ページ1:

th There exists.
วายร้ายมีอาว
ZeroDate:
No.
5. Fa, b, c € 7, a⋅cb+c) = b+canton histributive
1071 0 € Z, VaEZ, a+o
311 EZ. VaEZ,
7. Ha ez a
ca.
za
• a = √a
+ - a
D
4. Va, b, cez, a 70 and a. a.c
b)
and
Jo= c
O is called the
and live identity of Zon
1 is called the
6.1 70 € 7 ta ez, a+za
E
-a+oza for all α EZ
104029
for
Some WEZ
12 se, u+sa
minationibus a for some gez
multiplicative identity of
อกหัก หมอก
-a is called the additive inverse of 2
For ab € Z, a+ (-b) denoted by a-b
anh ab hendted by a.b
a + (-b)
Homily
2.
Home
2
(
(usi)
+, X
110i Kyo) G
uskimobun
☹ 24 modems wnnwoo®,
Theorem 11 for a bid c
17 Imb+a
= c +α, then b=¿
PE
17 Va € Z, a. Daa
37 ta & Z - Ga),
4. เย็นๆ ทาวน์เวอร์ AI -
47 α (-b) = - cab)
57 c-ax-6)= ab
67-(a+b) 2
..ยาม พาฝัน
EIJ.a
a+(-a) zo
a. 97121
10521 a
Fa € 7, 0·0 > Oa Zeno)
et
AEZ 144041201
40007024120
Then a⋅o = a.co+0=0+α
2
a²0 = α0+ao +60120
2
Since a + (-a) = 0 a is the inverse u.
VAL
Be
020
Ans
a·0+0=00+α0)
(3) ta e Z, - (-a) = a
"กษมมาดาม : HT
Ans.
+ α = IC+α = b²C
vs a b
PF Since ac-bs+ab
*
a ( ( b +
Dh
Of a
nadia MTH 10 วรว
Thas a 2-(-a)
'ff Assume 6+α = 4)α (-b) z-cab)
=Then
b = 16 +0)
2
Jo+ (a+(-a))"
Whamay = cb+α) + (-a)
= ((+α) + (-a))
ไปหาผมกลับ 2 C + C04Cq))
= C Ans.
2
C-a
20.0
b)
= 0, α(-b) = - cab) Arg
ac-b) +ab 20
9.6
Inaniu zo
L

ページ2:

ทีมมา สวน
1.5218
2. 94.5 5
W
3. q
2 C
60
4. 3M2 183
5.102217P
S2 126
Wz 72
d 2 540
Mz
bl
2
pz b
=
2190 Hาว
โลกา
ง น
าลคา
2.3+4 100
1. 8 7 2 5 2 6
2. h7 6 + 4 =
2.3+4 2 10
10
M7209
2 Y 1 6
2X
12X
ฟ้า 2 × X
9
2. ห่า
2
2
2 x + b = 2 x + b
X
29 -84582 -0,75X
Mi 5X-4
959
1
Y
ท) 5 × X 26X
9 7 5y - 4
2
- 5X - 1
2.M
12 19 -
1
2
1612
2
12
4
-4
61

ページ3:

Date:
No.
(5) (-a)(-b) =
PF c-ax-b
2
R
U
-((-which 4)
(-cab) Jcb 4)
ab clog 3)
-(-a),=)
แทย โลก คือ ได้
= (a+b) = (-a)+(-6)
Since [(-a)+(-b)] + (a+b).
20
Rau ot
e one
Awood Food/51001021
= (-a) + [c-b) + a] + b
2
(-a) + [ a+ (-b)] + b
[(-a)+a] + [c-b)+b6]
Z
2
7
0 +
of +
สมบท ก่อน คม
20, (-a)+(-b) = - ca+b) Ans.

ページ4:

Date:
No.
well-Ordering
minu
The well - Ornering Property m
Let IN = {1,2,5,... }
XE
"If S.SIN where stø, then there exists Br. S st. sex for all xes"
There fure if KE IN and k<s, then kes omalov
& kits
Theorem 1.1.2 Let SSN where SØ
Then a
least element in S is unique.
Theorem 11.3
1. Is the smallest element
in IN
Cordars
1.14 For k€ IN, there is not
ne IN st K<n<k+1
Theorem 1.1.5 12 is an irrational number
Theorem 1.1.6 (Mathematical Induction first form)
If SC IN where 50 and 5 satisfies these conditions
Theorem 1.1.9 S≤IN, S# 6. → Else S, SEX for all xes.
PE By Well-Ordering Prop, Is ES, [S≤x for all XES
เกรท 50 มม. สมมติ
there exists s'ES, s'ex for all XES
movatio
Suppose
We will show that
Since S≤x for all
S=5
X ES, SES
Since s'≤x for an XES, S'ES
Therefore 5-5 Anc..
18772282

ページ5:

Date:
No.
2. the
Sam
The IN, KES → K+1 ES, then S= IN
→>>>
Ex 1.1.7 show that for all n€ IN
i
2
lz
Ex 1.1.8 Let air, he real numbers where #1. Then for all nE IN
n
l
= are
Z
a tart...tar"
10
208"
- a
0+1
1-1
For prore Mathematical Induction, he can
Pon) be the statement for each nε IN
Tet Pens
P) is true and
2. WK EIN, PCK) is true imply Pck+1) also twee,
then we can conclude that Pon) is true for all neIN
Ex 1.1.9 For X, S ₤1B, show that
h-1
2
1-2 ^-1 )V for all nεIN
1-2 S + x = 3 g² + ... + x = 2 + y
x²² = (x-5) (x²+ + x^h-20
Definition 1.1.10 Define a sequence of number
Ex 1.1.11
by f₁ = 1, f₂-1 and
7
z
ne
fp = fg + fry for all beIN 5. Dz3
In is called the nth Fibonacci number.
Let, fn be the nth Fibonacci number Then show that
fx +44 +...+f₂n = fen+1 -1 for all nε IN
Theorem 1.1.12 for no E IN, let
where
2
] = { no, not 1, not 2,...)} SST S+ 57Ø
1. hoes and
2. Vke IN, kzno and KES →KTIES
Then 5
2

ページ6:

Date:.
No.
Another form of Thm
the statement
For nεIN st. nano, if
1. Pcnw is trae and
1.1.12
Tet h₁ = IN and Peni be
2. VKE IN 4 kana, Pck is trae Pck+1) also true
Then een is trae för all nano
Ex 1.1.13 Show
Show that >2n+1 for all neIN 4. nzs
Ex 1.1.14 show thatt 2^> Ŕ
for an relN H. ns
Theorem
If
1.1.15 C The strong induction)
SCIN where SØ and s satisfies these conditions.
7
LES and
2. the IN, if mes
St.
3. WhEIN, it mes Kemek
MES
4. the IN, if mεs for i≤m≤k, then K+LES
MES

ページ7:

h
MTH 2205
Date:
mwin, The well- Ordering Property
No.
2. 15/7
จาน
n
If SC IN where 5+ 6, Then there exists SESH SEX
for all X ES"
WS CIN SØSESSE
Therefore if KEIN and was, then K&S
nikon
Auno only
1.1.2 Lot SEIN where, St & Then a lepat element in Sis unique.
Photo SSIN Trust sz
71 SES, S≤x for, all XES
DE By hell- Ophering Boperty, Is ES, S,≤ X for all x ES
AS AS PAYS PUNAN SS, St ES
17
5
5,5
born ES MAS 5 Mon & Es
there exists S'ES
:•
Suppose
S'≤ X for all! X ES
We will ghow that 525!
Since SEX for all X ES, S. ≤ S'
since sex foran xES, 5 <S
+ b²ε C. (: u<b² < 1 and b² EIN as closed) 1
1.1.3 1 is the 1s med Me dement in IN.
by
PF Prove has Contrudiction, Suppose there exists a E IN
Let C = { & EN <<1}
st. ocaci
-Then C = of C: α = ¿)... lummig [there exists] (EC) Wangec.
Bg well. Ordering Prop., there the smallest, et in C, socs, b.
Thas 0<b<1, 50 of b² <b<D
town of
2
b²EC, 1:0 <12² <1 and BEIN by clasen,
a eta, since b is the smallest elt. centrodicts; in C.
Heve 1 is the smallest elt in INA
=
Ex.620.55 17.
b207852 785
Toooininko

ページ8:

Date:..
IN
No.
Complanty 1.14 is
For KEIN, there is not ne I st. Kanak+s
Lo KEW
Prove by Contradiction, Suppose there exists n≤ N of.
Kanak+1
Then
O<n-k<1 ARM Romiton , and p-k zo
clad
Thus, In-KεIN] amh n-k <1; C by coorKE Zand. n-k>0
a etch. C. by Thm 1.1.3
Hence there is no nEIN st. Kanak+1 #
+min, lolla
Theorem 1.1.5 √2 is an innational
12 EQ
(TEG)
number.
Fantor all laying
Q = { x a b € Z amb b z o}
Prove trg Contradition, Suppose that T₂ EQ
PF by
ป
Then Jmn & IN A. T2 sm (mn € 7 and √2 > 0)
where no
Let C = {NE IN 17 MEIN 7. 12m CN.
Thas C&C: REIN. =10")
Bg Well- Ownering Prep., 76 EC
st. b≤k for an KEC.
Sinee bEC, Ja E IN SH. 92 = a
JaE =
Theorem 18 theratieat Indvetter fingt farm).
Mo 1.1.5:
Sinee 1-2 <4, 1<√2 <2
Kangon Thus 14942, sob<a<zb
6
-bommpos.." o<a-b<b> solo <eb-a
(69226)
(0<26-a)
stree T2 = a, √2b> a
2
Rum Thus 2 bat so take-ad
Elf-abz a² - ab.

ページ9:

Date:.
Therefore [ bcsb-a) > aca-b) + Rink with was a Super
.. √2 2 a
2
2b-a
a-b
since ① and a-b>g, a-b€ C₁ the
|
a cth, since a-b2b and b is the smallest elt. in C.
Hence J2 is irrational buntor
2
Theorem 1.1.6 (Mathematical Induction: finst form)
574
]ť s Ċ IN where Std and 5 satisfies these conditions.
1.) IES and.
2.) NKE IN, KES KHIE S, then S=IN
Ex 1.1.7 show that for al nEIN,
n
Ziz n ch+1)
2
Ex 1.1.8 Let ass be real numbers.
Then for all nEIN
where r$1.
n
i
Zar
Z
120
a+ar+...+ar
zap-a
7-1
can
For prove Mathematical Imuction, we can
Pam) be the statement for each nεIN.
If 1.) PCD) is trae and
2.) VKE IN, Pck) is true jumple Pck+1) also true,

ページ10:

SUBJECT:
ทาง
ax+b20
สมา
CMA 2202
NO
DATE
รูปทั่วไม่
←
laxy
20
126 ab ineroosoo ; a to
การกรมฯ - ทานมาก ลมแรง
ตัวอย่าง
X-225
10 + 5
2 +5
115
215
->
->
7 -5 →
-7-12-10→7
ร
X27
2
ทร
·5
> -7
สมมา
3
mom
การทาดมา
hunon
หาก
Moocho
Im a zb nã₂ ba
a + b 2 c → c = a+b
X-32772x-3
• สมพิต ถ่ายทอด เมือ 4. 4
b now bad não a zd
ก๊า
azb
ติจาง X517 Az 5+72 12 na
2
7x2-34 AZ-34 20.5 não
x 20.5
X = 12
Double

ページ11:

SUBJECT:
2 0 45 19 มะทา
159
151 X
221
15 + λ
221
15+x+(-15)
15 + X
15+6
221+(-15)
z b
2
ข
2
21
21
2
ด
100419
ขิงท คม
าสทา
12
12+7
ทรวง
MOL
แนมทารอน อยู่เกิน
192
12
2
1227-7
T
-7
777±7
4-7
219-7
12
212
จากสมการ 4 n
0
2
2
บาท หมา 1. 40 + n 2 70
n2 30
2. p+ 82 2 142
D2 60
3.242x-5
X = 29
z
8
4. q = !-1
4
#z 14
214
5. 19213+5
526
NO
DATE : CCC