-c.IT/y2--X)ADBSc:doesnol]-×EZ7V-yEZy2≠✗
Borviceversa
≤
include intersection
Ex : A :{ 12+31,4/7
AAB
B-
831,415 -6}
E-
{ Haiti }
_
AXB
-
_
{ 435,6 }
answer> =
combinatorial
Cartesian
product
is the set
{ His):xEA,yEBf
proposition 12.4 :
]-×eztyEZy2≠×
ADBAL:{ 1,415,7}
,
Cartesian Product: the
-
-
=
disjoint
proof :
finite sets ->AnB={ 8=0
n
pairwise disjoint finite
Letnandbbefinilesets Then
.
2%
>
number
sets -7
§
,
/ AKI / disjoint impairs)
otsetswithk elements
IAIHBKIAUBHIANBI
n'↳ number oflistsotlengthkstromapoolotnelementswlorepitition
Addition : if / AnBl=0,the1A0Bl= / AIHBI
In)µ > number of lists length K fromapoolnelementswlrepitition
inclusive :
-
,
toranysetsA.B.IAUBt.LA/tlBl-lAnB1
if it's
an
not
n
number of
ways
empty
set
-
Quantities
name
used
:
everytime
an
object
needs
can be
arranged
look at the LHSSIRHS with combinatorial
proofs
a
Relations : awaytocompareasetof ordered pairs
Asserting universality :
"
✗ in
Ex: Every
A
,
.
.
.
"→V-
integer that is
Ex :{
✗ c- A
divisible
by
lois even
V-xc-Z.to/X--2lX
for some
negating relations :
*
✗
inn
"
.
,
1411,142%11121-41,1)tR )
denoted ☒
Existential :
"
we
n
Chapters :
.
for all
!=ln)n > number oflistslengthnfromapoolotnelementswlorepitionor
.
.
→
FXEA
is the relation 0nA :
☒ =/ AXA) -17
={ (a) b) EAXA :( a. b) ¢177
in other words.am/bittla,b)EAxA$
(a) b) -417
.
Definitions :
→
8=-2 / mod 6) since
reflexive _>V-atA,aRa / everyverkxhasaloop)
→
47=-23 / mod 8)
→
irreflexiue-V-aEA.am/altherearenoloops)
symmetric →
bRa) / every arrow
ta,bEAlaRb
isreverisbk)
antisymmetric ->ta,btAHaRb)MbRa) a=b
w/
transitive
in two
→
steps
,
Anda EA The
.
[a] :{ ✗
laRc)) / atrip taken
EX :
can betaken in one)
C-
0nF :
,
reflexive
,
equivalence classotaistheset :
EA :X
that
-
class
equivalence relation 0nA
-
same size as relation
-
-
,
antisymmetric
,
[ 81,271=881,2%{1,3%81,4}
transitive
,
transitive
[ { 1,334} ] :{ { 443,477
/ 3,3) }
→
of
⑧
→
3
q
→
on
the set A-
not reflexive
,
{ 1,334}
/ 4,414s
ifnisan equivalence
notirretlexive , / 3,3)ES
Ex : if the relation
→
yes symmetric , / 1) 2) (2) 1) $13,3) -313,3)
not
Inverting
all ordered
B-
'
=
not transitive ,l4Dts
,
,
:
denoted
pairs
$12,1)ts 141145
on
RT by reversing
ink
8lb,a) :( a. b) C-
,
,
would have
A/
-
,
~
:{ { 1,225,834,6} ,{ 5 } }
define
a
relation
E-
0nA
by saying
✗
Equivalence
belongtothesamepartotp :
Rt
✗
Éy 7PEP,XEPAyEP
Counting Anagrams:
two
Risa equivalence relation
relation
are
equivalence
relations
have same
letters indifferent order
Exitlowmanyanagramsotthename
Mexico
are
.
there ?
,
,
onanysetisan equivalence
provided
words
Mie ,X,i 40=6 !
This : reflexive , symmetric $ transitive
relation
disjoint sets
E- yitandonlyitxandy
]- PEP,fx,y}≤ P
Relation:
Ñ^°ᵈ
{ 1143,4 -5,67
given by
Partitions : Aisasetofpof nonempty , Pairwise
we
Pi'={ 1411,12111/3,413,444,3)}
$
A.it/swritHnasn-
unionism
17-811,111,2%11,371%3713,1)}
-
on
/ 4,6) , / 6,4) }
Ex:
=
relation
/ 4,4) / 5,5) , / 6,6) / 1,2) /2,1) , / 3,4) , / 4,3)
, / 2,2) , /3,3) ,
,
,
(3,6%16,3)
we
relations
inverse OTR
~={ 11,1)
antisymmetric , / 1) 2) c- 5$12> 1) ES ,
1=12
→
the
{2,3%{2,4%83,4}}
[ { 1,33 } ] :{ {1,73%{1,34%81,347,873,4}}
@#
} then
"
[ 17={81%827,837.84} ?
transitive
,
→
'
ofanysubsetisthesetofsetswiththesamesize :
00h0 : reflexive , irreflexive , symmetric antisymmetric
Ex :S :{ 11,2) , / 2,1) ,
0h28 " "
[ 01=80 }
antisymmetric
,
since 17-11=685+6
itthisthehas the
equivalence
Identity relations
≤ 0h27 : reflexive
81147-23)
Equivalence classes :
eachother)
ta ,b,GEAHaRbHlbRd
since
Let Abeaset , than
lnotwosidedarrowsotherthanloopsortwovertils
an arrow to
17¥11 / mod 5)
6118-2)
relation
Ex: How many anagrams of the
Gama,d,a=6 !
a
-_
3!
name
Canada
63,1=6×5×4
are there ?
whose
Explicit
An
formula
Letnskbe
Exinoinlegerisbotheuensodd
integers
!
(
n!
n
-14
:( µ,¥)=¥!=5%
>
so
,
=
Knt
35
so,
6×5×4
=
statement
-
,
is true
Claim now
-1k ,kzEz3n=2k,n=2kzH
⇐
5×4--20
-
after negation
Applied even
K,
suppose for the
-
kz=E
sake of
→
contradiction
contradiction that there
Where n=2k , and n=2kzH Then set them
Theorem
111=4%1+1%1
Ex : How
0< Kan
-
2- element subsets
many
then
tore there is
01-81,213,4 }
exist?
→
including the
.
Excluding
there
are
There
the
from
can take one more
like:{
are 3
4. You
from
842,3}
→
1. Observe that
no number that 's even $
it the conclusion
1,2%8137,8437
→
has
a
disjunction
'
/ !
makes it more concrete /
if the statement
"
-_
by proof by contradiction may help
"
if A , then B
contrapositive
"
__
if / not B) then /not A)
"
Fx :
212
a) itpisprime , then
if
2%2
is divisible
byp
.
→
original:"If
is not divisible byp , then pis not prime
"
converse :
inverse :
if B. then
"
contrapositive : "It / not B) then /not A)
"
A
"
,
induction
By assuming
"
it / Aomdnot B) then
/ something false) "
Ex :
'
.
slept: prove
.
"
if / not A) then /notts)
Contradiction :
slept
A. then B
"
→
that
Assume
11-21-37
.
.
-1-501-5 It
. .
-1-98+991-100
¥
Plalistrue / Basis step)
Plktistruesproveplktlistrue / induction step)
York ≥a)
to eachother:
Ki
kz=É There
odd .
.
.
' '
,
then this
-
.
⇐
101
jE%ÉÉk=5ohoD=l0%
sÉÉ k=i
or
that't
,
ifnotthisandnotthat'T
is existential ,buta constructive
contrapositive :
original
equal
/ contrapositive
it statement involves lots of negations
Contrapositive
.
117-(3,1+13)
,
kz)=
.
may eliminate some)
1,4%82,4%883,47
can take two
3. sets like this :{
together
→
4) you
-
,
a
which to use ?
→
842,3}
→
21k
214=2141-1
integers with
is
anumberthatisbothevensodd.letkiandki.be integers
4×3×2×1
Let risk be
odd definition
-
.
-
$
21k , kz)= /
contradiction
in words:
10×9×8×7=10×3×7--210
=
the
iioddl)
even
214=2141-1
3×2×1
10-61
(G)
Inis
-
Fritz niserensodd
6-32
=
27 ,
Proof : Assuming
!)
F×
(G)
negation
.
( %)=(=
K!
1k /
Ex:
>
C-
with OLKLN Then
proof
is
elusive
,
Review
Midterm 2
Se( 15s16:
Definition:
Let A be
nonempty, Pairwise
->
a
disjoint
nonempty: for all Ptp,
pairwise disjoint: For
->
partition of
set. A
sets
A is
a
Binomial theorem: Let
set pot
monomial:
whose union is A.
coefficient:
PIO
all P., P2,
tp, if
P.
binomial.
EPC, then
-> for
P,n4z 0
=
EX:(e + A
WP A
everything
pap
81,2,3,4,5,6E.Lep. 81,24,
=
P2 33,4,65,43 358.
=
881,25,83,4,45,8581
Let P
is
->
partition
be a
reflexive, symmetric,
-
The
equivalence
counting
and
many anagrams of
if the A's
there
called anagrams if
letters, perhaps in
x
are
=
YC, A.,
a
the
Canada
name
are
differently. However, for
"equivalent" anagrams
are 3!
that
are
6!
there?
anagrams
each
anagram
only rearrange
the
A's.
6!
6x5x4x5Xzx+ =
=
6X5x4
3xEX+
31
Sec it'
Definition:Let
n, KEN. The
K-element subsets of
a
set
symbol
(Y) denotes the
number of
of sizen.
Ex=(0) 1(1) 4(1) 6(5) 4(1) 1
=
=
=
Proposition 17.7: Let
KEN w/O-k <
combinatorial Proof: Let
does
->
X have
Answer 1:
->
=
=
n,
that are
x
=
size
501,2,
.
.
.
n.
(() (n 1)
=
Then
=
... How many subsets
k?
IP) by definition
Answer 2: Let B be a subset of x with size n-k. Let
A = x-B.
By
the
since there are
addition
(nK)
coefficient of
xby"in
1+
=
principle, (A)
subsets of
x
I
x'yn-"in the
expansion of
be
=
=
15
=
15.x.
4y": 60xy"
so the
=
(x1-1B1
=
n-(n-k)= 1
of size n-k.
coefficient
integers
w/o<Kn. Then:
6(j) (54) ()
+
=
17.2:
=
Let nsK be
(3)
integers
-x5x4
=
3x2x1
=
w/ Ock>n. then
5x4 20
=
x+yl
(x+yl"?
2 +3 + 4 + 5
(?) x" (y)
is
or x+
60.
n 10k
Ex:
they
different order
N, Ac, D, AE. There
counted
2(3)
-
10x3
formula:
p
Theorem
are
term. Ex:
+
=
Ex:
are
of
classes
is the
=
w/g-degree
x"y" is
one
(4) (() (Yi
makes it
A. which
grams:
have the same
solution: Let
on
stransitive.
Definition: Two words
Ex: How
relation
()
10(1)x"yn-k
=
is the constant factor. Ex: 10
Pascal's identity theorem:Let nsK
of A. Then:
equivalence
an
of
=
=
4ork
(x +1
of two monomials. Ex: xty
nsk,
=
the term
=
p pP.,P2,432
Sum
N. Then
expression of
monomial
a
each
n 6k
=
=
algebraic
t
Ex: what is the coefficient of
=
The union is
->
an
n
(i)
=
k:
-k):
Sec
21:
preserved by addition
order Axioms: order is
multiplication
by a positive integer: For all
of equals
well ordering principle
or
there is
a, b, and c
=>
lacb)e(c>0) =>ac<bc
lower
Lemma: There
Proof: Let
bound, has
of all
x be the set
suppose FTSOL that
between
1A,
x
->
0$ 1.
between OS 1
integers strictly
<0. Since
x
there exists
->
contrapositively, prove
empty
on
that the smallest
there aren't
any counterexamples.
Os is
there is
are
not
well-ordered,
any two distinct real numbers,
given
rational real number
a
large negative integers
there exist
between them.
x <I
Ex:
Every integer is
Proof: Let
-x
for
-x
E
M.
some
=
x & z.
Lemma: Let
CN, then
x be an
odd.
it
even
even or
x
=
either even or odd.
odd, It
x
x
*M, then
is even, then-x = 29
=
if
-X is
LH3
odd, then
PTHS -
-216 +11 so xisodd.
natural
is
exist
x is
1t0
pM=
number,
X-1
X-1is
odd, which
means
(Provew/w.0.P)
r-1
But for
be smallest with
n
pcm-1+1
=
0,Som20,
r-1
+22+r3..Hemyrel, add M
to both
sides
+r3+r3+...4 rM+wM
rM+WhtwMrut!
r-1
contradicting
7: there is
ordering
principle proof
no
counterexample.
template:
define the set of counterexamples
odd. If x-1 is
natural numbers that
are
neither
of all natural numbers that
x
XD.
X 11
and
X-1EA.
and smaller than the smallest element
is either
odd, by
even or
odd. If x-1
the lemma,
Therefore
Ax.=z
x
X
AsoxNixiseitherevenor oddfTAux=A
want to show x = 0, so that A
=
M.
is even,
X is even. In
=
0
by
either cases,
after all.
sAnx
3) Reach
a
contradiction
finding
are
the WOP, there existsa
By
even, OAX. Therefore
odd, If
X is
=
so: atm,
well
if X-1 is even, then
odd. By assumption
of X, x-1tM-X. Therefore
we
If
=Itr
+1is odd.
Let x I ↑ be the set
Since X-1 is a
m
+_I
·
Ex:
Un-N.P(n)
c=PEA
2) assume (+0. By Wop, have minimum element
EX. Since O
X is even or
x=- 26H1
wop. Let
1tr+r'+u3+...+=
odd
FTSOL that there
even nor
the Lemma,
-X
by
sums:
and Itr+r"+r3+...+
not both.
2 1-a),sox is even
and only if
integer,
Proof
even.
Proof: Suppose
x
x is
X is either even or
then is
smallest
odd, and
Geometric
Ex:
of X.z
for some bEE. But then
Ex: XE I is
neither
x
But then
E.
at
2b +1
even nor
if
even or
so-X is either
in all cases,
odd,
x'EX$x'<x. But this contradicts the fact
X was the least element
can't
well-ordered.
well-ordered,
I is not
a
x
that
counterexample
set
sets that aren't
on
·
leastelement X ofxLet x2. sinceofisanintegeris
x.xxx11-x X.50
then
W.O.4 can't be used:
minimum.
a
integers strictly
are no
↳
of integers
Well-Ordering Principle: Axiom Every nonempty set
a
any counterexamples,
exist.
-
which has
are
smallest counterexample."
a
↳we prove by contradiction
at (cbt
acb
states: "If there
or
ctC
with
by proving plm)
somehow
<< m
(math
s
not
me C.
reasoning
Pn)
principle of
the
Suppose
2.
for
Proof:
set of natural numbers
all
A
to Show A
N.
=
Usinpinduction
Albase case).
Of
to show KttA.
to
both
1=
3.3
3
We
1)
Now
n.
suppose KCA for
2.30 + 2.3'+..+
know
->
for
Then A =
number has some
other
property
1) invoke
=
is
an
property
property
hypothesis
P.
KH1
then K+16A:
that some number
work/algebra results
=
A
.
.
.
P.
P
and KEA, then KHICA
,
KESA, then (KDEA
.,
some number
=
I has
to show Kt has
property P
P
property
N
conclude A
->
integers: Do natural numbers
replace
numbers not
-
written as 49 +56 ford, but N
on
12+A, 13+ A, 14A
cases
We wish to show
(n2K,
n.
exists (such that
been
have
812, 13,
....
established
KEIA.
claim KHICA.
We
=
kit
s
172
if A
=
target
negative integers separately
real numbers -> not well-ordered
a
so
or
a
Proposition
from A
A
controlledby
size
when
- such
by
of domainstarget
=
=
(f) 11a,1) +f) =
b=
(
= 9: 5b, la,b) ef?
is the set imf
=
86:Ja,la,b) eff
a
function from A
to B
If:
A
+
B).
B is
a
codomain off.
be
and image included in
B*
bijective
EC.
domf $B1 imf, f is
or
is
that Va, b, c, (1a, b)
sets.
By Be
b"=BIA
it means, the set of all functions - we
B.B=Sf 1AxB:f:A-B?
24.10: Let A $B be finite sets w/ A
to B.
N
finite set
-
Sets of functions: let A $B
domain
exista, be
there does
Pa +56. Therefore KHCA.so
=
in A.
n 112 are
function is relation
image/range): f
K-
inductive
26:
function
a
By strong
that K-3=4'+56: observe that
important properies -> injective, surjective
-
well-ordered
such
all
domain: f is the set domf
->
a
PMISr,
functions: domain -> z,
not
kt
as bit
there exist
a'+1 and b'l such that KH
to show KH has
N
finished. If 171216, then 1-3212.
Inamely,
Sec 24
K
we
(K-3) +4=(4' + 5b) + P=9 (a'+1) + 5b.
Definition:
Inonnegative) even numbers: replace
rational
.
0x212:Ja,b E, n=4a +568.
magnitude
negative
->
=
if 12 2K +1215
P.
P
numbers:
2
conclude
can be
n = 12
strong induction
suppose there
element of A.
KEA,
->
->
nCA. Use
the base
induction work with other sets.
->
Proof: Let A
property P
OCA. shows that I has
hypothesis: Assume
to
Invoke PMI to
Every integer
EX:
N.
P
PMI
KEA, if 80, 1,
A, IGA,
property
property
↳) some -> changes
Prove if
property
use
add
OntN:n1 no?
inductive
has
we
KtM, KEACKHEA.
3) Inductive step:
·
4)
some K. We wish
2.30 + 2.3'+... + 2.3472.34+13411+2.3K+t
have
number to
each
Ot
otherwork/Algebra results
·use
by induction template: 1PMI)
2) Basis Step: prove
some
Prove of A. Shows has
each
satisfying:
numbers which have
·Inductive hypothesis: Assume
wish
2.3341). If
Let A be the set of natural numbers which have
->
for
*
23=0 and 30 -1=2,
Since
of natural
A be the set
3) Inductive step: Prove if
case)
191.s0K+IEA. Therefore by induction, A
prove every natural
To
of
natural numbers
set of
a
case
sides, we
proof
Regular
A N is
2) Basis Step:
8ntM: 2.3°+2.3'+... 2.3= 34'. We
=
inductive
=
1) Let
N, 2.30 + 2.3't...+ 2.34=34t'-
->
2.3
Suppose
satisfying:
KEN, KEA =>CKH A (inductive
nt
Let
we see
(PMI)
(basis case)
each
Ex: for
mathematical induction:
A LA is a
1 Ot A
Proof by induction template: (PMI)
Strong
Sec 27-23:
=
(B1 = b. There
are
"functions
inverse Definition: Let
R be
=81b, a):/a, b) CRE
GeneralinclusionsExeclusionmetM.Abeemitsets.There
relation, the inverse of the relation
a
infixnotation:a
a Rb
=
Proof: induction
one-to-one/injective: it
if
y
x
=>
f1x)
ffly)
x =.
And
-Base case
->
.
doing the
I makes sense when
·
fix)=fly), then
whenever
contrapositive of
-
set: RoS
B be
a
->
81x,y)
f
a
on
the
functions a
particular
y=1x18
2x2:
function from
f is
Because
-
and
187, but
to
z
and
is
onto N. Because for all xt*,
x
=
Let f:A
Bijection:
-
f:A -
->
there is a relation
↳ There
is
a
bijective
bijection iff "
->
B is
B is
on
a
function
is both an
s
injections subjection
domf"= B
sets AzB if there is f:A
+
B/ bjjection)
if 1A<
is associative: it is
B),
the
Identify functions: Let A
be
ida whose domain is A,
for all
if f
->
iff
is onto
counting
is not
->
Let ADB be
finite sets w/ 1Al
a b, the number
there is
no
of
number of
are no
b, the
number
bijections
a
identify functions
on
A is the function
CA, idola): a. That is: id =
a
under composition: Let
ABB be
8/a, ): aCA?
sets, $f=A-B
domains.
same
a
function.
Let at A. Then
both an at
->
A to Bis
(b) a.
subjective functions from
arb, there is
no
of bisections from
A to B is
to B is al if
Explain that
+ (x) fly)
=
.
It
B is
a
a
xb, there
To
function
prove
a
f:A -B.
statement
A is
3:
called
called "holes".
is is true
is not
template:
if
there exist
x,ytAw/xkY
$
injective)
IBIIA)
->
ShOw
->
invoke the
pigeonhole principle
injective
surjective
A
Describe sets As B and
"Pigeons" and
x1B1= b
injective
'HD" (3) (-) if
a
injective
=
b
functions from
If
set. The
Pigeonhole Principle proof
2) If
4)
to the set:
Sec 25:
B, then 1A) = (B)
a lb, the
equal
onto B
A to Bis
3) If
s
a
-(foide(a)flidelan-flies I
not one-to-one.
functions Summarized:
as b
are
relation from B to (
relations from A toD
as
=
=
I) the total number of functions from
if
(RoSoT Rolsot)
a
R
foidn f side of =f.
one-to-one, then 1AIIB
is
relation from A to B, is
a
=
=
contrapositively,
->
A, then R is transitive iff RoR
on
if foidn f side of ifs have the
if (A) > /B1, then is
relation from B toc, then
Proof:
Let A $B be finite sets $letf:A -B.
->
relation
a
a
A to (
Risarelation from (toD, then:
then
Pigeonhole principle:
->
is
identify functions
relation
equivalence
an
is
providedf
is the
Slad): aCA, deD, Jb,2, (a, b) eT, Ib,11 tS,K,d) ERE
(x)
imf.
so xt
relations: Let RBS be relations the composition of RSS
relation from
a
composition
1 to 2, but not onto E.
1A,
both
are
slide 24326 Handout pp. 51-53
Proof: Both (Ros) oT $Ro/soT)
function from
a
case:
left bright sides
is a relation from A to B, and R is
if R
->
codomain.
=1:
8 (a,2): 76, (a, b) cSn16,4) [RE
=
Rob is
Airf
-1
->
is
all be B there exists
=
subjectivity depends
Let t =
-> it
fla) b.
such that
at A
provided for
function f is onto B
n
composition of
=-
surjection:
Let f: A
inductive
on n
therefore, statements
is true.
to conclude
that I
cannot be
monotone
is
subsequence
A
list and
Ex: A standard (fair) six sided die
subsequences:
list
a
deleting
formed by
elements
from the original
the remaining elements in the same order in which
keeping
they
given sequence. For
Let I be
-> ux
-> dx
length
=
of the
length
=
is
is monotone if it is
subsequence
each
always increasing
or
at
a
uniform
probability function
5 for allstS
=
starting
imply that
appropriate.
PIs)
sequence
this
S=81, 2, 3, 4, 5, 6 8
decreasing
XCS, define:
longest increasing
sample space models
Experiment?
solution: the words standards fair
originally appeared.
A
rolled. What
is
probability: A spinner has unequal sectors
Non-Uniform
X
of the longest decreasing sequence starting at X
3
"
As
set,
a
the
sample space consists of all possible
marked
values the
on
it.
spinner
C
The function:
Letfis
->
Let
->
if
->
x
x
<
x+/ux,dX), we
AXA be the map
be two elements of 5 with x
By
y, any increasing sequence
claim that f
preceding in
at y
starting
is
the
can be
P(1)
sequence
if
x
<
y, any
prepending
->
in either
finite, so
his
at y
can be
A is a
24, and
is
lengthed by
to model the
dy
->
since
+27, then
121
=
2
,
is not subjective.
it follows the
->
pigeonhole
principle.
a
FTSOC that is
tB or
a
AB. if
then atfla)
since
a +
B
=
adflal
B, then
since fla)
fla) B,a cB.== Because
=
=
at
B, AB.
we have a
A.
Either
E
it
a
AB
contradiction in
cardinality
can be
IA) 1IB) if there is an
injection A + B,
-Cantors theorem
1A)
say
24), even
assigned
to
infinite sets
1A)= 1B1 is there is
for infinite
by saying
abjection
A-B
sets
A
⑤ Pi s -> A is a
->
->
function
a
pair (S,p), where
is a
finite, nonempty set,
PIs)10 for all st
Rs
Definition: it
is is
probability function
any finite set, the uniform probability
to bep(s)=
PIs) =2 T
:
2
those
=
is)
for alls.
Ts. "Ia
...
-
100
999,900
will test
on
3 sets the
given
the test. Find
an
apporiate sample space
sample space:
or
or
not
IN)
negative It
function. Suppose
->
sample 1,000,000 people
x
-
100 =
chosen.
$5
negative
999,900 will not have the diseaseof
0.95=949, 905 will test negative $999,900 x0.05:49.995
positive/false positive).
PIN, H
=
Plu,-=0.000005
0.049995
Cntion:
=
we
will have the disease. 95 will test positive
P14, +)=0.000095
Let
PIN,-)
=
0.949905
sbeasample space w/probability function P.
is a subset ALS. The
Probability of
SonP1a).
if Is,P)
->
PIs)=1
is
(false negative). 1,000,000
->
such that:
a
positive It)
probability
a
PIA)
samplespace is
patient
$14, H, 14, -, IN, H, IN, 78
event A
se(30-31:
Definition:
so:3
=
->
both cases B cannot be in imf.
infinite Cardinals:
P14: 8
experiment.
the test is
1,000,000
PxtA:x&f(x)8.
subjective. Then B=fla) for some
=
either the person has the disease (4)
next
Proof: suppose that f:A- CF. Define
Suppose
=
solution: Frist, we need
set, $f:A
proportional
EX: A disease affects (in every 10,000 indivinals. A test for the disease
lengthed by
case, lux,dx) #lwy, dy)
cantor's theorem: if
if
dxdyt
>
EP12) E,P13)
=
each outcome is
region containing it. The proportional angle of each
is 95% accurate. A
decreasing sequence starting
x. so:
s=81,2,3,98. The likelihood of
to:
to the area of the
injective.
prepending x.s0 Ux 1 Uy+13Vy
->
point
can
the
is
a
sample space:
elements s are the outcomes
the subsets are the events
an
event A is
An
combining
A-B
is the event A occurs B
A
Alset complement)
s
=
-
Ex: Rolla
number,
P(B) 0. the conditional
P(An B)
P(A(B)
does not
$B be the event
A be the event that
that the
B, AUB, A-B, $A=S
-
die roll is
a
the die roll is
prime
an even
number. Describe
A
4, 5, 62,
->
AnB =
->
AvB
-
A
82E,
->
82,3,4,5,65,30P(AUB) E
=
81,3,58,so P(A) E
S A
=
Probability
combined events: Let ASB
EsP(s)
S,P(s)
0
=
=
Since this is
an
empty
sum
P(E) = P(A vE) + P(A nA) P(s) + P(0) =
for
=
4)P(AUB)
IP(A) + PIB)
=
Ex:n people
Assume
(n-tuples)
are
365
chosen.
What is the
the same
birthday?
days in
take
-
There
Let
s'denote
a
As Bare
P(A)P(B).
sample space
=
PCA)
.
.
let be
s
a
positive
of all length -n lists elements
the set
p:s"-> B by PICs, s,
(5%)
a
are
year.
The
31,2,. . . 3658.
(365tn
a
.,
in s.
sn)) P(S)P(S2)...P(sn). Then,
=
the
n-fold
repeated trial sample space
probability
Solution:
consist of list
has
equal probability
ways
to
assign distinct birthdays
dif eredtrain
Probability thatallbirthdays areno
to
a
people.
function defined
on
a
probability space;
of
variable is
a
that
function x:s->
some set v.
3
x(s)
sample space
each outcome
that at least
is a
sample space, then random
flipped
3 times. Let X be the number
comes up. List the values
-
3654
->
Is,p) be
EX: A fair coin is
0,P(A)+ P(B) P(AnB)<P(n) + P/B)
two of them have
->
Claim:Let
is, if Is,p) is
-
P(AnB)
=
sample space.
nonzero, the equivalent condition P(AIB)
is
Definition: A random variable
3)P(n) 1 P(A)
+
P(An B)
in a
SeL33:
=
P(n)
A$B be events
by definition of probability function
1
=
P(0) 0
=
=
So
=
P(0)
if P(B)
Definition: we call
sample space, S A. Bevents. Then:
1) P(S) 1
2)
if An B 0
0
(S,.... (n)7SR
=
P(x)
If BLA
-P(s,..., (n) =
=
be
=
1
if PCA) is nonzero, the equivalent condition PIBIA) = P(B)
Define
be events in a
Sample space. Then P(A) + P(B) P(AUB) + P(AnB)
Let (s,p)
PIAIB)
integer.
=
=
-
->
=
might be true:
think
=>
=
-
PIAIB)
we
ndependent Repeated trials:
=
604,65,30P(A B) 5
B
->
->
=
=
B is:
P(B)
independent provided
P(AnB) 5
so
probability of A given
=
Definition:Let
the outcome ten
=
=
=
sample space (S,p) A suppose
Independence:
32,4,65,B 82,3,559P(A) P(B) =
-
in a
the definition allows two statements
A.
Solution:inthissample spaceSf3,
-> A
Definition: Let AsBbe events
is the event that A does not occur
six-sided die. Let
the events An
->
Sec 32:
spaces.
Blor bothoccur
A UB is the event either for
->
->
a sample
AnB is the event that both As Boccur
-
->
events:Let ASB be events in
HHH
3
of times that Heads
of x(s) for each outcome 3.
HHT
HTH
22
HTT
I
THT
TTH
1,
TTT
8
Random variables
Suppose (SP) is a sample space
events:
as
X is a
random
Algebra
random variables
Variable on 3.
->
->
for
each value
->
of X,
a
can construct events
"x >
have
P1x
>
3)
probabilities, so
P10(x)
=
flipped
event
Random variables:
Definition: Let (S,P)
card
is
$2,3,
be
say that
We
for all values
.
x 11
$4
.
.,
are
x be
"x11"
slet x
s
Product of Random
So,
B = 8C,3,
=
B4= b) =
a
=
a $
4= b) = P(x=d)
PS)=b)
5. Since
b
=
x be the
H, D&. Are
B, PH b)
=
=
X
rank of
by
F for
card
independent?
each
pair
this is equal to PIX = a) PH4=b), these
independent.
Sec 34:
Definition: Let
real-valued random
x be a
variable
defined
on
a
sample space (S,P).
The expectation
or
The variance of is the number
Ex: A die is
rolled. Let
solution: E(X)=
x be
random variable
defined
pair of dice
solution: Ex) =
=
resulting
number shown.
are
be
on
a
=
sample space Let
3. Then
E(x)
rolled. What is the
=
Find EX)?
=
3.5
=
x be a real-valued
x23X(SPIS= Epa. PIx=)
expected value
of the Sum?
2.5+3.E+9. E +5 +65+7.f +8.E
+ 9.
=
the
varix) Ellx-ExI()
18+2.f +3.f +4. f +5. f + 6f
proposition 34.4:Let(s,P)
Ex=sX(SIPCS)
value of X is the number
expected
A+10.
10011
+2
+
+
11.5 +12 To
3 +4 +
36
variables:
*
P1x11)=&=E
$4 be random variables defined
standard 52-carddeck. Let
a + A,P1x=a) . for each
=
st3, x(S)Y(s)
die s the value of the second die. What is EXy)?
solution:
=
ExY)=
Ex:
Find the
solution:
variance
(E)"
F(x-MY
of a die roll.
=
f/l-E1
2.542.34+
a
value at
Ex: A die is rolled twice. Let be the value of the first
the number of
find P(X11).
independent if P1x
10,6, 9, K, AE$4 the suit
to variables are
Ex: A
sample space (S,p), combine
variable value at st3, x(s) +4(s)
yrandom variable
=
drawn from
1a, b) Ax B, P1X
->
real-valued
Variance:
sample space
a
x
=
-X
Pla) about the
sb.
a
solution: for each
->
+ random
-x
t00·EX:
StS,x1332
PHTT, THT, TTH,TTTE
Independent
IS,p).
are
them:
8 PCs)
=
shown. List the outcomes in the event
=
of X,
"X= a
$4
P18SES:P(x(S(()8
three times. Let
an
in A
on the values
for any statement
P18st3:x(s1332)
=
solutions: As
Ex: A
that value. It's the event
other relations
using
set
we
Ex: A fair coin
on
give
on a
x
-x-y random variable value at st3, x(S)-4(s)
values of X,
heads
which outcomes
"
3"
These events
EX:
random variables: if
5) +7.6
=
252=7
36
4. D
+
2.F=25
12)H"" IEL"Y
Final
Exam Review
Se) 35:
Lemma: Let
Division theorem: Leta, bez w/b> 0. There exist
integers a Dr such that
bq
a
=
Definition: Let
remainder of
-v is the
Ex:
snbeintegers
a
quotient of
a is the
->
->
+ r0vcb
byn
swrite
byns
write
a
a
div 13=16212 mod
212
w/nso. we
13
a
write
fan v/0 erch.
divn=9
modn
a
a
->
=
->
Mtb7qUOers
Ex: Let
but
0,
divb= 0
a
S
a
<b(but a
modb
a
=
is
still a
nonnegative?
-
<b
a when 01 a
-
a
Ex: 18
36.3: Let
(Ialdion)
divn
it nta,
a
dion=-(1aldira)-1s
=
-
mod
a
is not the same as
a
modn=n- (Ialmodn)
Ex: use Euclidean
-ladion).
oo
Modular arithmetic: Suppose that
(ab) mod
n
=
modn)
[a]
+
on 2. There
[b]
integers
u
are
Operations
[a + b]s[a]x [b]
=
=
equal
+
$X
a
of the congruence modulo-n
In defined
sb
if
Ex: The
The
god of
god
greatest
9 5:1, 3,
Let
a
$6 be
integers.
an
integerd
is a common
divisor of
-x3
a
=
ged(30, 1st gcd (15,0)
=
sb be
a
of -6 5-16?
9,
god
2
answer:
of 210545
common one.
5, 15,95
exactly
by enumerating
the
divisors of
each 3
selecting
210:1, 2, 3, 6, 5, 10, 15, 30, 7, 14, 21, 42, 35, 70, 105,
gcd(210, 45)=15
210
15
integers
integers, not
be
set of Bezout
with
a
2b >0. Then
+
ax
both
zero. The
by, where
X
by
smallest
are
integers
Coefficients
=
2
a = 6s b = 16.
=
Bezout
coefficient
18 =16 + 2
gallons of
gadla,blic
what cis?
for
3x3-1bx1 2d
Question: what conditions
6 and 16?2
210 $95.
=
of the form
=
dlasdlb.
Ex: find the
the
Let
of
gcd(210, 45) g(d (95,30)
stop god (15,0):
6x1 6
[ab]
god
b<E.
mod
Ex: Find
by:
find
gld (95, 30) gcd(30, 1st
15, so
6x2= 12
Gefinition:
god?
gcd(a,b).
is
-
to
30, SO
Bezoutidentity:Let as
Sec 36:
a
we
positive integer
classes
on
1 x 30 +
proposition 36.5:
(b mod n)) mode
equivalence
4x 45+
nowb=0,
a, b, nt*Sn>0. Then
1amodnl/b modal) modu
Definition: In is the set of
relation
+
mode
(
modnt
congruent
=
Algorithm
30:2x15 +0, so
->
b/modn)
b are
210
d5 =
->
- -
11a
say about their
=
31-6
modulo
->
letc= a mode. Then
gcd(36,24) 12
->
as
=
&
=
=
35.8: Let a, b, nc&w/n >0. Then a
(n + b) modn
positive integers
420 $6796. What can you
a=
s
->
$b be
=gcd(420,96) 12
n 0
div7:2-18 div7=-3
Proposition
a
gcd(420,9() gcd(94,36)
-
a
1-a) civn
does not exist
420div96:4420 mod 94=36
<o?
Ex: -6 div3=-2s-6mod3=0 because
Note:
=
-96div36:296 mod 36=24
it na,
answer:
gcd(a,a) a
=
Ex:3 div7=0$3 mod7=3
question: what if
=
=
cremainder
answer:
positive integer:
1
=
gcd(a,b) gcd(b,c).
=
a
a
gcd(a,0) a
proposition
=4
divisor
Question: what if
gcd(a,1)
gcd(0,0)
v.
be
a
on
a, b, sc will
water with
it
jugs
ged(a,b)=1
a
we
gurantee that
$6
we can
make
gallons?
can make
a
gallons
no
matter
E
X:
Find integers
210=4x45
x
such
by
30=
+
that
+45y: 15.
210x
30:210-4X45
45 1430 + 15515:45-1x30
45
=
-
1x30 = 45
-
1x/210
5x45
-
Ex: Let
a
=
35Db
4x45)
(x,y
that
+
ax
by
d.
=
Ex. v
work backwords:
1 3
=
3
=
=
properties
1x(8 2x3)
-
-
1X8
30x
3/y
=
=
3x(35 4X8) 1x8
3x35 + 1 13)x8
definition: positive
Relatively prime
integers
relatively
$6 are
a
prime it
=
Sec39:
-
The fundamental theorem of arithemtic: Let
factors
primes
into
the
order of
is a
gcd(a,b) 2
sectar,
min
=
Ex: use formula to find
factor 210
=
2.3.5.7
god
=
min
14
45
See 47:
A
Ivertices) of G.
Ex: The
graph
E is
a
is
a
pair
visual:
o =(V.E) where - is
set of two-element subsets of
set (81, 2, 3, 4, 58, 891.28,
map
a
a
nonempty finite set
ledges)
82,38,62,18,33,46,34,525)
function fir->c
the Max Smin
->
G is
a
graph
coloring
v to (exists.
not
I
regular
a be
to
adjacent
US
3
son, retov, net
(v,E)
graph,
V, then
be a
degrees
BV
a
=
adjacent
(v,E)
it all vertices have the same
to
vid (v) IN,
=
End (v) 21E1.
=
graph. Then
of G are:A(x)
momact
a re
↑
neighbors:N(v) Sueviu-ve
of vertices
=
go
vertex of G.
neighbors
us are
the set of v's
v is
=
a
Definition:Let G =
=
=
be
a
graph.
maxd(v): v (Vess(4) minSd(): veve
=
degree. If
degree is
that
r, G
is
E
of vertices of G:
->
The order of
->
The size of G is the number of
a is the number
is
a
of G.
->
G is
complete
->
G is
edgeless
Ex.
graph
if every
edges
v
(()
=
IV(G)/
of a. c(x) E(G)
=
if
no
pair of two distinct
vertices
are
vertices
are
adjacent
adjacent
edgeless:
metal
28
or
18
04
50
se)48:
such
that f(v)
*f(w)
when vww.
is the size
of
the
smallest set
(such
a
Definition: Let GSH be graphs.
v(G) V(H)
s
subgraph
A
-> A
subset
A subset
->
from
of
v(() VsE(G)
+
*
The chromatic number of
3,
is the number
Regular
subgraph
is
of
-
tire
otoos too
coloring
say
SV, n=
adjacent
vertex
degree
arkS:
-a
su,r, EV. We
complete Sedgeless Definition:
min
=
graph Definition:
graph,
order Size Definition:
5
2335'70 15
no
r-regular
soI
5
23
2'3'5'7'
such that
210
210
of 210 545
a
the
=
.
....
a
since su, vz=oV, ut we have
neighborhood
plb.
o
minder, to
su, vz =
the
ze35e5...sb 2zf3zf5
2
=
ses, for
=
=
a
or
is
u
Max, Mins
of into
3.
prime it plab, then pla
95:335 23335'70
50d
n
with
-
Theorem 47.5: Let m
positive integer. Then
a
ab.*
primes.
positive integers
be
$b
a
be
a
primes. Furthermore, the factorization
suppose a, btzsp
GCD formula: Let
then
of
product
a
unique to
is
Lemma:
-
gardens
gcd(a,b) 1
n
if
->
-
f(v)
to
it su,vi+E
to
of vertex Definition: Let
Degree
-
->R" and each edge
adjacency:
symmetric -
13
-
-
( v,E) be
on
adjacent
to 2,2
of
irreflexive
-
function for
(81,2,32, 3361,25,662,323)
=
&adjacent
1X2
3x3
=
=
-
and write
adjacent
are
such
a
endpoints
intersect except at
=
is
joining flu)
in the plane
curve
a
graph
a
Definition: Let G
Adjacent
1x210
Findd=ged(a,b)
8.
=
-
-
to
edge-curve
=
15
S4, 08
e=
of
embedding
planar
A
->
of
E(G)
>
We
say
his a
subgraph
of Hif
E(H)
of H consists of:
of vertices of H
of the
vertices.
edges
of H, which
join vertices
in that chosen subset
-H.
He
->
=
(51.28, 881, 288) is
=
a
subgraph of
(81, 28, 351, 28,81,322) is
=
there's
(b)
(81,2,35,881,2881,35,32,328.
Ex:(e + a
edge
an
not
that is incident of
Spanning subgraph
Definition: A
spanning subgraph of
H
There
G
subgraph
a
a
of
of G of His called
provided V(G) U(H).
vertex
E(G e) E(()
graph is
edgeless
if
no
vertices
with a vertices.
graph
different subgraph.
called complete it
(b) How many induced
G have?
a
the vertices.
every pair
are
adjacent.
(a) How many
does
subgraphs
G
(32) 15
=
=
=
(b)2V(ku)
-
spanning subgraph
a
of a
connecting
weights of the included edges
all
cliques
-> the
is as
Definition: Let G be
a
->
graph.
1
=
20
-
1:
63
independent sets Definition:
set sis called
The sets is
Ex: Find
called
a
graph, $S-V(G).
clique of Gif every pair of distinct
independent
vertices in s
are
set it
no
pair of distinct
vertices ins
are
in G.
Cliques
o,
see
a
Let G be
in G
adjacent
-
-
including
of
=
adjacent
=
complete
a
does
A
each
some subset
(a)23(kx) 21532,768
v(( e) = v(()
-
a is
=
possible.
sedge deletion
of V,
graph is called
adjacent. A
are
spanning subgraphs
if etE(G), form m-eby
->
vertices
*notev(kx) bsc(kx)
spanning tree: Find
as
subsets
by
have?
the vertices such that the sum of the
small
a must be induced
2-1 nonempty
are
Question: suppose
=
Exa
minimal
subgraph of
complete sedgless graphs Definition:
of because
vertex that's not included
subgraph
A induced
calcliques of order
3:51,2,78, 82, 7,98, 61, 7,95, 51, 2,96, 83, 4,5
Ex.bcotafale
62.56,58,53,5,6,98.35,6,8,9
-
->
if VEV,
we can remove us
get a subgraph, but
we
also
have to remove
all
edges containing it:
v(G r) V(x)
=
-
-
EVE
E(G V) = E(G) BetE (G):vtee
-
-
(b) Cliques of order
(a) independent sets of order
↑
Ex:
draw G-807,88$G-7
7,05 is
*
·
the
edge joining the two vertices, not the
NoOne
induced
subgraph
nonempty
set
of
·agfo
subgraph
->
a
graph ALVCH)
of Hinduced by
A
is
graph w/overtices
is the graph
subgraphs? (b) How
sedges. (a) How
many
many subgraphs of are
subgraphs?
spanning subgraph of
"subsets
Gincludes all the vertices some of the
of edges which is the number of
spanning subgraphs.
->
a
=
spanning
the
clique
number
this number
E/H[A]) 85x, y8+ ECH): x, YEAE
is a
but
clique number, independence
Definition: Suppose that His
vertices. The
of Gare
Answer: (a) A
are 2
3,
-
question: suppose that G
induced
(b) each of
G 7
defined by: v(HCA)): A
subgraphs
set of vertices
HAT
The
we
these
contains
is one new one.
number
definition: Let
of his the
4:81, 3, 5, 65, 81, 5, 6, 8,
independent
83,6, 97
sets
of
cardinality
a be a
graph.
cardinality of the largest (lique. We
denote
by w/G).
independence
number of G is the size of the
denote this number
by a(k).
Extraisa
edges. There
4:12 or
W
is 4
largest independent
set.
complement Definition: Let
graph, the complement
a
connected to Definition: Let G=(V, E)
of G
connected or if there exists
graph defined by:v(t) v (G), E(G) Sxy: x,yeV(G), x #y, XY &E(G)*
is the
=
2 and
w
for
proposition
=
beagraph. Then c(k)= w(() w(G)
Let a
complements:
"Friends-Strangers Theorem":Let
48.13
w/l) 13
vertices. Then
or a
G be
graph
a
=
x
(6)
with at least b
(k) = 3.
on
c = ( V,E) be
graph. A
a
adjacent to
vertices with each
for all; svi-l-vifor each; from
* the
length
EX: Here
of
a
the next.
We
write w=
(vo,
...,
ve)
edges
walks
I
I
1w5-627
can
o
one
long
AAs
edge, it has infinitely
as
a
Definition: suppose that w. V0-v,....
=
is
has at least
graph
w,tw2
=
V0-v,-...
-
utov
Papath
let
in
a
a
a
the
Cyclic
graph
or
in G is
a
walk in which
(n,)-path is
graph, p
a
path
whose
edges
Definition: it 23,
=
=
an
no
vertex is
a
tob?
are
edge of
us
ends at
A
a
(n,wl-path
does not
repeat
any
vertices, spa path from utov.Thenpt
sr is
connected to
witwa is
a
w.
Let
(n, w)-walk. By
in G.
component of
class of the is
G
is a
subgraph of induced by
connected to relation
on
U(k).
a more than
we
are 15
graph in
are
are
paths
infinitely many
a
a
15
28
"I
paths
to b
⑧
path graph
on a
to b
verties:
Definition: Kn is the complete graph
on
excatly
one
Cutvertexscntedge
Definition: Let a be
is
a
a
graph.
cut vertex of Gif G-V has
called cutedge if G-e
has
more
more
components
components than
then G
5
·
b
star
·9
80
70
a
called connected if it has
⑨
there from
walks from
is
whose vertices component.
define the cyclic graph of ordern-> A vertex vis called
there from
from
graph
v
·b
graph Definition:
600, ,
=
.
.
.,
v(Pn) 81,2,
=
avertices:
.
.
.,
ny
E(Pn) 36i,i+ 15:11i n8
v/kn) 81,2,
=
=
.
.
.,
ns
E(kn) 8Si,j8:12ic;=n8
=
The star
graph
n= sedges 530, 18: 121nE
see 50:
:
path, it
a
v.
sovi, vae, sve, us?...SUn-1, Une
a-5-6-1)-b= 5x3
Definition: On is the
is
v
2, 3
repeated.
=
tob? (b) How many walks
are
99,0
vto u.
Lemma, there is
equivalence
01,..., n2sE(n) 331,28,32,38... Son-1, n, sn, 188-Anedgee
a
(b) There
avertices is the
on
Question: (a) How many paths
(a) The
88,7,6
us be connected
component Definition:
vesW2 w-...- WK
that starts at
does not traverse any
path graph
denoted (n.by: v((n)
answer:
2, 3, 4,
(n,v)-path, swea(v, wl-path. Then
connected Definition: A
graph:
2
(v wo)-w,v...-Wk
Su,,ve....,one, and
are
path from
W, be
Once
Path
77,8,6
=
definition: A path
path from
1, 2, 3, o
Transitive: suppose that his connected to v,
many walks.
walks with vt= No. The concatenation of W, Sw2 is the walk:
are
A
-
Let vEU. The walk
symmetric: Let
the
concatenation
1
8
vertices
repeat reversed
I walks can
EEo
66, 7,
4
nowhere
go
55
0,9
4, 1,
v.t
where
I
connected to
33,4,1,2
sequence(e)
traversed in the
->
->
06
sequence
in G.
0
o
Reflexive:
a
walks:
list (or finite
a
1.
1 to
walk is the number of
are some
walk in G is
that his
agraph. We say
(n,v)-path
a
·
I
dog
tinition: Let
be
vertex
EX:
49:
SeC
of
a be
Sn is the
graph wolvertices
G.
Definition: The complete bipartite graph
denoted Km,n, defined
Y
Definition: the
V(En)
=
501,2,
...,
1,
=
.
.
.,
are
Definition: if
n
3, we define
is
an
A
880, 18:12i
of
no
the
length at
=
on is the
graph wol
n}
leastthree in which the first
other vertices
are
repeated.
cyclic graph of order n, denoted (n
=
graph
a is
-
called acyclic if there
are no
cycles in G.
A
acyclic graph.
Definition:
A Tree is
connected,
a
cyclic graph.
Ex.
⑧
Definition.
leaf of
A
a
graph
Ex:agocoon
is a
vertex of
Vertices
depree 1.
1, 3, 718
are
leaves
is
Proposition:
a
Let T be
subgraph
*
a
tree with at least two vertices slet
vertex of 5. The T-v is
Definition: Let
a
a be
that is
subgrap
by:
E((n) 881,25,52,33,..., sn 1,n2, son, 123
=
Definition:
defined by:
T
a
a
v
be
forest.
agraph. A spanning tree of
his
aspanning
tree.
of G is
Cayley's formula: There
a
spanning tree if: T
are
contain
a
subdivision of 15
or
theorem: implies that
eg's
is a walk
=
Every planar map
no trees
on a
is a
can be
Kuratowski theorem: Let G beagraph,then G
0
the same, but
v((n) 3...,ny
avertices is denoted Ens
nsedges
Definition:A cycle
plast vertex
on
theorem:
colored with four or fewer
colors.
XUY
=
=
positive integer. The star graph
be a
80,
v
E(En)
nY
=
Apple-Haken
is
Elkm, n) 88Xi, 7;8:x>X, ytYz
SY1,..., YnE
=
<X,,... Xm5 v(km, n)
x=
edgeless graph
Definition: Let
vertices
by:
men vertices
on
tree, v(t): v(G)
distinctly labeled vertices
forest
clique
or
independent set.
K3,3
as
Gor it's
a
is
planar if
G
does not
subgraph.
complement G, must
have
a
large