.
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..
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is
wag
you
..
-
sm
M
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......
·
·
math
271
.
Thi
⑧
-
Dinh
SUMMER
NOTES
LECTURE
here.
.
..
.
.
.
IJan
~@°@°
.
is =π⑧⑦
⑤
'June 26
2023
To
Note: You
must
know
all definitions.
negation
Proofs
Cardinal
Proof:a n explanation
why
Statement:a
Example:A ll
that
in
or
You
false.
An
MATH.
These
combined to
be
can
make
are
new
a
statements
as
it
can
either
be true
or
Por
Q
i s true
than
statement.
To proof
a
Negation
false
Q
is
statement
P and Q
both
P and &
a re false
only
false
is
of P
P is
"If
P than
negation
0"
is true.
is
false.
kind
This
of
true
is true.
p
not
is
Q
true, but
when
proof
are
opposite
statement:the
of
a
Negation
when
true
both
false, the statement
P is
When
when
only
false only
Q
is
P
if
for all every, cach)
--
5:there
(is, are)
exists
Statement
Negation
not P
P and Q
Por Q
7
P,
Q
vietnamese
at
MATH
IP, Q
*
A
victnamese
a
is
feet
-xamples
of
There
exists
a
There
is
an
for
for
all
feet tall
Way
of
YP,
513,
than
Q
not
are
xandy,
number
integer
a
so
a
so
integerm,
and
rational
X
is
X
is
y
rational
an
so
integern
so
that
for
irrational
and
for all integers
that
there exists
that
is
is
an
m,
is
ntm
is
is
and
xty
irrational
intergermy
that
ntm
all
y
then
then
by
are
xty
irrational
or
xy
aven,
ntm
is odd
even
integern, ntm
is
odd.
prooving
what
you must
statement
if
P
real
a
if
andy,
x
integer
There exists
at MATH
a
victnamese
9
integer m, there
all
good
Nagation
real numbers
all
4, not
is not
All
tall
not Q
not
Q
A
vietname
good.
are
not a
or
P and
not
All
for
0
3
->
do to
prove it
Similar statements
Assumes P, Prove Q
Q
choose
a
P, Prova
P
Assume
suppose
Let
An
-
=
P and
XP,
false
statements
Q
All
true
false,
is
statement
a
we
prove
it's
trun
is
Six:
-
Suppose/start
with
to
want
what
you
prove.
true
either
is
good
are
true.
is
handsome
is
Statements
statement
a
sentence
vietnamese
Thi
P,
prove
↑
is
rational
is
called
"vacuous
truth"
I
know
integer
integer
n
itegers,
is even
n
is
real numbers
for
()
2k
n=
odd ()
n
2kH)
=
some
for
integer( x
some
integer
&x
is
odd because
-
5
2(-3)H
=
where
-
is
an
integer
Notation, Cab
Def
integer)
=
"all"means
b
divisible
is
alb
is
True
An
dividesb"
"a
reads
is even
n
412
2k
n=
=
214
true
012
false
210
June
multiple
divisor of 3)
a
of
integer
some
a
k
because
integer( x
some
not
=
an
4 2x2
integer false
I
where
=
is
an
integer
a to
because
2.0
and 0 2x0
where
=
0 is
an
integer
I
=
is
↑
4
statement
def
A
prime
def
A
composite
&
for
()
2 4xk
y2
214
a
is
false?
or
integer
a
b is
for
and ball
a to
a la
by
*
number
a
be
cannot
number
is
an
number
a number
larger
integer
is
an
is
even
than 1
and it's
than 1
integer larger
positive
and it's
divisors
not
a
are
1
and
itself
only
prime
Example
True
false?
or
For
integers
all
n,
if
n
then
it is
even
Ans:It is true
use
An
this
Way
of
integer
5P,
()
what
you must
than
YP,
is even
for
2k
n=
some
do to
prove it
integer ->def
prooving
statement
if
P
n
0
3
Q
->
Similar statements
Assumes P, Prove Q
Q
choose
a
P, Prova
P
Assume
All
suppose
Let
Proof
true
suppose that
Since
n
is
n
a
an
integer. Suppose
2kfor
=
some
Dink's
using
that
n
is
integer and
even
so
Bible
even
=
=2x2
↳
n
integer
↳n 2x
n
is
is even,
an
is
n
from def
+
of
even
2
in W2)
=
-
--
this is
even
2
as
number
12
n2= (2k)2=
2(2xY=2k
thus
no is
even
Example
&
True
for
Ans:It
An
integers
is
n
of
5P,
odd
no is
n
for
2kH)
=
integer
some
prooving
do to
what
you must
than
YP,
oddthen
is
i
odd ()
is
statement
if
P
if
true
is
integer
Way
own)
your
false?
or
all
Stry on
0
3
Q
->
prove it
Similar statements
Assumes P, Prove Q
Q
choose
a
is
integer
n
is
odd
P, Prova
a
P
2k + 1
n
=
n2
and
such
(2k 1)2
then
that
meal
either
and
+
=
n2 4k2 4k
=
n2
and
itasfar
and
in
1
+ 1
+
2(2k 2k)+)
+
=
-----
↳)
some
integer
n2 2k +
=
↳ prooved
is
n
Is o
n2=
2(2k2+2K)+)
&
True
for
negative
Proof:There
A
prime
Chat
GPI
choose
numbers
a
number
=
to
4). Then
a
=
is
41(H1 + 1 + 1) 41x43,
=
Thus, nothH
or
integer. Thus
an
is
integer,
1 + 1)
that
n2+nthl
so
than 1
larger
is
a
41x43. So
both
=
not
positive
integer
odd
he is
divisors
are
1
and
itself
so
is
n
and n2+n+4)
rith+hl
has
41
as
a
divides
41x43=1763
4171763
412+41+ 4)
=
and 1241<n2+n+1
divisor
prime.
an
integer
me
and
integer,
ntm=
ntn
ntur
im
:=n t n =an
so
that
ntm
Way
an
2n
=
is even
Choose
where
m n.
=
n
is
Then
an
m
is
an
of
is even
prooving
what
you must
statement
do to
integer,
if
P
YP,
than
Q
0
3
->
Similar statements
Assumes P, Prove Q
9
2k
only
than either
haha
be prime
21 and 43
true
->
=
some
composite
and it's
equation with y interceptc an
non-negative integer
so
is not
=
m +n
for
false?
Proof:Suppose
m R
odd, n= 2kH
negation
For all integers n, there exists
Ans:It
is
is
n
and
a1
Dind's Bible
+
=
it's
integer
quadratic
a
refer
an
21 +4174) 41(41
&
True
since
is prime.
non-negative integer
is
told me,
n 41
n
(2K22K)
where
as n2th+4)
We prove
exists
&>
choose
odd,
false?
or
all
an
Answer: It
i s false.
def
suppose his
integer,
suppose
5P,
Q
choose
a
P, Prova
P
prove it
or
43 =1763
ask andki
or
such
a h
=
that
n = alx
and
k
=
1
Proof:Suppose
is
Ans:It
is
n
integer,
an
choose
m
(m+r 1)
6 -n
=
=
true
-
ntm n + 6
=
True
n
2(3) 2(k)
=
3
where
=
is
integer, so
an
ntm
is even
false? (false)
or
There exists
Negation for
-
6
=
integer
an
integers
all
m
is
an
n
is
an
m
thatfor all integern,
so
m, there exists
an
ntm
that
so
integern,
is
even
that
h i m is
so
odd
integer
integer
m
+3 n+1
2m
1
+
2(2m) +1
An
integer
ProofSuppose
def
odd ()
is
n
n
is
he
an
"FA
12
if
I
The
does not
converse
Contrapositive
for
asand C,
all integers
bjc
an
integer
and
ntm-4m+1
where
in
is
an
integer, so
-
"If
of
as
as
original statement
p
a
then not
not
truth values
same
truth values
same
if all
then
and als
original statements
Infact, they
are
logically
equivalent
al
Way
b at
c as
statement
=
=
Some
of
if
P
integer
attas
YP,
a(t t s)
=
=
some integer
->
=
-
and alb-c
5P,
a(t s)
as
=
-
Q
prooving
do to
what
you must
than
t
b -c at
albtc
integers
are
alb
b+c
is
n
"V A, if
Q
then "
is
have the
have
always
=
"If Q
them P"
PthenQ"
always
Then
n 3m+1.
false?
or
a
integer,
prooved
2x+ 1
integer
choose
of "If P then Q"is
The contrapositive
True
some
PthenQ"is
of "If
converse
for
2kH)
=
0
3
->
prove it
Similar statements
Assumes P, Prove Q
Q
choose
a
P, Prova
P
prooved
True
for
all
is the
for
false?
or
asande,if all
integers,
asand C,
integers,
a
bc
a
lbtc
alb
b -c
=
2 al
=
b
albtc
and alb-c
then
all
and
a c
-
c
as
=
b
aL b
+
=
if
int
are
b+c aL
-
aldtc and all-
this true?
of
converse
all
and ale then
-Ease
as
=
for
2b a(Lts)
odd integers
we
=
b
=
false
I
L
odd integers
for
we
choose
0,5 1
=
b
ia
=
->not
a 2
12)
an
=
b
integer I
-
=
a
decrease
=
may of
false
c
a
=
-
=
)a
2
=
y
a
decreasing
mag ot b
false
I
want
and
a
b
to
such
integer
that
so
we
Is=false
choose
a
to
be
even
atm
is
odd
We
for
choose
example,
Then
negation:"There
it's
prove
a,bic
are
btc
a
=
albtc
c
that
albtc
and
alb-e,
but
a
by
oral.
1
=-
True
a.0
True
-
a.
=
als
and
allow
and
and
1
=
=
=
=
so
as
and
integers
2.0
0
=
B -c 2 2.1
Thus,
2, b
=
bands
integers
are
2 XI
because
Example.
For
all
This statement
is true
The
"For
is
converse
the
is:
and
all
proof
is
then
aldbtc
and al3btze
easy.
asand C,
integers
and ale
if
and alsbtze
aldbtc
them all
and
ak"
true?
converse
Ans. The
all
and C, if
as 6
integers
true.
is
converse
Working
alabtc
3b 2c
+
=
b 2B
21
-
B
=
3D
=
-
=
=
B bax
=
i at
=
+
2b +c B
c
Y ak
al3b 2
-at
a(2k
=
-
t)
bat-bak aldt -3k)
=
=
↳I
alb
al
Proof: let
asbyc
are
integers.
Theb 2(2b+c)
=
=Gat-bak=a (2t-3k)
Thus
Def
I
all
and
rational
where
-
Suppose
and
alabic
(3b 2) dak
+
=
2k-t
and
2t-34a re
number:A
number
that
can be made
as
examples
number:An
irrational
terminating
decimal
of
( 2(3b 2c)
=
+
3(2b
some
c)
+
irrational
numbers
number
is
as
fraction of two integer
b 0
3
or
a(2k-t) and
=
-
als
-
irrational
-at
for
and 3b+2c-at
26+a= am
integers.
rational-a
Def
al3btes; thatis
a
real
number
252/pi/culers
that
cannot
be
expressed
as
fraction
of
two integer
integers
I
and
t
Quotient
integers
and
g
called
a is
that
a
when
quotient
I
for
Theorem
so
r
the
I
11
~
Remainder
-
remainder
and
gdtr
=
divded
(L
26
I
02
a
and
a
aso,
where
there
exists
a
unique
v2d
by ds
-
is
a
l
integers
any
I1
4
-
d
the remainder
a
mod
a
dird:the
a
20d
=
20
3g 6r
Proof thatfor
el
an
in
floors
let
so
is
we
x3
and
0213
2 cases.
have
with
is
for
and
integer
some
with
nen+1)
=
ax(n+1)
=
where
indirect proof
kinti)
even.
and nth n(n+1)
integer
some
integer,
an
with
so
n(2k+ 2)
=
=
is
22()
=
even
Ceilings.
and
be
real number
a
LX) the
=
x7
(-7)
=
+1
is 2
3
is even
integer. Then
a
for
n=2kH
x
20
0223
odd
is
u(KA)
n
n, nt
-
by
and
even
integer,
Caseh
is
n
max
-
is
because
all integers
Proof:Suppose
20=6x4+ 2
divides
20
when
Z mod3=1
caselh is
=
because
=
cased
2
=
mode I
by do
divided
is
a
by d
divided
is
a
quotientwhen
=
the remainder
-
when
=
the
=
floor of
ceiling
X
of
X
Definitions
(x) n
a
=
Mx7
E)
=
a
is
is
an
an
and n=x<n+1
integer
integer
and
a-txn
Ex
(201)
=
L-201]
True
or
=-3
because
numbers
the statement is
Negation There
I is
because
-
an
is
integer
an
and
integer
22012H1
and
1
-
-
2012
-
2
=
-
3H
false?
for all real
Solution
2
exists
a
false,
real
x,(x) (Y)
=
we
prove
number,
it's
such
negation
that
LxJF(x]
-
Proof
choose
211
and
Solution
Negation
I. Then
X
23
the
for
all
the
statement is false.
there
exists
choose
numbers
real
x
real
a
1 is
=3C2,
an
integer. Than
Lx2)=14):a
because
I
is
an
integer
2x4 2 +1 2x6.
=
and
x
1
because
(x)=LE1=1
=
=
y, Lxty
and
x
numbers
LyJ
+
=
prove that
to
Easy
2x6
3
that
Lxty/f(x)+Ly)
such
y,
yE
=
=
=
42 3) (3)
3
=
=
+
~
(x) +
Lyb:(16
Proofchoose
and
3
x
=
31324. Then
(x)
and
111.322,
There
exists
(1) 2
+
=
y
=
=
21.5)
A
=
& is
1
131
is
an
1
+
=
that
so
Lyty)
where
(x6+Lyb
hence
real number
a
Then
3.
3
where 3
2.
(xd
Thus,
=
is
+
hy)
integer and
an
121.5<2) Ly)
integer and
that
so
x
=
21.5)
=
=
1
where
I is
an
integer
2 x + y) 2 +3
=
=
(x2) (x)2
=
Ans:True
ol
choose
LOS
L0012
=
-
20001)
2
0
Proof
L001
=
0
0 0
=
2x12 2001
There
exists
Ex
Proof
prove
we
likely
use
or
when
we
Prove
Suppose
this
a
is
an
that
so
we
kind of
fail
that
to
and
integer
an
an
and
integer
0
0
<0.0kR
=0.1). Thus
(x2)
=
Lx)
an
integers
it
this
a
contradiction
=
x
(27) and 2x4 2x12
=
and then
conclude p.
-
what
we
prove
when
means
"
"
not
directly.
for all integers
is
proved
0 0
=
fin
is get
suppose not
proof
prove
integer. Suppose
n
not
is
x
in,
We
even.
odd. Since
if
it is
n
is
even
them
o n is
that
prove
-
Suppose
is
O
is
o
by contradictions
contradiction,
by
because
0
=
because
=
True
real number
a
Indirect
proof (proof
P
=
82
false?
or
Loo 20001) 0
=
=
=
True
To
(x2)
x=001, then
choose
then
=
=
even
odd. Since
n
a
is
even-
by contradiction
is
odd, n 2kH. for
=
-
-
some
is
integer
an
1. Then
integer
thus
even
er
(2k+)2 4k2
=
his
CIn
=
odd
which
2(2k27av)H
+4kH
where
=
contradicts
the
assumption
that
222
it
+2*
is
even,therefore,
N
is od&
=>
Proof (by contradictions
(TA)
Suppose
there
exists
an
integer
a
so
it
that
is
even
and
i
is
odd.
n
is
by
if
31m...--
then
it
falsesiose netbut
he
n 3t
a
suppose there exist
=
~
-
-
-
r 3t
S
=
n
3t
-
-
=
35 2
or
+
where
3 An
) 3t
3
(x
+
=
3924733
+
3ab2
a
(3x)3
211x3
+
+
7
ak
=-
- -
3
n
272+1
C
31r
=
C
contradiction
30
Thus
3
Definitions: A
real
-
x
where
=
A
real
of the
Ex:SI
r
->
wir
even
I
->
Thus
in
221
2x2
II
!
have
no
common
m
is
t
is
r2
is
have
even
dix, kE2 4x2
-
=
=
use
prive
factors.
da
=
n
won't
2/nts
n2
is
rational.
rational
is
m,
in
not
proof (by contradiction]
I
a,m,at
=
is
ESX
numbers:
=
2m
no,
proof
of
the
Suppose
12
and
irrational.
is
Sketch
irrational
rational Es
2-we
irrational
of
Sketch
is
integers
a re
is
x
->
X
number
on
number
examples
+
+
-
-
3
3(3x)(1)
3(3k)1
+
an
=
even
irrational
->
mis even
n
in tager
=3kH
1
case
3Yt!
n
common
factor
of 2
is
an
them
Ex
is
irrational
is
proof (by contradiction)
Sketch of the
suppose is
53
rational
is
,nE2,
a
=
n
myn have
mtm
3
=
31m2
o
no
prime factors.
common
3r
=
31m
->
have
m
bkx,kx =129k)
=
factor of
common
3
3r)
=
-
n2
31n2+
3x2-
=
Thus
II
3 In
irrational
is
↳ How did you
conclude?
-
I A
for is,
we
use
Suppose
a
Fuclidean Algorith:
-
is
all
an
then
integer,
3:kan
is
the
31n"
integers n, ifbind
Ei
then
gcd(6,9)
31n2
for
that
Homework: Prove
Proof
"if
where
that
compute
algorithm
suppose
n
then
3
is
31m
kn2
=
some
integer
gad(asb)
3
=
q2d(6,0) 6
=
Divisors
6
of
Divisors
0
of
gcd(0,0)
are
are
no
=
11, 12, 13, 16
all
common
non-zero
2cd:Definition:Suppose
1)
2)
integers
divisor.
asbe
both
so that
n ot
and glb, and
gla
For
any
Algorith
de2,it dla and dlb
uses
following
repeatedly
lemmon
lamuna:if
a
gbtr
=
then
gcd(a,b) gcd(b,)
=
them
dig
a
and b
are
zeros.
Then
ged(a,b)
is
an
integer
g
so
that
Ex:find god (32)123)
using Euclidean Algorithm.
=
+
21gcd(22,21)
48 27(1)
=
+
+
=
6(3)
3gad(6,3)
+
=
6 3(2)
0g2d(3,0)
+
=
↳
use
R
123
YS
48
->
6
Thus
-
->
47
=
18
-
y
47
+
=
+42(123) 3.
=
cuclidean
the
Use
to
13
18
(8(32))
S
8
3
x
method.
3
D
5
-
find integers
a
344
=
344
235
109
and
x
y
3 235
=
25(1) +09
=
1092)
=
so
compute
that
god
largar
alb
-
H7
i 3(2)
((3)
=
+
1
+
70.9cdC(,0]
qcd (1,0) gcd(344,23S)
=
=I
(344,235)
(344,235)
q(d(235,109)
(7(6) + 7
=
god
smaller
=
=
3
to
algorithm
7(2) 3
17
v
andy
x
2
=
2
->
integer
N
n
27
the table
some
⑧
⑧
21
for
=
can
zero.
3
gad(ab) axtby
we
3
=
remainder is
when
=
=
3d)
->
stop
We
This gad (32) (23)
3
gcd(2),6)
6
21 21))
Theorem
g2d(48,27)
27
73 48(1)
or
gad (TS,48)
75(1) 48
+
=
=
21
gcd (123,75)
+ 25
321 2x123
123
and
use
n 2
23sy
=344x
NO
344
2350
109
17
↑
M
-
2
#
->
2
13
3
-
A
69
69
=
y c101
=
3
=
19
28
=
x
that
+
+01
+
lemma:asbigs of
a
if
,
then
bgtr
=
Proof:Suppose
a
b oth
not
g2
best
It
is
ga
easy
r
r
is
gcd(b,r)
g1292,
prove
integers. Suppose
easy
exists.
by
proving
that
gally
to
prove
Let
both
not
a
and
b
are
eros,
not
by contradiction that
god (asb)
both
exists.
b and
r
g2 g(d(bs)
=
gi is
and
common
divisor
of
3 and
r)
galv.
gcd(a,b)
=
and
gala
r
to
are
It
=
prove
cros
gad(as) gcd(b,5)
go=ged(ab).
firstwe
a re
=
let
exists. Let
and b
a
-
q
=
gilb
ba
git- glIg
=
gx(t 48)
-
=
Next,
proof
we
prove
gaegy
by proving
go is
a
common
divisor
a and 3,
of
and this
is
easy.
Zeros,
so
ged(b,r)
types
Three
proof Direct proof,
of
indirect
proof, induction
induction:
iP
2 3
i
+az
a1
=
ai=a
22
Ei
2 +2
4
Principle
let
①
②
a3
+
ay
s
+
+
at
+
+
+
32+ 4 s
+
+
+
(a (2)
induction
of
+
+
2 2 2
+
=
=
4 5
+
+
=
P(n) be
a
where
sentines
n
is
integer
an
P(a),
For
all
P(r)
then
-
Ka,
integers
for
1,0,1,2,
all
if P(X) then D(k+1)
integers
nc
a
5,5,5, 2,8,,xo,
Template
To
prove
Base
1)
P(n)
for
case: Prove
2)inductive
all integers
is
a
we
prove.
P(a)
step: Suppose
KI
a
is
an
integer
(IH)
Suppose PCK)
we
want
4(k+)
prore
Thus 6 P(r) for
Ex
P(K+1)
to prove
Prove that
all
integers
for
nent's
=
12a
all
a
integers
Solution:
:Ii
Basccase:
1
=
z 4 4)
=
=
=
Inductive Step:let K:1
Eni
we
want
to
be
an
integer. Suppose
k(IN)
=
(IA)
2
2)
-i (1
x
=
prove
2
ii (i)
=
1x1
+
|x(kx 1)
((+)
+
+
by
(IA)
2
(x(1+71)
+
=
2((x+1)
2
2)
(2x+
2
I
Thus
Ex
Prove
for
=m)
that
as+2n
is
divisible
by
all
for
3
integers
all
(
rel.
n=
0
integers
by
induction
Solution:
Base
=
Inductive
step:
want
we
6
case:( n 0)
-
to
3x0
0
=
divisible
is
is
8
an
integer.
(IH)
by 3
divisible
(k+1)+ 2(KH)
prove
where
=
(x+2Ki s
Suppose
(x+
2k 3t
2x0
+
by 3.
0
=
Now,
+1)"
(kx
2(k+1) (x3 3(2
+
3k
+
+
=
=x3 + 3k
2
1 2(k+)
+
+
2
3k+ +2k +
+
(x 2k 31x2
-
3k2
+
3t
=
b(t xx2
=
where
Thus,
Ex
n8+2n
Prove that
Base
+
(t+k+K+1)
is
is
an
divisible
5" 4ns1
case: P(0)
is
kx
3
divisible
be
want
Now
an
1),
(k+*+2(kH)
so
all
by
1
=
integers
3
-
-
is
for all integers
so
1
where
8(0)
0
=
=
IY-4(-)
5**! U(K+)-1
4k-4-1
5(54)
-
=
8(5t
where
integer
thus
integer.
divisible
4k
5(8t)
is some
an
by 8 (IH)
by
5* -4k-1 bt
=
8
4-5
=
=
is
is divisible
=
5(5*
O
is
=
5th
divisible by 3
480
integer. Suppose
to prove
5(54)
3
+
+
for
50-4(0)-1
k2O
Inductive step:let
3k
integer,
by
+
+
+
=
we
3k 3
+
+
+
-
1)
-
16k
+ 16k
2)
+
5"*- 4(xH)-1
is
divisible
by
8
for
all integers
n 0.
Prove
is it
Solution
Base
induction
by
on
for
him
mining
that
in
all integers
Basice
na
case
2
E=
Jn+1
52 1
+
=
E
>
52
52
↳
-
52
↳it
because
CLI
2
to
Since
211
then
2
+C
2
I
I
I
prooved
KId
Inductive step:let
be
an
k
integer.
Suppose.I
N
,
ist
it
/x+
it
in
It
Thinking.-
prove:
Ei= if t
*
I
>xxx
see
1
by
+
fix+
then
+
e
sixis
x
!
x
-
Thus,
1
0
for all integers
+
2x
1
+
1
+
Tx+
n72
+
-
Since
Ext1)
(
Fix
Il
Proof
by
tan
n
let
divisible
is
be
a
-(r 0)
by
for
3
all
integers
no
integer.
an
3q
a
=
example
cases
=
n3+2n
(39) 2(39) 4g3 6g
=
+
=
=
+
(r 1)n 3g 1
case 2
=
=
+
3
. . . .
Case
3(3a2 3a)
+
3(r 2)
a
=
3
+
3g + 2
=
- -
n
4
3g
=
n
ifa 1
3
+
=
24ifg 1
3a +)
=
+
=
3a 2 35ifg 1
n
=
-
+
This
take
&
all
integers
all cages
covers
Strong
if
=
...
Principle
Induction
PCa), P(at), P(b)
nz
arb,
where
for all ixcb, if 4(m)
and &
for
all
integersm
a
Template
Basecasa (n a,
=
Prove
.
.
.
.
b)
P(a), PCati), ...,P(3]
Inductive step: let
we
to
want
kLb
be
an
integer. Suppose P(m)
for
all
integers
D(K)
prove
Sprove PC17]
Thus,
Ex
P(n) for
all
intergers
R29
be
let
a
0,al92,......
a0=12-a1=29
by
ak
5ax-1
-
=
Base
-
for
and
sequence
(by
cases
2
for
strong induction
(n 0,7)
on
of
=
B
Next
0.16213
90
an
12
=
29
=
3x30
=
7x2
+
5x3'+ 7
=
defined
integers k22
that
a n =5x34+ Tx24
Prove
Proof
bax
the
x
27
all
integers
n20
on
where
a
zmck
(IH)
where
a <mcK
then
P(KA)
than P(n)
for
Now,
definition
by the
12
since
of
the sequences
Jax-1-bak-2
ar=
SISx3+ 2x2
*
25x3*
+
")-6(5x** =wx2")
35x24
=
(3")
-
+
13)
3
Thus,
1)
-
1 )a"
(7)2"
+
5x3"+7x2"
prove
solution
(n
s
1
=
3
2
=
3x) 2x)
=
where
0
-
co
1, 4
=
8
=
and for
120
2=
2-1)0
be
an
-
integer. Suppose
CMCK
that
(=
3x2*- 2()*
K22, by definition
2k (k -) + 2ck
of
sequences
2
+22k 2
-
= 2
(x 3x2*
-
=
M
(x
k
prove
Now, since
=
-
=
lat
to
want
3x20
=
3x2"
2
step
m) Fol
=
putting
n)
=
+
=
(k ( k -)
on
for all integers
2(-1)"
=
by
=
=
inductive
recursively
22n-2
3x2
=
defined
(n 0,1)
8 6
we
+
1
2
n =
the
scavence
(strong induction
=
=
(n
=
that
(n
Base cases
20
for all integers
be
Co,Cl, C2
n=2
2)2
-
an=
Let
2
+
5(3*)
x1
3
0)3"+
15
-
- 2*
(2")
-
IoH
by
42x2
-
-
+
1 0)3"
Ex
30.3*
+
3x3
=
2
*
1)
*
* 2(3x2*2 1"-)
-
*
-26
-
4-l)"
(m
3
=
x2-2L-1)"
for
all
integers
integers
A2L-1"
x 3(2)
=
3(2)"
aix
=
4"
-
x2"-2(-)*
3
an
-
2) 1
=
=
Thus,
4
-
for
all
integers
nzc
Sets
things
elements
Set collection of
=
of the
set
(members)
"5"
How
do
"means" is
describe
we
Ans: Justfell
the
Using
A
Ex
B
12
set?
a
what
are it's
us
"
21 is
elements.
of
the set
all
integers.
curly brackets [3
[1,2,3, Dani] by listing
2
IR
of
element
=
=
↑
is
an
2(x2 x >103.
=
x
+
rational
set
of all
the
the set
of
all
is
Complex
=
numbers.
real
numbers
numbers
subsets
"I
means"
is
Edition: A,B
a
of
subset
sets
are
A
A C BL=) if
x=
then
x =
B
(fx =A,x B)
z
%=the empty set
let B [1,2,33
set
that
has
-
elements.
no
=
all subsets
list
they
①
P, [13,[23, 233, (123, (1,33,22,3)
a re
B E)
if
Definitions
0
than
x=
A
This set
(1,2,3]
set
31,23
Chus):If
it has
Hlef: the
has
Afictions
def
8
4
on
subsets. They
subsets.
than
elements
an
A,
of
sets:U,
I,
X =
B
X
ARB
=
2x 1x A
=
=
2x
=
A lB
3x
=
/
1
or
XGA
x
EA
B3
x =
and
and
x=
B3
B3
x =
are
0, 913,[23
and
91,23
sat
of all subsets
of A.
a"subsets.
by PCA), is
A,B sets
A
UB
are
and 21,2,3)
They
A
has
denoted
Al BES Fx =A
,
operation
21,2133
B
31,23,22,3391,33
has
set
power
xt
and
BEVxEA, XE B.
I
8,313,323,333)
This
of
B.
the
AUB
E)
AnB
=
E)
x =
x
A-B
CE
we
X= A
E)
sets
see
XEA
o r xeB
and
xt
A
and
X =
using
venn
B
xAB
Diagram
IIIIII
E
A
AUB
AnB
It
i
An
(AUBJ nC
B
A
da
Me
B
A
t
Mugsee
B
+
-
c
we
C
2
Au(Bre)
W
A 1B
-
A
B
-
c
At
CA-B)
-
C
C
Jrue
for
or
false?
all sets
This statement
"There
for
is
false.
It
We
prove its
sets A,B and so that
example, choose
Since
So
are
A, B and C, AUCBrc)
A,
IDCAVB)
so
A
213
=
I
B
c
=
Therefore
point
negation.
A
CAUB(nC
0
=
but
10
Au(BRC)
M
-
have
Au(Brc)
IEAUCBUC),
RC.
(AVB) MC.
I
=
c
1CAUBSMC-
Aure)
Merce
u
We
B
62AUB) nC
True
false?
or
For
A, B and C, ifA BUC
all sets
A-B
then
EC
W
I
IIIIImin
-
got
A BCC
-
Se
statement
A
C
BUC
true
is
suppose
6D
+
-
A and
*B
X =
-
X
GB
W
XC B
A B
t
direc
e
C
This
B
A
True
exist
x
I
B
A
x
XEC
or
F
-
G
x
Proof
C
letA,B,C
be
X= A
and 2 *B.
Thus,
sets
so
Since
that
ALBUC.
and
XEA
We
ACBUg
to
want
We
know
show
that
A-BCC. let
XEBorXEC,
but
xAB
A-BCC.
True
or
false?
for
all
sets
A,Band C, ifAs Buc and
BI CUA
then
-
-Next*A_Buc
>
A
/
↑
1
-
·
choose
and a
cun
R
e
point
2
c
B
A
[a3
=
=
[3
E3
=
C
x =
CCAUB
A-B;ie,
so
xEC.
for
then
02
=
True
for
false?
sets, A, B
or
all
↑
but
[13.
=
CUA,
I AUB
=
=
CAUB
but
=
C, if Al(BUC) 0
and
C
=
=
and B 0
Buc
ICC
because
[3, B 33 and
A
chose
example
A
thn
=
AIC =
B.
4
A((BRC) 4
Sketch
=
↳
=
x
AK
and
X E A
4C
x
9
This
is No
there
means
#
Me)
&
x
=
A
B
Since
I
L
⑦
A =2
0
=
&
that
means
A
T
-
and
cB
x
o
x
BUC
B
or
x*C
B
=
x
C
always nullset
is
a
subset
every set.
of
: Suppose
A,Band),
thatA/C=B. Suppose
Since
-
XE
contradicts
which
June
we
such that
AlCBRC)
sets
XEAK. We
know
XEA.
the assumption
that
want
Since
to prove
XEB,
ALKBUC) 0.
=
sets, A, B
and
2 C
C, ifA/B,
then
-
B
A
XE Brco
Thus, XEB
A
-
Ak
Since
and
x=A. and
Suppose
prove
XEB.
XBnc,
x=
ALNC)
AKCB.
B.
BEC
Truc
e
C
I contradiction
C
AlB=
suppose AlC
=
I
and
XEA
&X G A
and
-
C
x
x
.
B
A
-
-
x
-
↳
X
by contradiction.
-
I
f
55
want to
v
I
-O
XEB
0. We
=
false?
or
for all
Alc,
are
G
B
I
-
↑
↑
As B and
roof:Suppose
prove
and
Since XEA
thus
True
AUB
x4C.
Since
XEAK,
XEAIB
Since
therefore,
that
Alc-B.
So
x=B.
know that
we
and
we
A
or
x
XE
or
B
and xEC
xeBre
y
=
[BrC)
(AUB)nC,
know
we
ye
AUB
and
y
= C.
yEAUB,
Since
know
we
yGA
or
y EB.
cases,
ye
ye
A.
B.
Then
y=AU
(BUC)
ytBnc
yeB
yeswe get
(AuB(nc (AU(Bre)
Since
Thus,
and
and
for
so
ye AU (BUC).
-
E(2,33
A
all
A,B,
sets
(and
=
B
D
2
x1(x)
(ADB) xCCID)
=
D
[2,33
=
24,5,6)
=
[4193
↳14↳(5),(ss)
=
↑
↓
!
"
·
W Es
B
-
A
-
A
B
we
true
is
)
<I
(A-
L
-
D
-
·
&
2
>13
3
S
=>
sats
(Ax) ((BxD)
I'
true
Ax2
&
↳
&
2.
2(x)
④
8
·
⑤
(2x)
8
I
⑧
⑳
⑧
~
&(2)
biss
L
4*s
C
There exists
choose
Not
Y
A,B, C
and D
2x6)
-
(
-
6Ax(BXD
such that
(A1B) x(C(D)
B
LBXD=
C]
=
4). CAB ( xCCID)
but
(1,2)
=
CAXC) (KBXD)
x=AK
gat
AlB2C,
A/C
proved
CB
↓
=
Au
ProofSince
casel:
Suppose
X-A
and
OC
I
cased
we
want
to prove
=
AB and C, (AUB(nc Au(Brc)
statemement
This
2
sets
all
Suppose XCB.
that XeAIB.
know
contradiction:XtC and
false?
or
for
a
that
ABC. We
such
by contradiction.
XB;
have
we
sat
are
that
XeB
we
XE
I
Therefore
we
have
x4C
x
=
Counting
When
objects",
count
we
A
recipe looks
of
to
way
a
which
them,
make
is
called
a
recipe.
like:
StepDo
Step 2
think
we
this
,
DO
done
be
can
in
ways.
me
n2
3)
~
"
Step 1
Do
A
recipe makes
A
A
recipe
a
when
Ex
How
Ans
9x10 xoxo
Ex
goods
is
digit
I digit
-2
Choose
2nd digit =
loways
How
to do
object
to
the
get
make
number of
want.
what
we
a
does
step
in
exactly
the objects
objects
is
depend
not
one
(using
nixhax...*nk.
there?
are
4 ways
1
(" digit [ loways
positive
are
there?
4-digitintegers
are
digit
4
number
even
x10x10x5
How
many
even
positive
there so that
digitsare distinct?
the
X
It
-
-
5
=
0,2,4,6,8
if
"O"
I
choose
W
-
then
for
my
not
So
divide
we
choose
4th
Choose
st
choose
and
choose
god
1x9x8x2/4x8x0x7
Ans
1x9x8x7
+
4x8x8x7
se
here
options
if
them I
not
1
10 options
not
a
I
have
17
over
for
's".
my
worka s
number
independent.
recipe
into
Hence
two
of
have
recipe
steps
on
previous steps.
way.
the recipes
and digits loways
choose
many
we
ways
numbers
choose
4
of
↑
every
I
ps choose
Ans
makes
and
works
positive 4
many
number
works, the number of ways
recipe
a
end
at
the
the
(i)
iso it
recipe
When
if
sense
works
recipe
Ex
&2
*
does
are
is
n ixnax...... nk.
Ex)
How
2"because
:
A step
-
-
in
in
of
choose
How
many
nx(n-1) x
Arrange
ways
we
a choices (chose
have
it
or
not)
objects.
a
objects.
can
we
of
it
arrange
objects?
a
.x(n (x+)
-
...
..
3 of
8
many ways
let
() be
Ans]
set,
objects
8x7x6
As
find
we
have?
elements
a
be:
can
of
a
k
setof
a
for each elementof the
recipe
a
Arrange
How
does
subsets
many
a
can
#
the
we
of
choose
of
to choose
ways
for (2)
formula
a
objects?
I
objects
from
the
by counting
a
objects?
number of
ways
to
arrange
in of
in
objects.
a(n-1).... (n kH)
-
differently,
Now, count
I of
((n=)
objects
stepl
choose
Step
Arrange
those chosen
the
(*) !
an
k objects
(k!
answer:
thus:(n)k!
n(n-1)
(k) h(n-is
so
(n -k + 1)
...
=
(n -k+1)
....
=
k!
(9)
letA
How
98x
=
84
=
31,2,3,4,5,6,7,8,9]
=
many
subsetxof
↑x8x7
A
that(x)
so
C
|x(2)
3
and
3 eX
S25,23
->
1x
8x1
=
number
·2
↳
number
i
can
of
ways
2
-
of
ways
choose 3
t hence
i
can
you
element
choose next
can
make
than
this
once
setm ore
not
a
good
recipe.
Counting:Recipe,makes
A
step
in
a
can be:
vecipe
kof
arrange
sense, works, good.
objects
n
->
A))
ways.
1x terms
(2)
How
is
atleast
one
are
there
I?
using
the
digit1,2,3,4,5,6,7,8,9
9.x.9x9x9-8x8x8x8
Ans
wrong popular
1:choose
2:fill
!11
so
not
exactly
4x8
to
onetoslot
0
st
put
·He
WI
11
fill
4:
Ax9xax9
answer:-
A
3. Fill
Ans
-
4-digitpositive integers
many
there
ways.
x)!
meness......
k!
-
that
(*)
-
=
Example:
so
objects
of
I
choose
'I
you can
a
one
good
11
/I
make
1
in
digl t 9
9
·
a
I
11
CL
"9:
this
multiple ways
recipe.
exactly
twol's
exactly
(24).+(8)
three is
8x ())
+
exactly
x
4
four
Is.
4-digit positive integers
and
at least
one 1
atleast.
a
sample:How
that
there
Answer:
-
what
many
is
94th
One
using
the
digits.
1,2,3,4,5,6,7,859
2?
don't
want
is
don't. No
we
there
are
I's
No
or
⑧
D
-
A
NO
(AUB1=(Al+IB)-
want
84
for
City
the
total,
84
+
do
the
-
not
of
set
lANBD
44
9th
have
Whatw e
digit.
LetAbe
-
we
2's
No
Is
B
such
want
integers
as
thathave
integers
no
a
I's
choices
or
for
each
of
the
nod's.
is
no
B
25
what
we
/A1
have
thathave
intgers
is
we
don'twant
8as
=
we
&B1 8411
12
=
PARB1=14
have
8
AUB
is
for
choices
11
each
12
21
of the
13
1
4
digits.
111
ALBY-By-
Now, AAUBI
DAUBrc1 DAD
=
+
DB1+1CD
-
DARBD-1Anc-1BreD+DAninc
B
A
A
M
↑
·
A
A
C
↑ A
UBUCUDD=
-
+
-
DAD+ ABD
ADCD
6
doubles
4
tripples
the
AADA
quadrable
so
Pascal
identity
(ii) (e) +(+)
=
Got(algebraic
(i)
+
(in)
-
Easy
is in nek-ll!
-
!
-
-
I
-
(x!
((x
1)!(n x)!
=
+
-
-(i)
Binomial
xty)
idea
Theorem
(i) inyo
=
of the
proof:just
do
it.
(xty)=a()
.....(xty)
n
times.
functions
from
Def:A function
every
A
XEA,
domain
:
=co
B
there
exactly
is
A
one
to
a
yet
satis
is
a
subset
oft
of
AxB
so
that
for
(xy)ff.
that
so
of
f.
domain
of
=
(xy)=f
set
a
f.
y f(x)
(=
underf.
the image of
=
=
"f:AtB"means f
is
a
function from
A
to
B".
1:IR - IR
f(x) x2
all XCIR
for
=
V
This
f
means
3xx,x)1x=IR]
=
which
&
A 31,2,3,43
Let
and B
=
t
is
EC1,2)3
=
[((2),(2,2) (3,4), (4)6))
[(1,4), (2) (, (3, 7, (4,
a
=
-
a
=
Recipe choose
functions
the
image
Choose
the image
choose
the image
choose
the
image
of
of
are
which
there
17
2
+
31
of
of
a
no
may
from
be
not
what
is
a
the
image
sat
at
A to B
3choices
43
choices.
so
from
A to
that
(sy) It.
Be Ans:Yes.
1)
3 choices
3 choices
IRXIR.
B?
yet
function
set
f(x) the image
many
by falling
function
A
to
exists
yes
How
from
=
f function
⑦
function
1
define
subset of
[2,4,6]
If
We
a
=
2FA, butthere
because
Nos
a
is
·
of
each
is
in
the domain
We
visualize
can
:
2
0
-
⑧
a
o f AxB
subset
using
or
diagram.
⑧
IR
(i
when
X
When
A
is
rational
is irrational
X
B
I
i
arrow
from
Igoes
to
of 1
image
24
2
3
C
4
the
diagram
arrow
of
a
functions
1
2
must
have
we
2
-
-
x4
3
function behaves
A
arrow
⑧
O
x
fex-
In
A to a s
234
W
f:IR
from
⑧
X
O
o
X
function
a
like
exactly
fo
soucen
8
image
X
of X
0
a re
o0
0
o
0
->
⑧
0
Let:
f:
AEB
is
Xajb
=>
1:A ESB
Here exists
one
to
one
7A, if
one
to
feal f(b)
=
then
a
b.
=
many
ajbt A,
so
that
fat=f23)
arrow
from each
x
lights ource.
a
one
but
all.
i n the
element
domain.
f:At B
E Xb=
f()
onto
is
"
=
- 08
B, FazA,
O-
b.
=
o
-jo
B everything
A
I
2
5
->
-
B has to be used.
in
(co
2
46
domain).
-
-
810
12
A
B
f [(x2),(2,4),(3,6), (4,8)(5,12)]
=
How
many
from
one
XSx4x3
functions
one
=
Ans:6
A
to
[1,2,3,4,53
A
there
are
22,4,6,3,10, 123.
to
x 2.
[1,2,3,4]
=
I
2
-
30,
3
~
IAM
B
IBI
=
DAME
(=>
7
one
7
onto
EC
to
functions from
one
from
functions
to B.
A
A to B.
R 2x2
=
↳site
offunction
How
sets
many
from
A
functions
21,2,3,43
=
-2"
>q
Total
whatwe
((2)(24
3"
Let
fXIR-112
-
+1
3 x
=
a
one
is
22,43?
⑮
=
-
don't
Ans
f(x)
B
2
=
Ex)
to
there.
a re
-
2)
3]
-
+
defined
for
to
want.
each
one?
is
by
X
EIR
fonto?
Solution:
-
fi s
one-to-onco
foot: useact.Supposeateein
f(x) f(b)
=
2at
2
a
f
is
2bH
=
b
=
onto.
Proof:
Suppose
b5117.
Choose
a
b
=
then
acIR, and
2(b)
f(al=2a+1=
+1 b01+1 =
=
f:2
2
-
f(x) x2
on
+
=
5:
Thinking.
choose
a
so
=
one
onto
not
2 1
+
f(x)
that
f(al
to
x
=
Not
2
+
one
to
one
not
one
b.
2at1 b
=
a
EX
2-21
let
7:
Yes
to
one
defined
one?
to
one
is
5)
=
fx= f(b)
E Xb=
f()
which
fibl-1
is
-
2
3.
235-=
which is
-
onto
is
We must
prove that
·
-
is
at
2
f:AtB
2.
x =
NO.
that
Then
=
each
fonto
Is
9,3t21. Suppose
Suppose
for
=
true
one
a b.
f(x) 2xH)
by
->
definition.
B, FazA,
b.
=
B t1A
(false)
arto
Negation:there
exists
1
=
fzal -b·
for all acA
By so
roof
exists
There
b
2200se
b,
2.
suppose
=
2a
then
for all
so
a
a
ifa
b
=
t2,
E Xb=
f()
=
B, FazA,
b.
f(a) b
7-1.
1
but
=
is
I
integer
not
for obo.
7: 2
Consider
is
Suppose
show
=22,
asb
that
by
xt2x
one?
to
it
one
&
given
-
a
such
that
feal=f(b). We
b.2a 2b
=
=
implics
to
want
a b=
onto
is it
b
7
=
G
choose
not
b
]i
=t
such
·
Vaz
that
Suppose
a
II
21
fafb
then
Ga
is
is
what
an
to
we want
integer
and
show
thus
hence
onty.
7: Qt 21
Consider
Ex,
Show
choose
one?
to
it
one
is
x
f(11)
Rs 1.1<2
XEx".
but
fext=fax;
10)= 1.2
bacause
=
No
such that
x=
f(x) LX)
by
given
(10) 1 21.2)
and
1
because
fax)
=
=
=
=
but X*x
101+(.2.
as
1.2<20
=
it
onto?
is
&
21.
be
suppose
choose
a
Then
b.
=
fix togiven by
Consider
is it
feal=b.
7aGPsuch that
*b=21,
show
is
integer this
x+2x
one?
to
one
d
f(a)=La)=13). Because
-
such
Suppose X, x=&
2x 2x" implies X x 1.
that
fax)
f(x").
=
to
want
show
X
x".
=
=
=
ontoh
it
is
-
X, x"G.
Suppose
Choose
a
Now
=
b
=
2a f(a).
=
Ex
f:
let
fone-to-one?
IS
defined
2+ 41
Is
by tex)
for each
xn+2x
x=21. Is
=
f
one-to-one?
Is
forto?
Solution:
t
(E)
not
one-toxone
is
roof:choose
Ii s
and
x0
=
f(x) f(y),
xxyz -s
-
=
y=-
2.
(7yz2 x 2s
onto
=
not
Rof:choose y=-2. Then for
any
Ex
f:4-1)
let
Then
xyt2,
xty)
but
and
f(x) f(x)
=
!
f(x) +y)
0
=
f( -z) f(y)
defined
by
fix)
=
=
=
x
y.
Eab =
x=1, f(x) x2 + 2x (x+12-11-12
=
but
=
-
2
2x2 +x for each
fexity.
so
y
=
f
one-to-one?
for
any xc2,
Is
XE2.
Is
f onto
solution:
fi s
to-one
one-
f(x) f(y)
proof:Xxxy GA,
=
then
2x2 + x 2y
+y.
2x2
2x2
X
2y2
-
2
-
2y
2(x2 yz)
-
x
x
-
2(x y)(x y)
-
x
x
-
x
y(2(x y)
+
xy
+
heace
x
=
y
y
=
Proof:(by
f(x)
completing
the
square)
y=-1. Ten yet
2x2 +x
=
x
2
=
+x
2
x2 x2
=
+
and
+x
+
2xx1)2|t)2
0
x2
=
-
y
0
=
1
=
Y2
*
x(x
1)
+
0
-
=
+
=
=
=
y
=
=
+
2(x y)
y
0
y 0
1)
+
=
=
y
+
+
=
-
+
+
y
-
x
choose
=
+
f onto 7
is
Notinteger.
so
f(x)
by.
-
-
,
fonto?
is
b.
Composition
fiAtB
let
Ex]
functions:
of
and
LetA S12,33)
=
7
let
g
be
git)
B
function. Then
a
function
from
defined
to
if
got(x)
by
=
g(f(x))
for
[(2,1) (41), (6,2)
Then
3.
f
is
from
function
a
A
to B
and
is
g
a
function
from
E
14
I
2
-
3
3
6
- >
got
I
I
fog?
-
=
-
xy
IS
Es-a E
I
f
Ex
g =)
=
They
f:IR then
g is
same
gilKeR
fog(x) (2x+)2
=
is
Ars
the
S
not
is
have
A
and
B
and
x=
=
=
I
each
[2,463, [1,2,3]
=
[CI, 2), (2,4), (3,6)3
=
goti s
domain
and
for
every
element
=
gof(x) 2x2H
=
False, because fog() 973 gof(l)
=
in
X
f(x) x2
by
fog:got?
=
and
onto
is
one
to one
onetoome
defined
and
f
onto
fo
I
for
the
and
each
domain,
g(x)
=
XEIR.
f(x) g(x)
=
2xH
for
each
XER.
B
to 3
and
got 3 (1,1), (2,1)((3,2)]
=
d
IMPORTANT
Ex
a)
f:
Let
AtB
f and
if
and
I
then
one-to-one
are
functions.
be
giBtC
goti s
True
one
false?
or
to
one.
An
=
⑧
Co
B
A
C
gotTrue intratively
proof:Suppose and
and
by
and
I
want
we
to
can
you
to
show
see
that
one
XsYE
where
x=y
=
g(fx)) g(f(B))
one
are
g(x) gay) then
=
picture
to
f(a)=f(b)
than
is
B.
want
We
to prove
be
false.
ab
9cb
where
=
got
is
one
7
to
A.
Suppose
one.
2,BEB.
where
B
a=
not
it's
possible
g(f(x) g(f(b))
=
f(x)=f(B)
Since
b)
If
because
f(x):f(B)
know
we
got
is
to
one
and
of
since
f
then
one
is
is
g
to
one
1-1
1-1
is
x
B
=
&
one
Not
got
o
possible again
intution will
·
-
-eno
- >
o
6
-
-
5
A
9
B
·C
got
mustprove
got
is
one
must
one-to-one. We
is
f
i s one-to-one.
We
pint
not
possible
&
-
Jone:
Suppose
in
0
me
C
win
innatever
--
Suppose
a
to
one,
B
I
L
=
gof(a) g(f(a)) g(f(3)) gof(b)
=
get
we
-
f(a) f(b).
Suppose
asGA.
Since
b.
=
prove
f
D
that
and
=
=
ab
got
2)
If
got
This
is
statement
is
-
got
Choose A [1,
=
is
23, B
f:A +B
choose
goti s
g
is
as
suppose
g
is
one-to-onco
f: AtB, 7
Negation: 7 sets A,B, C, 7
g: Btc
one
to
fin):1
g(1) 1
one
then
one
one
[1,2,33)
because
not
to
=
gof(l)
to
and
not
is
g
I)
and
f(2)
g(2) 2
=
g(z)
and
1
as
one
gof(xl:gofcys
hence
I
2
=
=
-itse
18
32.52:
-
( 21,23
=
=
9:BtCas
choose
Now
false:
that
such
but
one
2
=
got(2)=2
g22) gC3) 2
=
=
case/
g(f(x)= 1
then
casch
g(f(x)
then
goti s
1-1
2
=
x
y1
=
x
=
y 2.
=
=
3-
I
d)
If
one-to-one
is
got
This statement
is
Proof
and
got
that
be B, Jaz A, fca)
is
is
one-to-one. That
xaP, 72x=A,
y =Bs 7
g(feel=g(f(n)
then
=
b.
we
g
is
one
to
one.
show
where
1
=
g
is
is
ec1=A. Suppose I
is
one-to-one. That
onto
gexi-gcy)
we
must snow
X
=
y.
X
that
such
6(xy) Y
=
=
is
want
to
2
=
g(f(x) g(f(y)
got
Since
then
that
f(x)X
such
<pEA,
have
we
onto
is
true.
Suppose
is
f
one
to
one
we
because
have
g(x) g(y)
=
[x=dy
so
x
y=f(dy)
f(x)
=
=
showed
we
x=y.
Inverse
Definition
f: A-B
is
g
is
invertible Ex There exists
called
fog-IB
that
giB -Aso
of
f
inverse
and
got:IE.
ft.
g
=
Theorem: F.AtB
f:2-21
Ex:Let
f
Is
one-to-one?
invertible
is
the function
be
fis
Es
defined
by
one
to
fixi-ex
and onto.
one
for all
x=4.
f onto
Is
~
Yes
Thus, Ii s
defined
f:A
not
by
gof=In
No
invertible,
find
a
function 9:2- 21
I whenisever
=2CI)
X
=
so
inverse
of
f
in
P
this
function
and
one-to-one.
-
onto
B
is
can
this
case
is
both
i
y,
7 6
↳
-ener
2
-
fog2X B.
=
-
3
-
4
-
-
S
3
6
A
-
I
↳
3
-
-
=
i
I
-2
4
-
g: B
for example:let
g:2+2
ii
-> - -
=
that
got:I
gixs=
-
g
but
we
-
-
3
4
-
S
S
-
6
9.
be
the
function
Relations
Def:A
relation
to
related
is
X
sat A
a
on
relation in
the
in
g
Ry
x
AXA
when
subset
of
is
R
is
a
relation
on
As
we
say
(xxy) GR
when
(x,y)ER.
>
T
0
4
X
88
⑧
3
0
.
I
⑳oo
O
-
734
Ex
(etA
1
[1,2,3,4]
=
E(1,2), (1,1), (3,4))
is
=
How
many
2
Ans=
We
can
relation
visualize
on
a
A
relation
a
91,2,3,43
there?
arc
=
A.
on
using its directed graph
relation
&·
&
3
R
Def:Let
R
R
R
Eb
Fxxy, z EA,
is
1,
on
=
reflexive
1
=
D
and
What
relations
is
=
antisymmetric:if121
for
same
then
then
1
2
/
and
transitive:i f
122
relations
what
is?
reflexive:1<A
false
antisymmetric:if
then
yRZ
y.
XRZ.
3, 43
if
4=4
then
4=4
true
1=1. True.
1-1. True
"s"?
than
Symmetric:if122
x=
True
21, 222, 313, 4-4.
symmetric:112
and
[1, 2,
=
and D=1
=
1
then
yRx
=
3,4 4
=
then 1=1.
D
antisymmetric:if 1
=
yRx.
and
xRy
FxcA, XRx.
reflexive ES
"="?
Symmetric:if 1=1
transitive:1
A
Set
2 2, 3
=
if
is
then
Ry
XxcyEA, xRy
let R be relation
what
relations
x
if
Ex
is transitive
A.R
Sat
on
Axcy= A, if
Ex
antisymmetric
Reflexive:1
3
the relation
symmetric
is
is
be
and
then
I <1
transitive:if223
241
and
and
not
symmetric.
124
213
true
than
then
1=1. True
113
·
True
false
ASD.
3
C4
True.
then
Vaccous
truth.
224 true.
First
"if
weareashere
2
or
it
reflexive becaus
not
is
iti s
symmetric because
not
xcys2=2.
Suppose
and
4/2.
214
(1-1
Xly
Suppose
and
yext
have:
then
2=
1 xtK
but
it.
and -11
y/1.
and
then
we
prove x12.
must
and (E2.
ykwhere
the 2
where
=
nonce
Ex
X0.
transitive:proof
is
we
but
0
because
iti s
not
antisymmetric
it
OC21
X/L.
LetA E1, 2, 3, 43.
=
3(k1), (2,2), (3,3), (4,4)3
[D,2), (2,4)) (3, 4),
How
many
[4,4), (14)].
reflexive relations
2'
because
Ans:
reflexive
is
4
on
are
reflexive
not
31, 2, 3, 43
A:
fixed.
arc
because
11.
there."
(161), (212) (3,3), (4,4)
I
&
4
·
O
3
0
O
⑧
O
X
X
2
0
I
⑧
X
A
0
/
↳
↳
·
O
L
⑧
O
O
23Y
Ans: 2k
you
Ex
either
choose
or
not
choose
from whatever
is
left.
[1,2,3,43
LetA
=
S(1,2) (2,4), (2,1) (4,2)] is symmetric.
S(1,2), (44), (2,1) 3 is symmetric.
relations
many symmetric
How
Ans:
20
4
X
⑧
&
&
3
2
D
X
⑳
⑧
⑧
A
· a
⑳
*
·X
②
23
⑳
0
y
are
there
on
A
11,2,3,43?
=
so
2
k
il
yes
this
is
reflexive.
Ex
51,2,3,4]
LetA
=
SC1,2)3
is
antisymmetric
Q [(1,2), (2,1) (1,1), (2,2)]
a
[(1,2)s(1,1), (2,2), 24,3]
=
&
this
->
vacuous
is
truthfirstcondition
because
not
antisymmetric
is
:
bacause
1Q2
antisymmetric. For
does not
not
true.
was
but
IF2.
and dQ1
C1,2)
and
(4)3)
case
it
is
vacuously true.
on
to
has
-
·
34
antisymmetric
(
antisymmetric
as
not
not
antisymmetric
How
24* 3)
but
x
yRX
fy.
of
having
relation
antisymmetric
many
and
its
there exists
not
as
a re
there
31,2,3,43?
on
AskTA.
Ans:
Ex
Jxsyt A, XRY
[1, 2, 3, 43.
A
Let
=
is transitive.
*
2(12)3
R
[(1,2)) (2,1), (1,1)]
because
T
transitive.
is
is
=
2R1
IRL
and
not transitive.
2Rd·
but
[(1,2),(2,4),(14), 434),
=
(3,4) 3
is
transitive.
⑳
·
2
↓i
(12) DR2 and 2R4than
1R4·
=
(234) 2R4 and 4R4 then 2RL·
(1323:AR4and RL them DR4
=
(3)2):3R4
and URL than 3R4·
CHSU)=GR4 and 4R4them 4R4.
Ex
let
A
[1,2,3,43
=
Is
there
a
relation
R
on
[Islec 3,43
so
that
R
not
reflexive, symmetrics antisymmetric, but
is
Ans:No.
Proof
(by contradiction]
Suppose there
not
transitive ->
R
is
that
x
Ry
that
xRL
Suppose
is
such
a
relation R.
7xcys2, xRy
antisymmetric then
xRy
and
and
then
XR2
yR2
and
yRX
yRZ butxR1.
S
and
which
x=y.
is
We
a
see
contradiction.
transitive?
1A
reflexive, symmetric, antisymmetric, transitive?
R
IS
21,2,3,4].R relation
=
=
R is
R
is
R
is
R
R
21
on
defined
R
is
[23
y
then
=
=
x
yo2:223 +4
XRX.
must
snow
we
then
31x+2x
B
XJyt1.
Suppose
by
+
Suppose
that
xRy
3
is
=3(x+2y)
must
we
x+
x
yRx by showing 3Dy+ 2x
not antisymmetric.
Suppose xsyt2.
R is
that
is
x
319H2=3121
Suppose
transitive.
3k-
dy
and
XRy
yR2
choose
Xt2y
=
x
9 and y 6.
=
-
2y
=
y
=
=
* 6.
We must
=
show
-
bk
+
-
4y
3y+ 6x
3(-yA2k)
that
XRML.
-
YA22=3t.
is
that
yR
and
a
y+2x yAd23k-2])
Suppose XRy
=
=
but
sys2t2.
where
3K
=
2 3t:
and
then
316H18:3124
and
3AXA2y,
that
is
xRy
=
2y 3 k
3k
=
·
prove
R is
y
-
2
Man
2L
x +
3xx-2y
+
x
+
=
2k 3x
x
xy+37 y
-
-
=
24
/
+ 2 =3kx 3+
+
-
3y
x722 3(x+ t-
~
=
3)
A
31,2,3,45s
is
R
is
Choose
and
then
Notations
gal)=1
Relations:An equivalence
=
A
is
easy
to
A
Transitive:Let
gRf
and
on
equivalence
by
f,gt F, fRg()
than
a
set A
is
relation divides
gaxs= faxs.
g [(1,1), (2,2)
=
(332), (4, d)
-79
but
f(l)=g(I)
Suppose
number
is
=
R
that
xyzPCA)
fex)=gax)
=
relation
I
defined
on
fex)=fext.
Suppose
E1, 2, 3, 43. R relation
·
is
Suppose
xcy) 2EPCA)
an
Nx1=AX1,
Reflexive:Let
x=P(A). Since
Symmetric:Let
the relation
is
a
and
relation
A into
gal) KCI)
·
=
on
Since
g2K K21)
=
it follows
f(l) K21).
that
A
equivalence
classes.
a
=
prove
R
of
of
Example:Let
It
class
equivalence relation
number
A
to A.
1xRa]
(9.7 3x=A
equivalence
ktt-y cry.
f [1,11, (1,1), 23, 1),
(MD] and
fig
reflexive, symmetric and transitive. This
The
·
figskff.
transitive:Suppose
Equivalence
are
fog ff
antisymmetric.
not
falll=1
R
them
Suppose f. If
symmetric. Suppose
is
where
from
=
reflexive.
R is
y)
f
f set ofunctions
=
R
XRY
so
ty.
2, xRY (b1x
then
1
x =
=
symmetric.
21
x=
where
xdy x + 2x 3x
=
Xny:[1370 and
Then
=
and
xxy
by
Suppose
reflexive.
is
y [1, 23,=323.
YRX.
so
&2 0.
X
Letx=[1,23
yux=xnyto,
than
Xny*.
i.e,
=
XRY.
hence
relation
is
XRY;
=
because
xRYE6 xnyEG.
d.
dnd=
since
Proof:Let
x [1 3
but X*2
antisymmetric.
not
xny1d
②
xy GPCA). Suppose
not
transitive.
R
is
by Axxy =PCA),
Let
symmetric.
YRL,
and
defined
4GP(A), but$R0
because
reflexive
not
PCA)
on
·
equivalence
classes.
of
on
PCA)
defined
=
P(A), xRY
=>
NXD
=
AYD
relation.
XRX.
XRY, i.e, NX1=1Y1. Then MYD=
1X,
Suppose
by:xxxy
XRY and
YR2,
thati s
so
YRX.
Ax1=MYD and MY1 12D. then Ax
=
1Y1=121,
=
so
XR2.
=
f(l) gCN).
=
where
-
yt2kzY.
Example:Let
A
R
Easy
How
to
many
Example
127A
the relation
is
R
that
prove
equivalence
Two
Ans 4.
31,2,3,43,7
=
is
classes
functions
related
when
defined
PCA)
on
↓x,y
=
fel
b
g(N)
=
relation.
have the
they
B
[1,2,3,4,5,6,7,8,93,
=
fig tf, fRg
by
there"
=
R relation
from Ato A.
functions
all
set of
a defined
on
equivalence
an
are
are
=
image
same
of
So
1.
the
possible image
number
of
of
which
1,
be
can
1,2,3,4.
[2,4,6,83
=
by
PCA), XRY
DXUBD=DYUBD
ES
How
there?
classes
are
many equivalence
r6 # of possible sizes of XVB'
I
=
#
Congruence
Modelon
on
21
where
no
2
xcyt2,
x=y(modn)
tr
to
congruent
x
is
nAx-y
E)
y
modulus
n.
Proof
Reflexive:Letx= 21. Since
Let
Symmetric
Transitive
and
n
xxy=2.
X-x 0 nx0 where
=
=
Suppose
xEy(modn), i.e,
and therefore
y x (mode).
2)
and suppose
x=y(mode)
site 2.
Now,
Lot
xy, 2=
for
y-2=nt
Of
21
so
ndx-y,
so
(mode).
nk
for
x-y-
some
key. Now, y -x =-(x-y)
x-2
and
y 2 (modn).
=
(x-y)+(y-2)
=
Then
1 2 +nt
=
adx-y
n(stt)
=
and
where
ny- 2,
3
x 0
=
(mod3)] (x= 2130x)
=
[d] 2x42139x 13 [
=
-
=
-
5,
-
[ 6,- 3,0,3,63-remainder
=
-
2,14,7,....3-remainder,
(2) (x 21131x- 3 Ea Esis
=
when
n= 3
=
we
=
have
3
congruence
↳ remainder d
classes.
6
=
I
·
S
!
O
B
2
4
S
38
4
S
S
⑧
S
01
0
&
389
=>
2
2
3
34
so
SttE2,
=
=
=
-
nk
n(-xx)
=
=
some
[0] 3x =21
n
x=
x
x-y
so
as
=
X=
2
(mode)
where
KEY,
so
aly-x.
n
6
=
X
O
2
02
↑
3
03
4
0
S
2
34
4
02
4
O
3
3
O
·Y521
5
No
n=
⑧
00
O
0
R
5
234
I
0
U
2
0123x5?
X
6
0
B
0
2
0
3
⑧
2345678
9
2
4680246
8
3
69
↑
⑧
08
I
303
!
6
25814 7
260
826
4
05
0S
0
S
628406284
07
41852963
⑧
7
8
⑧
q
D
642
64208
8
987654321
ad
n
6
I
⑧
=
2
↓
0
N
2
I
3
-
ac
Theorem
Modular Arthimatics
if
a b
(mode)
=
Proof:
and
b
a=
(mode) and
Stt 2. Then (atc)
=
d(modn)
USCA but
(moda) and
-
n(sc
=
bd
ac=
d(modn),
c=
(bAd)
+bt),
(modn)
bd.
=
then atC =
b+d(mode) and
c=
Easy
Suppose
5
4
3
that
is,
(a b) A(c d)
-
-
=
where
8+tE2/
na-3
ns +
=
and
ac
bd (mode)
=
and nced.
nt=n(stt)
sctbtE2,
and
so
Thus,
az
-
arb=ns
and
for
crd-nt
bd acsbc Abc -bd (a b)c
natc)
=
-
=
(b+d)
and
-
nac-bd- thus,
some
+
b(c d)
-
atc= bAd
Def:
An
integer
Theorem
is
a
modulo
invertible
gcd(ayn)
=
Sketch of
is
a
n
Ex
there exists
an
b
integer
that
so
ab=/(modn).
n
1
=
proof
a
invertible modulus
71
=
Ib =
a
21,
(mode)
ab=1
nbab
Ex
7b
E-
73jkty),ab-1
=
modulo
is invertible
a
1
=
nix
=
7b,k= 2),abtnk 1
(
=
=
gcd(a,n)= 1
3
abtrk 1
=
g gcd(a)n)
=
Sketch
gla
-
of
and
gl
gD
=
=
gIn
proof
ab tak A
=
y
gcd(a)n) gla
=
and
+
gen
I
gatg D
=
gak
n
-
9
3
-
8
Y
=
7
-
-
6
-
3
-
0123
whenever, we
4
0
have
-
3
-
1
abtric: 1
2
2
we
-
1
3
⑧
1
2
can
0.
conclude
↑
b
34562
230
↑
is
inverse
&
of
2
⑧
3
9
O
N
mode.
a
1000 -1(modll)
=
Ex
Is
96
compute
invertible
modulo
dT1?
Find
21
-
=
96 1x79+
96
17
=
+ 11
29
17 Nx)) +
11
4x17
=
=
11 1x6
=
(x S A
=
6
1
S 5x1 0
n
-
5
-
=
96
is invertible
modulo 2410
A
-
17
271x( k)
-
2
=
An
43
-
G
19
S
+
0
D
8
1D
+S
gcd(221,961=1. Thus,
D
271
=
6
that
271x+96y 1
so
gcd(271,as)
27 2x96 A79
79
xxy
-
96x48 1
+
96
of
inverse
271
is
=
modulus
48.
14
1]
t
31
21117
28
E48
48 (mod
=
-
271)
48
(mod 271)
t 48 27kc
=
+
t
-
48 27k
=
t 271 48
=
x
1
=
+
if2a 23 (mode)
this
true
is
b(modn)
then
=
a=
modelo
I is invertible
when
n
↓
this
True
a)
ab
=
3)
(a)
c
3 3
b
if
=
al
even
ba=3b
a
·
·
a
=
3
=
=
-
=
is
a
the remainder
=
27
24
6(-4)
=
-
=
nence
3a =
3b
1, b 9
b 1
=
(mode)
=
=
a
b
a=
ncyt.
=
and
then
a
b
6,
=
39-3b
then
(mode)
exist
ab,
n
(mode)
bd
but
Proof:choose
3b
=
=
false.
statement
is
Negation:there
if
is
=
-
What
i
3a 36.
and
no2, if
Fasb
then 3a
(mode)
b
a=
then
Ed
=
This
2,
nt
ifa b
a
when
false?
or
X
not
true
is
where
-
4E7
9
8
where-1-S #2
6(-).S)
b
2243 when iti s divided
of
by 9?
ab
=
23 1
=
=
80
*
(23) =(-1)Y
273
x
9)
then as (mode)
2
2
=
=(2)
a
->
from theorem.
qk b*
=
=( 1)9
-
=
So, the
-
1
answer
is
8.
-END