ページ3:

Subject:
No.:
A
Date
......
def
A^B
171)
XXEA and X EB}
AnB
X = A^B
XEA and XEB
.
X & AB >>
X&A or XB
⑤
Complement (7)
=>
A° or All or A. the complement of A
A 的補集
:
U. universal
set
(宇業)
» A - U \ A = { X\ XEU and X & A}
Ā
VA
• XA XEU and X&A
A^ A = AA = U
index set 索引集
[Note] I
:
1 = {1,23
:
1= {1,2,3] 7
La
L=
UA
=
AvAz Ai
=
AEI
=
NI Ax = A₁VARVASI Ai
Û
Los Ai = Ai
J = N
I = R
=7
NEI Ai
=
無法表示
Double A
N

ページ4:

Subject:
.
Ai
(聯集)
{X X = Ai for some I & I}
No.:
Date:
As
Jos Ai =
-
⑥
=
A1
A3
{x
{x
XEL
st XAx3
t
XEU.Ai
MEI
XE Ai for some t & I
JEI SA XAi
X = Ai for all i = I }
t
x x = Ax. Vi t I }
X & Ai Xt Ax for all to I
X & Ar
Vi I
☹ De Morgan's Low (isk Akh RV)
(迪摩根法則)
A.Bi, NEI Sets
:
(1) A^ ( Bi) = LA^Bi
NEI
ex: An (Buc) = (A^B)u (A^c)
(2) Av (Bx) = A AUBE
ex: Av (B^c) = (AUB) ~(AUC)
(1) (2) prof
職業的交集交集的聯集
Double A

ページ5:

Subject:
Reason for (1)
No.:
Date:
X & Ac (13x) XEA and X & Bi
AEL
XEA and X = Bi for some ÑEI
XE AnBi for some ÌEI
=>
XE LUI Anbi for som it I
Reason for (2)
XE AV (B)
UA
itl
Ai
XEA
XE A Bi
XEA or
XE Bñ for all reI
X & Av Bi for all Ì EI.
X & AvBi for all it I.
A
(聯篥的補集=補等的交集)
Atl
U
JJ Ai
(交集的補身-補業的聯集)
④poof H home work.
Double A
3

ページ6:

Subject:
Properties of Integers (1)
Well - Ordering Prinesple ( A )
No.:
Date:
Every numempty set of positive integers contains
非空 检
a smallest member
一個最小元素
每個由正整數所形成的非空集合必包含一個最小元素
eg 5= {3.5.7.9.2.4.63
>
min 5=2
Well Ordering Principle:
US ≤ N S # =>
Note
①良序原理是數學上最基本
T
s.ta≤s Vs S
/
Us
Ja ES sit a = mins
的“公設,不需證明是“直觀上”公認接受的結果
②良序原理等價於“數學歸納法” Muhematical Induction
或“選擇公設" Axiom & Choice (分析常用
* "Zom by 31 12" Zorn's Lemma (π #n
Double A

ページ7:

Subject:
Defintion
Divisor
Multiple
12/
倍數
a.b, dz
od is called
稱似
d+0
Prime)
質數素數
a divisor for a
if there exists
CtZ
的
a
No.:
Date:
存在有
C
to t
by
In this
如果
denoted by
such that a = c.d
17 13
dla dy a by 17 $2
dp a b 19 #7, d 17 a
a=cd
case, a is called a multiple of d af d 18 th $7.
Note
o a,c,d z z
a
g= c.d
denoted
da
12
by
12715
② 因數d不能是口
③
X, VxEZ
1
為所有數的因數
XD
VX = $
○為所有數的倍數
d is
a
of
d/a
d is
a
if a/c and
and blc
0 = 0.X
common divisor of a and b
公因數
and d/b
da A B A d Z b by Ax
common multiple of a and b
公倍數
ca c ¼ b by ft + y
la
& 11 b
alg ble
alab blab
Double A
4

ページ8:

Subject:
The
gentest
Common
最大公因數
No.:
Date:
Advisor of a and b
denoted
by
god (a, b) is the largest
common divisor of a and
be
• gcd (a,b) = max {dtz/ d. a common divisor of a and b}
max { db #
dla and d/b}
The least
Common
multiple of
a and b
denoted
by
lam (a,b) = min
Icm (a, b) is the smallest positive
CON C. a common multiple of a and b}
= min {CEN/ a/c
common multiple of a and b
and b/c3
(Note)
1/a, 1/b
=7
ged (a,b) |
.
A +0,
=7
9/19116
alallbl
b|lallbl => 0</ab&N
2
/
lallbl ☆ a HA b by -11 I by là jouer ox
Note
a, b = Z
a²+b² + 0
0
a
b
7
delal
。
@ged (a,b) = N
Tcm (a,b) N
③gcd
god (a,b)/a
ged (a,b)/b
0*
aam
④ 線性組合性質
arbid & Z
dla, dlb
Jam (a,b)
xb] 1cm (ab)
=7
d/xa+yb, xyEZ
Double A

ページ9:

Subject:
④ abid & Z
Reason
No.:
Date:
dla, dlb = dl xa+yb, VXY EZ
dla, dlb
by desas a-mid, b=nid for some min Ez
・ xa+yb = x (md) +y (nd) = (xm+yn).d,
21 ryb, beyz
=7
5
a and b
are said
to
⑥
PEN, P is called
相對的質數
Vxy + z
be relatively prime it god (a,b) = |
互質
a prime (or a prime number)
質數
(1) P=2 and (2) the only positive divisions of P are I and P
(PEN,P是質數為P2且P的正因數只有1跟自己
eg: P. 2.3.5.7, 11.13.17.19
Question ab
Ł Z
a=4,
p's primes
How to find god (a,b), Icm (a,b) ?
a4
Athey α-4. b-6 >α-4-14-2-2
fu:輾轉相除法
-(1)(4)-(-2)(-2)
b = 6.1 = 7.3
(-3)(-1) (-2)(-3)
a 的倍數:
a b 127 ±1.12.14
15197
7
b by 7±1, 12, 13, 1b.
IT, 12
ged (ab)
=
4.1
4-2
4.3
18
→ Jam (a,b) -12.
(Euclidean Algorichm) b 63 13 % 2 = 61, 62, 63
Double A
S

ページ10:

Subject:
No. :
Date:
%/25 Well - Ordering
Ordering Principle
HS CN
S+ & RES
St
as
Vst S
E
a
+
min
S)
Theorem o.! ( Divison Algerschen for 2, 8) 8
整數除法原理
本质数
Let a,b tz
b70
Then there
such that
exists
unique
q, r = Z
a = q⋅ b + x, where
r
0
obrab
In this
case
t
上稱為
Q=14
is called the
✓ is
(除数)
Y
a被占除的商
called the remainder
(1757) 4-6- (1)
a 4 b 15 by 15 #2.
3/14-01419 $3) 14 = 4.3+2
g.
7
b=3
12
2
-Ya $)
quotient
of
a divided by be
a
divided by b
a=qb+r
front (By Well - Ondering Principle)
Jet S = {a_xb| x+z and gab 30} = {0} UN
then a-xb = a-(-1a/>b = a+/alb
Take x = -1al; then
So a-xb zo if x = -α = z
Hence
a-xb & S if x = -)a|
This implies that S+
boba+la)-1=a+lal
NO
Double A

ページ11:

No.:
Subject:
(Existance of of and x),
"
存在性
SNU {}
S
* P
Date:
.. By Well - Ordering Principle, S contains a smallest element
This is
+
(a-qb
32
ES
Ł Z
s.t
and a-gb
a-gb & S and
min S)
=
a-gb ss, √SES
S
Let
Y = α-gb
Then
α= gb+r
Clam orab
IT
··
=
r-α-gb&S Nu {o}
..
Consider
=>
20
>r-b= (a-q;b)-b- a- (2+1).b
bo
.rb <r=
=
min S
.. r-b & S = {a-xb xoz and ax-b zo}
9-19+1)-6
:. a- (2+1). b <0
r-b
..
r<b
..
a = qb +r
+ for 68
<
Double A
6

ページ12:

Subject:
唯一性
Note
No.:
Date
Uniqueness
Suppose
that
3.A
→
off and r
3 b, b*, V, Y* 6 z
a-q-b+r
a-gb+r=q=b+r*
E
and
6 + 16 (670)
0
(8-8)b-r
*
{0}UN
=
- <b
712-2*1
=> b= q
a, b & N
azb
a-b zo
a-zbo
b
M
NN
= 0
+
=
Y-y*
{
a = *b+r*
7
a
Y
+1
⑫
Y b
0=2<3
14.4.3+2
-14= (-4)3-2
a = (-4) 3 -3+3-2
=
= (-5)=3+1, 06/<3
& b
Y
=
r-a-gb
0-(9+1)-520
≥0
Double A

ページ13:

Subject:
Theorem 0.1
*
Hab Z
No.:
Date:
=> 7! q, € Z
Reason
5.t. a. gib+r, where o Ev</bl
b20b70
+
* A= q(-b) + obr<b
= (-1).b+r, oεr</bl
0684161
Recall
a,b,c,d
#
a = c.d
d/a
#
ged (ab)
lem (a,b)
Theorem 0.2 (GCD's Linear Combination Bezone's Jolanercy) 13.
prot
最大公因數 線性
組合 見祖 恆算式
趕本
any
For any ab Z with
a+o
b+0
=
there exist xiy 6 B such that xaxy. b god (a,b)
Vaib + B 72xy + 3 st. xaxyb = god (mib))
(用良序原理
14 * f AB B sh. Euclidean Algordham)
Double A
7

ページ14:

Subject:
No.:
Date:
........
ged (a,b) = Za + Zb
Note
E Z
©
ged (a,b) - mn
{xa+yb]
E Z
Xa+yb
70
(Reason ged (aub) & N, ged (a,b)/ xa+yb, vx.y 68)
③
α=2,6=3
ged (a,b) = 1
1 = (-1).2+1.3
ged (mb)
xx
2.2+(-1).3
Questim
=>
How
to find
god in
(a,b)
How
to
find
t
(FA 17 =th Enclidean Algouehm)
3. t xa+yb = gcd (a,b)
Exmple (to 4 i t tk kg An j k z h )
a=172b-20
Find god (a,b) = xatyb
a
172
20
b
8b
160
12
9-86
0-86
12
8
b-(a-8b)-7b-a
96-06-(0-86)
88 2 (za-17b)
⇒ za-17b = 4 - god (a,b)
(a-8b)-(96-a)
4'
0 (9b-a)-2/24-17b)
11
2
-17
20-176
Double A

ページ15:

Subject:
Gorellary
bez
ged (a,b)=
1 (a,b) =>>
6互質
No.:
Date:
#x₁y +8 st xa+yb = 1
(16 Za +25)
Lemma 0.1 (Euclid's Lemma ) P4
歐基理德引理!
(1)
a.b.c
E Z
到
cab
c/ 0.6, god (C, α) = 1
a)
=
c/b
Prot
(2) a, b 6 Z, P&N a prime, pla.b pla
Use Theorem 0.2 Bozant's Identity
or
P/b
Note
a b B, PEN, a
a prime ged (a,p) or P)
Every integer greater
than
Theorem 03 (Fundamental Theorem of Arithmetic for B ) 3 4
整數的算數基本定理
(整數的質因數唯一分解定理)
I can
每個大於1的整數
be uniquely
a
factorized into
可以被分解為唯一集後的質數
product of
primes.
where Pi's, bi's Primes
HEN
n > I
PEPP
q1q2=.
then k=l and Px = bi ti = 1, 2, 1x k
=
①(存在性)
o P. Pz. m, Pk, Primes st. - P. P. Pk
© (• #-#) If no P₁ Pz. P=.
Double

ページ16:

Subject:
Prot
10 & it not then #P: a prime S.A plu
再利用數學歸納法第二型(存在性),
9 AN IA Euclid's lemma (n-1)
存在/
Eng
eg. n. 300
2/300
21150
-7
n = 2.2 5.5.3
55
525
=
2² 3.52
3
No. :
Date:
Note
ab & N
W
a,b> 1
a = P P P where he's distinct primes
b = P P P
0
M²
2
Mk
ged (a,b) = P
{n₁, mi}
Mi Mi N{0}
.
2
.
min [1,M]
min [ns,ms)
P
PK
Icm (a,b) = P
max (nimis
may [nz, M
12
max {3.M
. (Il.
PK
max {nk,mks
②
god (a,b). Iam (a,b) = a.b
a-b
3
lom (mb)
=
ged (a,b)
eg
ged (172, 20)
lam (192, 20)
Exercise O Prove
=
= 4
12.20
god (172,20)
172.20
=
4
that it a = x²=y³
b
then az
a. z6 for
Some
Prove that
for
Some
王CZ算款基本定理
HEN
n-lis
Tational
W2
哈
Double A

ページ17:

Subject:
No.:
Date:
# Complex Numbers (14 2 Theorem 0.4) 811
• * * 15 C = { a+bi ab & R} = R₁ + Ri
***
.複數運算:
a,b,c,d op
i² = -1
to a+bi = c+di def> a=c and b=d
①相等
@A
=(a+bi)+(c+di) def (a+c) + (b+d) i
@h, (a+b)·(c+di) det (ac-bd) + (ad + bc) i
③乘法,
④除法,
0+01
=0
1 R-plane
z = a+bi (a,b)
c+di o
C÷0
ord+o
+0
Sid alf
C²+d²=
=0
-**
C+di+0,
a+bi
=
C+di
a
=
a+bi -di
C+di C-di
(ac+bd)+(bc-ad) i
C²+d²
ac+bd
Cod") + (bc-ad)
⑤共軛函數一: C → C
-
z = a+bi z data-bi
=
+
-
JZ, Z2 = C
C²+d2
z=a+bi
對X軸的鏡射
/
Z₁ Zv
03
Double A
z-a-bi
9

ページ18:

Subject:
⑥複數的長度(絕對值
No.:
Date:
z = a+ bx 6C 121 det
Ja+b
z = a+bi > (a,b) = /R+R
the length of z
z的長度=王到原點。
121 Jab
的距離
a
1/9 Note
#
D
=
INI
到
| Z₁ Z2 | = | Z1 |· | Z2 |
Z
=
到
⑦d. C x C
:
V21, 226 C
21-22 = 211.122 √zzi ol
个
det
Pu {0}
d( 21, 22) / 21-221, V21, 22 6 C
⇒
(c,d) a metric space
l ≈
R² 拓撲同構
⑦複數的極式(用長度角度表示)
121-J5762
z = a+bi
121- Jasp
A
Ju7b, Sing=
z = a + bx
= C
abcp
=
√a+b²
a
va²+be
2
CSA+
長度
Ja+b
及
十
b
Ja²+be
i sin A), where
i
0
a
1:王到原點的距離
A:向量庭與X軸正向的夾角
Double A

ページ19:

Subject:
No.:
Date:
⑧複數極式的乘法(長度相乘,角度相加
去
6 C
Z₁ = | Z₁l (as A₁ + i sin A₁)
Z2 = 22) (Az + sin Az)
Zz
7 Z1 Zz
=
Observation
複數極式
乘法
驗證此
(105(A1+A2)+ùn(A+A2),
cos (A,+ Az) + ì sin (A1 + A2),
(cos α + i sin x) (cos B + i sin ß)
Cos (α+ B) + i sin (x+B)
(sub-singing) + (as a simß sind wasps) i
(a+ba)(c+dx)=(ac-bd)+(ad+be)i
千
=7
>> cos (a+B) = as a cosy - sin a sing
}
合角公式
公式
sin(x+b)= sinaass+sinBasx
複數極數乘法基本上是正餘弦的合有公式
Observation
複數極式
乘法
複數乘法
(94) = x²+377 +3774>
(as A + i sin A) ³ = ( us A + i sin A) (DSA + i sin A) (OSA + isinA)
cos (SA) · i sin (34)
i²=-1
coss A +3 costa i sina +3. COSA (i Sina) + (x Sin A)³
=
(Cost A - 3 CSA Sin³A) +
(3Q5" A sinA - Sin³ A) ī
Double A

ページ20:

Subject:
cos 3A Cos³ A - 3 cos A sin² A
=
=
A-3
Sin 3A = 3 csA SMA - Sin³A
大學複變數函數論
en
C&A - Cos A + i sin A
No.:
Date
正餘弦三倍角公式
At a = ar
laya"
Az
(A1+A2)
=
.
(exM)) = ex(36)
"
9 # * * £ 1 (De. Moivre's Theorem)
Z = /2/ (COSA + i SMA)
=2" = /z/" (cosna + isinna,
^=z)^(CosnA+
到
幫助計算C的次方z^
幫助解王,的n次方根,解五方程式,
Question
Given
ZEN
+
how to And x = C
t
(解複數204
such that xn = Z ?
的n次方根)
Fact
Z & C
REN, Solve X EC, X" = Z
Write
z = /z/ (cos A+ isin A)
Then
2
Double A

ページ21:

No.:
Subject:
Reason
0+2471
Date:
Then x = 12) * (as +264 + i sin Atthx), where kool, m, n-1
肝
{x". z
=
Z
=
Z
/z) (Dos A + i sin A)
x= |x| (cos + i sin d
> x = 1x1 / and + isinn ()
n
|| (snd+ind) = |=| (OSA + i sin A)
|x/" cos no = /2/ ces A
1x/" sin n = 12/ sin A
|x)" = |2|
Cosud = COSA
Sin 2
上有一個可能
1 x = 2 = W/Z
np=A+zkπ for kez
Exercise
=
sinn sing
X-25
=
=
+2
n
K = 0, 1, 1, -1
z=3+2i & C
2 Z = −2+2Ń & C
Solve Z
solve z
Double A

ページ22:

Subject:
No.:
Date:
#
Theorem
Fundamental Thanem of Algobra
Algobra, FTA, Grass) π tox y
Every
non constant
complex
每個
非常數
74142
polynomial
多項式
must have a
必
A
- 119
complex root
1 複數根
#
t
✓ f(x) + C [X], defix) 21
#
flα)
= C [X]
day fus=1
# Mathematical
7BE C
St. f(x)=0
On & C
Sit f(x) = B(x-α1) ... (X-XN)
Induction (the best best that ith I, I IB)
Theorem 0.5 (First Principle of Mathematical Induction, FPMI) Piz
=7
=>
Let S be a
是
Set
-
of
positive integers
些正整數形成的集合
Assume that s satisfies the following
滿足
two conditions:
假设
0
G Si
@ It nes
then
U+1 = S
Then S = N( 5恰為所有正整數的集合)
Proe (用良序原理去證
Observation (The 0.5
: 165
By 0, IES • By @, 1+! ES.
I+ 6
2 ts
7
"265
By IES
-7
S = N
Double A
13 ES

ページ23:

Subject: #(高中版)
No. :
Date:
Theorem 05
Let Pin) be
a statement about Postive
敘述
有關
EM Ingers
Syguse that
PLI) is
true
and
if P(n) i
true for some
GN
then Plu+l) is also true
Then Plu) is
true for all EN
Example
Prove that
1+ 2+ + n².
n(n+1)(n+1)
6
Un EN
Phot
Let S =
NEN 1² + 1 +² =
(UH) (2H)
6
}
N
The
goal
is
to show
S = N
1² = 1
1 6 S
=
6 1-2-3 = ((1+1) (2+1)
(condition D)
6
Suppose that
n
= S
(n+1)(24+1)
"nes
i. 1+27 +4² =
6
<= 1 + 2 + 1" + (n+1)² = (1+1)(24+1)
+(n+1)²
結尾:
:. S satisfies conditioms © ©
=
6
(n+1) (n (2n+1) +(n+1)
" By Theorem 0.5 (FPMI)
=N
=
(n+1) (2n + n + b + b) = (1+1) (→ 24 17
n+ba+b) ²
24²+74+6
6
(n+1)(n+2)/2n+3)
(n+1)((n+D+1)(2(n+1)+1)
=
6
n+lbs (condition 2
Double A
12.

ページ24:

Subject:
No.:
Date:
Theorem
0.6
Second Principle of
Principle of Mathematical
Induction SPMI) P13
S&N
Assume
S
satisfies
the following
two
conditions:
@
1 6 S ;
it Ins
where nal
then + ES
Then
S = N
Examples Assume
defined by
that { Anno
01-1
02=2
is
a
Sequence
t
Un N, 133
au = An-1 + anz
+
Brave Ame
that
An < 2"
+
Let
5 = {n
EN
an < 2" }
≤ N
The goal
is
to
show S = N
=
2'
1 ES
265
a₁ =122
"A₂ = 2 <4 = 2
1,2 ES
(condition ()
Suppose that 1, 2, 1", n-1, n&S
1, 2, 1, -1, NES
i: a. <2', A2 <2², !",
An-12241
h
Double A

ページ25:

Subject:
小
an+1
=
An + An-12"+ 2+
n+les ( condition
ci
s satisfies
conditions
@ and
No.:
Date:
2 (2+1) = 2" 32" 4-2
y
detinal
by
By Theorem 0.6 (SPMI),S=N
Exercise Assume that {
01=1
d2=3.
•An = An+ + Au-z. Hut N, 433
{au3na
a
Sequene
prove that
An 2
(4)", Uno N
t
# Functions (Mappings
Maps) P1 $
①
(Detivition) A function from
函數中
乃
a
rule
指的是 規則法則
a
of A
-
that
exactly
#中每一個元素~ 恰好
a
set B
# 到集合,
assigns to each element
分配
one element
b of B
B中的一個元素b
a set A to
denoted
by
by b-p(a)
(一個由集合廾到集合乃的函數中
指的是一種將A中每個元素a恰好
个指
(對) 符號上b=p(a)
分配到B中一個元素b的對應關係)
Double A
B

ページ26:

subject
No.:
Note
Date:
代表
函數 $
A
→ B denotes
a
function
....../.
(1)
區域
A. the domain of 中定義域
from AB
ABUB
(2) B the
:
co domain of
對應域
(3)
(A) = { p(a)
at A}
the
range (image) of &
中
A
(4)
C&B
p" (c) = {RGA P(A) 6C},
the
prejmage
f
of & on C
前像
$19
· P(A) ≤B
• C&B, & (C) ≤ A
SE P(A)
t
t
$(c)
史
Example
A
a
b
C
C = {z, w} $(c)
y
2
W
B
=
5. (a)
(a) for
Some
at A
(+) & C
E
P(A) A-{a, b, c]
{$(a), (b), (c)}
"
{y, w, x}
{b} => $" ({23) up" ({w]) 163.
165
=
Double A

ページ27:

Subject:
@ (Equality of Function) My 10
Two functions
pei
A B
and 4: A-B
No.:
Date:
are
called
equal if play = pras, VaGA,
中二4 (中興中在定義域中每點的函數值相等)
denoted 6 = 4
③ (Compositions of Functions
Suppose
that
My to 197
$: A > B and 4: BDC are functions
The dental by 4; f
compet of P and þ
AC detinal
is
the
function 4.4 : A+ C
det
(4.6) (a) & (p(a)), HAGA
A
B
B
by
Circle
You A L > B + > C
a (a) ((a))
400(0)
4(4(a))
Note
凼數合成滿足結合律不一定交換性
"FA-B
:
g. Bch c→D functions
ho(gof)。(hog)of(函數合成結合律)
=>h
A
-B
→D
ho(gof) A
:
(hog) of A
C
D
D
Feason
Double A
14

ページ28:

Subject:
Reason
holg.
ho (g⋅ f) A-D
(hog) • f
functions
:
A
→ D
No.:
Date:
HasA, (no (got)) (a) by both h ((got) (a)) dat hy (tm))
((hey) of) (a) by the (hog) (terms) by ((Fra)
ho
def
(k₁ (goti) (a) = ((kg) of) (^) Hart
*ho (got) = ( hog) of
(2) FA>A,
J
:
AA function
get fog
+
→
•Example f⋅ R+R
:
R-R
L
f(x) = x², X = R
E
9xx) = x+1, √x ε p
(got) (x) + g (f(x) = g(x) = x²+1
(fog)(x) tf(g(x)=f(x+1)-(+1)=x+2+1
VXR
t
=7
got = f.g
④
One-to-one Fundion)
A function $: A+B is called
one-to-one
if for
(a) = (a) implies a₁ = Az
任意兩點a1,a2經由中對應到同一點中(a)=f(a),
則a=d21必為同一點)
-
Note
3
Double A

ページ29:

Subject:
No.:
Date:
Node_
A-B
a function
中
one-to-one
論證)對!
by dot;
def
For
any
6
↗函數比較好
a₁, az A, plais- & (as) implies α = Az
a.
12點經中對應到同一點,原2點(4)相同)
的方法
It ai, az & A with
a. + az
then p(a) ≠ 中(az)
~10~P
+
one -
to -
one
function
=
1-1 function
Example :P
> R
a
Function defined
by
(相昊2點經由中對應到也是相異兩點
射
injective function
f(x) = x+2x R
=
中(x)=1+2
Rove that is
one - to-one
Reason
Suppose p(x) = $(x+) for Xl, Xx & P
def of
X1+2 = X2+2
7 x₁ = X2
中
one-to-one
Ones Auction) 07 $124 $X
A function : A+B is callal
onto
it for
bB
any
there exists
at A
such that f(a)=b
(對應域中每個元素b,皆存在A中一個元素a
/
經由中對應到b=中(0))
a function, p: onto
Note
中:A→B, a function
by de Hb B
PIA) B
t
+
AGA
B = a≤A} =
sto play tin
{p()/act}= $(A) 中的對應域中的值域
Double A
15

ページ30:

Subject:
No.:
Date ........
#
:
a
onto function
f a surjective function
A
B
Example
d: RR
a function defined
by
Prove that is
onto
$ (x) = x³, VX EXP
(x²= b = b+0
(14 / 71 - - 119 12 AR
Reason
Hb & R
t
choose
a
= 36 = R
ER
(3-8=-2)
⑥Identity
(a) = a³ = (36)³ = b
Function
Gidentity
The Blemity Auction
M
a
Set
F
onto
伯域
整y軸
A is the function id+ A+ A
defined by
Jda (9)
= a
Has A
+
(由4到A每一點都對應到自己的函數idu.
Note
& A→B
a
function
ida: A-A, identity function
ids: B-B
。
idA •p = p
poids = p
(任何凼數跟單位出數合成不變)
(Invertible functions)
J 142 141*X
A function f: AB is
callal invertible if there exists
a function go
In this case,
g
is called the inverse function of f denoted by
可使
B
B>A such that got idd, fog=idz
etabub
by 9

ページ31:

Subject:
A±B
# A.
>B I>A I >B got - ida
Note
f: A→B
:
a function
an invertible function
TL
Exercise of AB
I
B-C
Prove that if t.g: H
it
Grove that if f.g.
A>B
It got - Ida"
:
BA
4
No.:
Date:
togid
f a 1-1
and
onto function
加上 onto
functions
then got
1-1
onto
then got.
onto
functions
onto
↓
then f. H and g
# Modular Archmetic P5 LAT 1 $x
The theory of
Congruence 17 13 150
Deturcion (1) 19 Gauss,
NEN
a
Fixed
integer
a.b Z
T
n
a and b
Ja-b
ab是n的倍數
對模數月餘
are
said to
congruent
modulo
n
st
模n同餘
denoted
三
a = b (mod n
Note
名词
形容词
module
modulo
模(度量單位
congruent
JAA
congruent
Double A
b